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|---|---|---|---|---|---|---|---|---|---|---|---|
pBN5sjfn3CUEa9mv | physics | motion-in-a-plane | relative-motion-in-two-dimension | A particle has an initial velocity $$3\widehat i + 4\widehat j$$ and an acceleration of $$0.4\widehat i + 0.3\widehat j$$. Its speed after 10 s is: | [{"identifier": "A", "content": "$$7\\sqrt 2 $$ units"}, {"identifier": "B", "content": "7 units"}, {"identifier": "C", "content": "8.5 units"}, {"identifier": "D", "content": "10 units"}] | ["A"] | null | Given $$\overrightarrow u = 3\widehat i + 4\widehat j,\,\,\overrightarrow a = 0.4\widehat i + 0.3\widehat j,\,\,t = 10s$$
<br><br>$$\overrightarrow v = \overrightarrow u + \overrightarrow a t $$
<br><br>$$= 3\widehat i + 4\widehat j + \left( {0.4\widehat i + 0.3\widehat j} \right) \times 10$$
<br><br>$$ = 7\widehat... | mcq | aieee-2009 | 12,050 |
7tfsktGFYJZKOBp6KaCtV | physics | motion-in-a-plane | relative-motion-in-two-dimension | A man in a car at location Q on a straight highway is moving with speed $$\upsilon $$. He decides to reach a point P in a field at a distance d from the highway (point M) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach... | [{"identifier": "A", "content": "d"}, {"identifier": "B", "content": "$${d \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${d \\over 2}$$"}, {"identifier": "D", "content": "$${d \\over {\\sqrt 3 }}$$"}] | ["D"] | null | <p>Let the distance QM = l and distance RM = x.</p>
<p>Time to reach from Q to R is $${t_1} = {{l - x} \over v}$$</p>
<p>Time to reach from R to P is $${t_2} = {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$$</p>
<p>Therefore, $$t = {t_1} + {t_2} = {{l - x} \over v} + {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$$</p>
<p>On differen... | mcq | jee-main-2018-online-15th-april-evening-slot | 12,051 |
IEMsrMGJtZxDTzgVgsHBJ | physics | motion-in-a-plane | relative-motion-in-two-dimension | The position co-ordinates of a particle moving in a 3-D coordinate system is given by
<br/>x = a cos$$\omega $$t
<br/>y = a sin$$\omega $$t and
<br/>z = a$$\omega $$t
<br/><br/>The speed of the particle is : | [{"identifier": "A", "content": "$$\\sqrt 2 \\,a\\omega $$"}, {"identifier": "B", "content": "$$a\\omega $$"}, {"identifier": "C", "content": "$$\\sqrt 3 \\,a\\omega $$"}, {"identifier": "D", "content": "2a$$\\omega $$"}] | ["A"] | null | Given that,
<br><br>x = a cos $$\omega $$t
<br><br>y = a sin $$\omega $$t
<br><br>z = a $$\omega $$t
<br><br>Velocity in x-direction,
<br><br>V<sub>x</sub> = $${{dx} \over {dt}} = - a\omega \sin \omega t$$
<br><br>Velocity in y-direction,
<br><br>V<sub>y</sub> = $${{dy} \over {dt}}$$ = a $$\omega $$cos $$\omega $$t
... | mcq | jee-main-2019-online-9th-january-evening-slot | 12,053 |
EX0cmRwaf0Nbuoj1pnaQd | physics | motion-in-a-plane | relative-motion-in-two-dimension | A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60<sup>o</sup> with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is : | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$v"}, {"identifier": "B", "content": "$${{2v} \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "v"}, {"identifier": "D", "content": "$${v \\over 2}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266495/exam_images/nowpqlkptjtdzln4xnnn.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Physics - Motion in a Plane Question 60 English Explanation">
<br>AB =... | mcq | jee-main-2019-online-12th-january-morning-slot | 12,054 |
sdkRni7JNbzv7DlroNM0I | physics | motion-in-a-plane | relative-motion-in-two-dimension | Ship A is sailing towards north-east with
velocity $$\mathop v\limits^ \to = 30\mathop i\limits^ \wedge + 50\mathop j\limits^ \wedge $$ km/hr where $$\mathop i\limits^ \wedge $$ points
east and $$\mathop j\limits^ \wedge $$ , north. Ship B is at a distance of
80 km east and 150 km north of Ship A and
is sailing ... | [{"identifier": "A", "content": "2.2 hrs"}, {"identifier": "B", "content": "4.2 hrs"}, {"identifier": "C", "content": "2.6 hrs"}, {"identifier": "D", "content": "3.2 hrs"}] | ["C"] | null | <p>Considering the initial position of ship A as origin, so the velocity and position of ship will be</p>
<p>$${\overrightarrow v _A} = (30\widehat i + 50\widehat j)$$ and $${\overrightarrow r _A} = (0\widehat i + 0\widehat j)$$</p>
<p>Now, as given in the question, velocity and position of ship B will be, $${\overrigh... | mcq | jee-main-2019-online-8th-april-morning-slot | 12,055 |
UkRHjiSuqelinbY9pF7k9k2k5gya6di | physics | motion-in-a-plane | relative-motion-in-two-dimension | A particle is moving along the x-axis with its
coordinate with the time 't' given be<br/> x(t) = 10 + 8t – 3t<sup>2</sup>. Another particle is moving
the y-axis with its coordinate as a function of
time given by y(t) = 5 – 8t<sup>3</sup>. <br/>At t = 1s, the speed
of the second particle as measured in the frame
of the ... | [] | null | 580 | <p>For a particle ‘A’, its position along the x-axis as a function of time $ t $ is given by:</p>
<p>$ x(t) = 10 + 8t - 3t^2 $</p>
<p>To find the velocity $ v_A $, we take the derivative of $ x(t) $ with respect to $ t $:</p>
<p>$ v_A = \frac{d}{dt}[10 + 8t - 3t^2] = 8 - 6t $</p>
<p>At $ t = 1 $ second, the velocit... | integer | jee-main-2020-online-8th-january-morning-slot | 12,056 |
FallPqJDbo23S96oK07k9k2k5hhu0w3 | physics | motion-in-a-plane | relative-motion-in-two-dimension | A particle moves such that its position
vector $$\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j$$ where $$\omega $$ is a constant and t is time. Then which of the following statements is true for the velocity
$$\overrightarrow v \left( t \right)$$ and acceleration $$\overrightarr... | [{"identifier": "A", "content": "$$\\overrightarrow v $$ and $$\\overrightarrow a $$ both are perpendicular to $$\\overrightarrow r $$"}, {"identifier": "B", "content": "$$\\overrightarrow v $$ and $$\\overrightarrow a $$ both are parallel to $$\\overrightarrow r $$"}, {"identifier": "C", "content": "$$\\overrightarrow... | ["C"] | null | $$\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j$$
<br><br>$$\overrightarrow v = {{d\overrightarrow r } \over {dt}}$$ = $$ - \omega \sin \omega t\,\widehat i + \omega \cos \omega t\widehat j$$
<br><br>$$\overrightarrow a = {{d\overrightarrow v } \over {dt}}$$ = $$ - {\omega ^2}... | mcq | jee-main-2020-online-8th-january-evening-slot | 12,057 |
ZxjUBxMWqDW8EEUc0z7k9k2k5l8xvqh | physics | motion-in-a-plane | relative-motion-in-two-dimension | A particle starts from the origin at t = 0 with an
<br/>initial velocity of 3.0 $$\widehat i$$ m/s and moves in the
<br/>x-y plane with a constant acceleration
$$\left( {6\widehat i + 4\widehat j} \right)$$ m/s<sup>2</sup> . The x-coordinate of the
particle at the instant when its y-coordinate is
32 m is D meters. The ... | [{"identifier": "A", "content": "40"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "60"}] | ["D"] | null | $$\overrightarrow u $$ = 3.0 $$\widehat i$$
<br><br>$$\overrightarrow a $$ = $$\left( {6\widehat i + 4\widehat j} \right)$$
<br><br>$$\overrightarrow S = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}$$
<br><br>x = 3t + $${1 \over 2}6{t^2}$$
<br><br>= 3t + 3t<sup>2</sup> .....(1)
<br><br>y = $${1 \over 2} \... | mcq | jee-main-2020-online-9th-january-evening-slot | 12,058 |
hDRWmi5w2yihs2aXhHjgy2xukf6dfz1n | physics | motion-in-a-plane | relative-motion-in-two-dimension | Starting from the origin at time t = 0, with initial velocity 5$$\widehat j$$ ms<sup>-1</sup> , a particle moves in the x-y plane
with a constant acceleration of $$\left( {10\widehat i + 4\widehat j} \right)$$ ms<sup>-2</sup>. At time t, its coordinates are (20 m, y<sub>0</sub>
m). The
values of t and y<sub>0</sub> ar... | [{"identifier": "A", "content": "5s and 25 m"}, {"identifier": "B", "content": "2s and 18 m"}, {"identifier": "C", "content": "2s and 24 m"}, {"identifier": "D", "content": "4s and 52 m"}] | ["B"] | null | $$y = {u_y}t + {1 \over 2}{a_y}{t^2}$$<br><br>$$y = 5t + {1 \over 2}(4){t^2}$$<br><br>$$y = 5t + 2{t^2}$$<br><br>and $$x = 0(t) + {1 \over 2}(10)({t^2}) = 20$$<br><br>$$t = 2s$$<br><br>$$ \Rightarrow y = 10 + 8 = 18m$$ | mcq | jee-main-2020-online-4th-september-morning-slot | 12,059 |
vqJtPg0eqiAK91QJmL1kmioyfet | physics | motion-in-a-plane | relative-motion-in-two-dimension | A mosquito is moving with a velocity $$\overrightarrow v = 0.5{t^2}\widehat i + 3t\widehat j + 9\widehat k$$ m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s? | [{"identifier": "A", "content": "$${\\tan ^{ - 1}}\\left( {{\\sqrt {85} } \\over 6}\\right)$$ from y-axis"}, {"identifier": "B", "content": "$${\\tan ^{ - 1}}\\left( {{5 \\over 2}} \\right)$$ from y-axis"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}\\left( {{2 \\over 3}} \\right)$$ from x-axis"}, {"identifier": ... | ["A"] | null | $$\overrightarrow v = (0.5{t^2}\widehat i + 3t\widehat j + 9\widehat k)$$ m/s<br><br>At t = 2 s<br><br>$$\overrightarrow v = (2\widehat i + 6\widehat j + 9\widehat k)$$
<br><br>Direction cosine along y-axis,
<br><br>$$cos\theta = {{(v.\widehat j)} \over {\sqrt {{9^2} + {6^2} + {2^2}} }} = {6 \over {\sqrt {121} }} = ... | mcq | jee-main-2021-online-16th-march-evening-shift | 12,061 |
pluW1uWZzhcGoy2coR1krpltdw4 | physics | motion-in-a-plane | relative-motion-in-two-dimension | A butterfly is flying with a velocity $$4\sqrt 2 $$ m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is : | [{"identifier": "A", "content": "$$12\\sqrt 2 $$ m"}, {"identifier": "B", "content": "20 m"}, {"identifier": "C", "content": "3 m"}, {"identifier": "D", "content": "15 m"}] | ["D"] | null | The given situation can be represented as<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxujviik/355fde85-a369-4675-acbc-a7ab4aec0fc5/3524cbc0-6a4e-11ec-8561-951bc0cc527f/file-1kxujviil.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxujviik/355fde85-a369-4675-acbc-a7a... | mcq | jee-main-2021-online-20th-july-morning-shift | 12,062 |
1l6npzmwm | physics | motion-in-a-plane | relative-motion-in-two-dimension | <p>At time $$t=0$$ a particle starts travelling from a height $$7 \hat{z} \mathrm{~cm}$$ in a plane keeping z coordinate constant. At any instant of time it's position along the $$\hat{x}$$ and $$\hat{y}$$ directions are defined as $$3 \mathrm{t}$$ and $$5 \mathrm{t}^{3}$$ respectively. At t = 1s acceleration of the pa... | [{"identifier": "A", "content": "$$-30 \\hat{y}$$"}, {"identifier": "B", "content": "$$30 \\hat{y}$$"}, {"identifier": "C", "content": "$$3 \\hat{x}+15 \\hat{y}$$"}, {"identifier": "D", "content": "$$3 \\hat{x}+15 \\hat{y}+7 \\hat{z}$$"}] | ["B"] | null | <p>$$x = 3t \Rightarrow {a_x} = 0$$</p>
<p>$$y = 5{t^3} \Rightarrow {a_y} = 30t$$</p>
<p>$$\overrightarrow a (t = 1) = 30\widehat y$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift | 12,063 |
mG1tloVLkEn6UfkpdErqQ | physics | motion-in-a-plane | river-boat-problems | The stream of a river is flowing with a speed
of 2km/h. A swimmer can swim at a speed of
4km/h. What should be the direction of the
swimmer with respect to the flow of the river to
cross the river straight ? | [{"identifier": "A", "content": "150\u00b0"}, {"identifier": "B", "content": "120\u00b0"}, {"identifier": "C", "content": "60\u00b0"}, {"identifier": "D", "content": "90\u00b0"}] | ["B"] | null | Draw velocity diagram<br>
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263699/exam_images/q6ox0gx23a27tfo4repk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Morning Slot Physics - Motion in a Plane Question 58 English E... | mcq | jee-main-2019-online-9th-april-morning-slot | 12,065 |
qESm4glJNeyyZYmUeM1kmiq39no | physics | motion-in-a-plane | river-boat-problems | A swimmer can swim with velocity of 12 km/h in still water. Water flowing in a river has velocity 6 km/h. The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his starting point is ____________$$^\circ$$. (Round off to the Neare... | [] | null | 120 | <p>The situation is depicted in the following figure.</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ky4h8o5o/f083b4a7-d200-4281-9536-a0c2edcd1cb3/be20a8c0-6fc3-11ec-8887-d75613c1bf3a/file-1ky4h8o5p.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ky4h8o5o/f083b4a7-d200-... | integer | jee-main-2021-online-16th-march-evening-shift | 12,066 |
anTz96GHNUDdELsewo1kmkrt6h4 | physics | motion-in-a-plane | river-boat-problems | A person is swimming with a speed of 10 m/s at an angle of 120$$^\circ$$ with the flow and reaches to a point directly opposite on the other side of the river. The speed of the flow is 'x' m/s. The value of 'x' to the nearest integer is __________. | [] | null | 5 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265114/exam_images/mqpv4fbsyq9pbdpryafy.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Morning Shift Physics - Motion in a Plane Question 44 English Explanation">
<br>$${V_R}... | integer | jee-main-2021-online-18th-march-morning-shift | 12,067 |
1ks1alku2 | physics | motion-in-a-plane | river-boat-problems | A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30$$^\circ$$ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle $$\theta$$ with the line AB should be _________$$^\circ$$, so that the swimmer reaches point B.<br/><br/><img src="data:im... | [] | null | 30 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264657/exam_images/wmd9d6rh3hvrodiyd00q.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Physics - Motion in a Plane Question 42 English Explanation"><br>Both velo... | integer | jee-main-2021-online-27th-july-evening-shift | 12,068 |
1ldpmvb3w | physics | motion-in-a-plane | river-boat-problems | <p>The speed of a swimmer is $$4 \mathrm{~km} \mathrm{~h}^{-1}$$ in still water. If the swimmer makes his strokes normal to the flow of river of width $$1 \mathrm{~km}$$, he reaches a point $$750 \mathrm{~m}$$ down the stream on the opposite bank.</p>
<p>The speed of the river water is ___________ $$\mathrm{km} ~\mathr... | [] | null | 3 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1legtl5mi/0cbc31d7-4d97-4528-a713-75b055d39281/e4bc61a0-b350-11ed-a9df-eb6c254c9c33/file-1legtl5mj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1legtl5mi/0cbc31d7-4d97-4528-a713-75b055d39281/e4bc61a0-b350-11ed-a9df-eb6c254c9c33/fi... | integer | jee-main-2023-online-31st-january-morning-shift | 12,069 |
DM0Ps3OxhqEIO1tJC8e0I | physics | motion-in-a-straight-line | average-acceleration-and-instantaneous-acceleration | The position vector of a particle changes with
time according to the relation
$$\overrightarrow r (t) = 15{t^2}\widehat i + (4 - 20{t^2})\widehat j$$<br/> What is the
magnitude of the acceleration at t = 1 ? | [{"identifier": "A", "content": "50"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "40"}, {"identifier": "D", "content": "100"}] | ["A"] | null | $$\overrightarrow r = \left( {15{t^2}} \right)\widehat i + \left( {4 - 20{t^2}} \right)\widehat j$$<br><br>
$$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \left( {30t} \right)\widehat i - \left( {40t} \right)\widehat j$$<br><br>
$$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = \left( {30} \righ... | mcq | jee-main-2019-online-9th-april-evening-slot | 12,070 |
1ldsqh453 | physics | motion-in-a-straight-line | average-acceleration-and-instantaneous-acceleration | <p>A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2s. The average acceleration during contact is ___________ ms$$^{-2}$$.</p>
<p>(Given g = 10 ms$$^{-2}$$)</p> | [] | null | 120 | The speed of ball just before collision with ground is<br/><br/> $u=\sqrt{2 \times g H}=\sqrt{2 \times 10 \times 9.8}=\underset{\text { (Downwards) }}{14 \mathrm{~m} / \mathrm{sec}}$
<br/><br/>
The speed of ball just after collision is
<br/><br/>
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 5}=\underset{\text { (Upwards) }... | integer | jee-main-2023-online-29th-january-morning-shift | 12,072 |
1lgyqeux5 | physics | motion-in-a-straight-line | average-acceleration-and-instantaneous-acceleration | <p>Given below are two statements</p>
<p>Statement I : Area under velocity- time graph gives the distance travelled by the body in a given time.</p>
<p>Statement II : Area under acceleration- time graph is equal to the change in velocity- in the given time.</p>
<p>In the light of given statements, choose the correct an... | [{"identifier": "A", "content": "Both Statement I and Statement II are False."}, {"identifier": "B", "content": "Both Statement I and Statement II are true."}, {"identifier": "C", "content": "Statement I is incorrect but Statement II is true."}, {"identifier": "D", "content": "Statement I is correct but Statement II is... | ["C"] | null | Area under velocity time graph gives displacement of body in given time.
<br/><br/>
Area under acceleration time graph gives change in velocity in the given time.<br/><br/>
So Statement I false but Statement II True | mcq | jee-main-2023-online-8th-april-evening-shift | 12,073 |
lvb29gxb | physics | motion-in-a-straight-line | average-acceleration-and-instantaneous-acceleration | <p>A particle moves in a straight line so that its displacement $$x$$ at any time $$t$$ is given by $$x^2=1+t^2$$. Its acceleration at any time $$\mathrm{t}$$ is $$x^{-\mathrm{n}}$$ where $$\mathrm{n}=$$ _________.</p> | [] | null | 3 | <p>Given the displacement of the particle $$x^2 = 1 + t^2$$, we want to find the acceleration, which is the second derivative of displacement with respect to time, $$a = \frac{d^2x}{dt^2}$$, and we are given that the acceleration at any time $$t$$ is $$x^{-n}$$, for us to find the value of $$n$$.</p>
<p>First, let's f... | integer | jee-main-2024-online-6th-april-evening-shift | 12,074 |
1ldofog2s | physics | motion-in-a-straight-line | average-speed-and-instantaneous-speed | <p>An object moves with speed $$v_1,v_2$$ and $$v_3$$ along a line segment AB, BC and CD respectively as shown in figure. Where AB = BC and AD = 3AB, then average speed of the object will be:</p>
<p><img src="data:image/png;base64,UklGRo4GAABXRUJQVlA4IIIGAACQOgCdASoAA4gAPm02mUgkIyKhIpRKEIANiWlu4XShG/OB8Z9jX6u8017N8tplt... | [{"identifier": "A", "content": "$${{{v_1}{v_2}{v_3}} \\over {3({v_1}{v_2} + {v_2}{v_3} + {v_3}{v_1})}}$$"}, {"identifier": "B", "content": "$${{({v_1} + {v_2} + {v_3})} \\over 3}$$"}, {"identifier": "C", "content": "$${{({v_1} + {v_2} + {v_3})} \\over {3{v_1}{v_2}{v_3}}}$$"}, {"identifier": "D", "content": "$${{3{v_1}... | ["D"] | null | $A B=B C=C D$
<br/><br/>$$
\begin{aligned}
\Rightarrow \text { Average speed } & =\frac{\text { Distance }}{\text { Time }} \\\\
& =\frac{A D}{\frac{A B}{V_{1}}+\frac{A B}{V_{2}}+\frac{A B}{V_{3}}} \\\\
& =\frac{3 V_{1} V_{2} V_{3}}{V_{1} V_{2}+V_{2} V_{3}+V_{1} V_{3}}
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift | 12,075 |
ldquqh93 | physics | motion-in-a-straight-line | average-speed-and-instantaneous-speed | A vehicle travels $4 \mathrm{~km}$ with speed of $3 \mathrm{~km} / \mathrm{h}$ and another $4 \mathrm{~km}$ with speed of $5 \mathrm{~km} / \mathrm{h}$, then its average speed is | [{"identifier": "A", "content": "$3.75 \\mathrm{~km} / \\mathrm{h}$"}, {"identifier": "B", "content": "$4.25 \\mathrm{~km} / \\mathrm{h}$"}, {"identifier": "C", "content": "$3.50 \\mathrm{~km} / \\mathrm{h}$"}, {"identifier": "D", "content": "$4.00 \\mathrm{~km} / \\mathrm{h}$"}] | ["A"] | null | <p>Average speed</p>
<p>$$ = {{\mathrm{Total\,dis\tan ce}} \over {\mathrm{Total\,time}}}$$</p>
<p>$$ = {8 \over {{4 \over 3} + {4 \over 5}}}$$</p>
<p>$$ = 3.75$$ km/h</p> | mcq | jee-main-2023-online-30th-january-evening-shift | 12,076 |
1ldr33bvy | physics | motion-in-a-straight-line | average-speed-and-instantaneous-speed | <p>A horse rider covers half the distance with $$5 \mathrm{~m} / \mathrm{s}$$ speed. The remaining part of the distance was travelled with speed $$10 \mathrm{~m} / \mathrm{s}$$ for half the time and with speed $$15 \mathrm{~m} / \mathrm{s}$$ for other half of the time. The mean speed of the rider averaged over the whol... | [] | null | 50 | <p>$$ \Rightarrow {t_1} = {{{S \over 2}} \over 5}$$ ........ (1)</p>
<p>Also, $${S \over 2} = {{10{t_2}} \over 2} + {{15{t_2}} \over 2}$$</p>
<p>$$ \Rightarrow {t_2} = {S \over {25}}$$ ....... (2)</p>
<p>$$\Rightarrow$$ Mean speed $$ = {S \over {{t_1} + {t_2}}}$$</p>
<p>$$ = {S \over {{S \over {10}} + {S \over {25}}}} ... | integer | jee-main-2023-online-30th-january-morning-shift | 12,077 |
1lduhkejs | physics | motion-in-a-straight-line | average-speed-and-instantaneous-speed | <p>A car travels a distance of '$$x$$' with speed $$v_1$$ and then same distance '$$x$$' with speed $$v_2$$ in the same direction. The average speed of the car is :</p> | [{"identifier": "A", "content": "$${{{v_1}{v_2}} \\over {2({v_1} + {v_2})}}$$"}, {"identifier": "B", "content": "$${{2{v_1}{v_2}} \\over {{v_1} + {v_2}}}$$"}, {"identifier": "C", "content": "$${{2x} \\over {{v_1} + {v_2}}}$$"}, {"identifier": "D", "content": "$${{{v_1} + {v_2}} \\over 2}$$"}] | ["B"] | null | $$
\begin{aligned}
& \text { Average velocity }=\frac{\text { Total displacement }}{\text { Total time }} \\\\
& =\frac{\mathrm{x}+\mathrm{x}}{\frac{\mathrm{x}}{\mathrm{v}_1}+\frac{\mathrm{x}}{\mathrm{v}_2}}=\frac{2 \mathrm{v}_1 \mathrm{v}_2}{\mathrm{v}_1+\mathrm{v}_2}
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift | 12,078 |
1lgoye1mh | physics | motion-in-a-straight-line | average-speed-and-instantaneous-speed | <p>The distance travelled by an object in time $$t$$ is given by $$s=(2.5) t^{2}$$. The instantaneous speed of the object at $$\mathrm{t}=5 \mathrm{~s}$$ will be:</p> | [{"identifier": "A", "content": "$$5 \\mathrm{~ms}^{-1}$$"}, {"identifier": "B", "content": "$$12.5 \\mathrm{~ms}^{-1}$$"}, {"identifier": "C", "content": "$$62.5 \\mathrm{~ms}^{-1}$$"}, {"identifier": "D", "content": "$$25 \\mathrm{~ms}^{-1}$$"}] | ["D"] | null | The distance traveled by an object in time $$t$$ is given by the equation $$s = (2.5)t^2$$. To find the instantaneous speed at a specific time, we need to find the first derivative of the distance function with respect to time, which gives us the velocity function:
<br/><br/>
$$v(t) = \frac{ds}{dt}$$
<br/><br/>
Differe... | mcq | jee-main-2023-online-13th-april-evening-shift | 12,079 |
luyita7s | physics | motion-in-a-straight-line | average-speed-and-instantaneous-speed | <p>A particle moving in a straight line covers half the distance with speed $$6 \mathrm{~m} / \mathrm{s}$$. The other half is covered in two equal time intervals with speeds $$9 \mathrm{~m} / \mathrm{s}$$ and $$15 \mathrm{~m} / \mathrm{s}$$ respectively. The average speed of the particle during the motion is :</p> | [{"identifier": "A", "content": "9.2 m/s"}, {"identifier": "B", "content": "8.8 m/s"}, {"identifier": "C", "content": "10 m/s"}, {"identifier": "D", "content": "8 m/s"}] | ["D"] | null | <p>Let's denote the total distance covered by the particle as $$2d$$, where $$d$$ is the distance for each half. To calculate the average speed, we need to find the total distance traveled and divide it by the total time taken.</p>
<p>For the first half of the journey, the particle covers the distance $$d$$ at a speed... | mcq | jee-main-2024-online-9th-april-morning-shift | 12,080 |
lvc5838s | physics | motion-in-a-straight-line | average-speed-and-instantaneous-speed | <p>A train starting from rest first accelerates uniformly up to a speed of $$80 \mathrm{~km} / \mathrm{h}$$ for time $$t$$, then it moves with a constant speed for time $$3 t$$. The average speed of the train for this duration of journey will be (in $$\mathrm{km} / \mathrm{h}$$) :</p> | [{"identifier": "A", "content": "70"}, {"identifier": "B", "content": "40"}, {"identifier": "C", "content": "30"}, {"identifier": "D", "content": "80"}] | ["A"] | null | <p>To find the average speed of the train for the duration of the journey, we need to know the total distance covered by the train and the total time taken.</p>
<p>The train accelerates uniformly to a speed of $$80 \, \mathrm{km/h}$$ over time $$t$$, and then moves at this constant speed for $$3t$$. The average speed ... | mcq | jee-main-2024-online-6th-april-morning-shift | 12,081 |
dLNi3PAPf18R3gA5 | physics | motion-in-a-straight-line | average-velocity-and-instantaneous-velocity | The co-ordinates of a moving particle at any time 't' are given by x = $$\alpha $$t<sup>3</sup> and y = βt<sup>3</sup>. The speed to the particle at time 't' is given by | [{"identifier": "A", "content": "$$3t\\sqrt {{\\alpha ^2} + {\\beta ^2}} $$"}, {"identifier": "B", "content": "$$3{t^2}\\sqrt {{\\alpha ^2} + {\\beta ^2}} $$"}, {"identifier": "C", "content": "$${t^2}\\sqrt {{\\alpha ^2} + {\\beta ^2}} $$"}, {"identifier": "D", "content": "$$\\sqrt {{\\alpha ^2} + {\\beta ^2}} $$"}] | ["B"] | null | Given that $$x = \alpha {t^3}\,\,\,\,$$ and $$\,\,\,\,y = \beta {t^3}$$
<br><br>$$\therefore$$ $${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$$
<br><br>and$$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$$
<br><br>$$\therefore$$ $$v = \sqrt {v_x^2 + v_y^2} $$
<br><br>$$ = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2... | mcq | aieee-2003 | 12,082 |
1krwa2r8z | physics | motion-in-a-straight-line | average-velocity-and-instantaneous-velocity | The instantaneous velocity of a particle moving in a straight line is given as $$V = \alpha t + \beta {t^2}$$, where $$\alpha$$ and $$\beta$$ are constants. The distance travelled by the particle between 1s and 2s is : | [{"identifier": "A", "content": "3$$\\alpha$$ + 7$$\\beta$$"}, {"identifier": "B", "content": "$${3 \\over 2}\\alpha + {7 \\over 3}\\beta $$"}, {"identifier": "C", "content": "$${\\alpha \\over 2} + {\\beta \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over 2}\\alpha + {7 \\over 2}\\beta $$"}] | ["B"] | null | $$V = \alpha t + \beta {t^2}$$<br><br>$${{ds} \over {dt}} = \alpha t + \beta {t^2}$$<br><br>$$\int\limits_{{S_1}}^{{S_2}} {ds = \int\limits_1^2 {(\alpha t + \beta {t^2})dt} } $$<br><br>$${S_2} - {S_1} = \left[ {{{\alpha {t^2}} \over 2} + {{\beta {t^3}} \over 3}} \right]_1^2$$<br><br>As particle is not changing directio... | mcq | jee-main-2021-online-25th-july-evening-shift | 12,083 |
1l6p41lxu | physics | motion-in-a-straight-line | average-velocity-and-instantaneous-velocity | <p>If $$\mathrm{t}=\sqrt{x}+4$$, then $$\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)_{\mathrm{t}=4}$$ is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "zero"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "16"}] | ["B"] | null | Given,
<br/><br/>$t=\sqrt{x}+4$, Squaring on both
<br/><br/>$$
\begin{aligned}
& x=(t-4)^2=t^2-8 t+16 \\\\
& \frac{d x}{d t}=2 t-8\\\\
& \text { at } t=4 \\\\
& \frac{d x}{d t}=8-8=0
\end{aligned}
$$ | mcq | jee-main-2022-online-29th-july-morning-shift | 12,085 |
1ldtxhh99 | physics | motion-in-a-straight-line | average-velocity-and-instantaneous-velocity | <p>The distance travelled by a particle is related to time t as $$x=4\mathrm{t}^2$$. The velocity of the particle at t=5s is :-</p> | [{"identifier": "A", "content": "$$\\mathrm{25~ms^{-1}}$$"}, {"identifier": "B", "content": "$$\\mathrm{20~ms^{-1}}$$"}, {"identifier": "C", "content": "$$\\mathrm{8~ms^{-1}}$$"}, {"identifier": "D", "content": "$$\\mathrm{40~ms^{-1}}$$"}] | ["D"] | null | $$
\begin{aligned}
& x=4 t^2 \\\\
& v=\frac{d x}{d t}=8 t
\end{aligned}
$$<br/><br/>
At $\mathrm{t}=5 ~\mathrm{sec}$<br/><br/>
$$
\mathrm{v}=8 \times 5=40 \mathrm{~m} / \mathrm{s}
$$ | mcq | jee-main-2023-online-25th-january-evening-shift | 12,086 |
lgnxvqd2 | physics | motion-in-a-straight-line | average-velocity-and-instantaneous-velocity | The position of a particle related to time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$. The magnitude of velocity of the particle at $t=2 s$ will be : | [{"identifier": "A", "content": "$14 \\mathrm{~ms}^{-1}$"}, {"identifier": "B", "content": "$16 \\mathrm{~ms}^{-1}$"}, {"identifier": "C", "content": "$10 \\mathrm{~ms}^{-1}$"}, {"identifier": "D", "content": "$06 \\mathrm{~ms}^{-1}$\n"}] | ["B"] | null | The position of a particle as a function of time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$. <br/><br/>To find the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$, we first need to find the velocity of the particle as a function of time.
<br/><br/>
The velocity $v$ is the time derivative of t... | mcq | jee-main-2023-online-15th-april-morning-shift | 12,087 |
1lgvr2x57 | physics | motion-in-a-straight-line | average-velocity-and-instantaneous-velocity | <p>A person travels $$x$$ distance with velocity $$v_{1}$$ and then $$x$$ distance with velocity $$v_{2}$$ in the same direction. The average velocity of the person is $$\mathrm{v}$$, then the relation between $$v, v_{1}$$ and $$v_{2}$$ will be.</p> | [{"identifier": "A", "content": "$$\\mathbf{V}=\\mathbf{V}_{1}+\\mathbf{V}_{2}$$"}, {"identifier": "B", "content": "$$V=\\frac{v_{1}+V_{2}}{2}$$"}, {"identifier": "C", "content": "$$\\frac{1}{\\mathrm{v}}=\\frac{1}{\\mathrm{v}_{1}}+\\frac{1}{\\mathrm{v}_{2}}$$"}, {"identifier": "D", "content": "$$\\frac{2}{\\mathrm{~V}... | ["D"] | null | <p>The average velocity is defined as the total displacement divided by the total time. Here, the person travels the same distance $x$ twice, once with velocity $v_1$ and once with velocity $v_2$. </p>
<p>The time to travel distance $x$ with velocity $v_1$ is $t_1 = \frac{x}{v_1}$, and the time to travel distance $x$ w... | mcq | jee-main-2023-online-10th-april-evening-shift | 12,088 |
kEn7LbtD3nafqy6u | physics | motion-in-a-straight-line | constant-acceleration-motion | An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 m.
If the car is going twice as fast, i.e 120 km/h, the stopping distance will be | [{"identifier": "A", "content": "60 m"}, {"identifier": "B", "content": "40 m"}, {"identifier": "C", "content": "20 m"}, {"identifier": "D", "content": "80 m"}] | ["D"] | null | Assume $$a$$ be the retardation for both the vehicle then
<br><br>In case of automobile,
<br><br>$$u_1^2 - 2a{s_1} = 0$$
<br><br>$$ \Rightarrow u_1^2 = 2a{s_1}$$
<br><br>And in case for car,
<br><br>$$u_2^2 = 2a{s_2}$$
<br><br>$$\therefore$$ $${\left( {{{{u_2}} \over {{u_1}}}} \right)^2} = {{{s_2}} \over {{s_1}}}$$
<br... | mcq | aieee-2004 | 12,091 |
FJLr8Z4hwAqCn4QY | physics | motion-in-a-straight-line | constant-acceleration-motion | A car starting from rest accelerates at the rate f through a distance S, then continues
at constant speed for time t and then decelerates at the rate $${f \over 2}$$ to come to rest. If the
total distance traversed is 15 S, then | [{"identifier": "A", "content": "$$S = {1 \\over 6}f{t^2}$$"}, {"identifier": "B", "content": "$$S = ft$$"}, {"identifier": "C", "content": "$$S = {1 \\over 4}f{t^2}$$"}, {"identifier": "D", "content": "$$S = {1 \\over 72}f{t^2}$$"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263437/exam_images/rulc8xik6beuixultxbg.webp" loading="lazy" alt="AIEEE 2005 Physics - Motion in a Straight Line Question 103 English Explanation">
Initially car starts from rest so u = 0.
<br><br>Now distance from $$A$$ to $$B$$,
<... | mcq | aieee-2005 | 12,092 |
s5pd1xyXZdDWWlbt8dedb | physics | motion-in-a-straight-line | constant-acceleration-motion | In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed '$$\upsilon $$' more than that of car B. Both the cars start from rest and travel with constant acceleration a<sub>1</sub> and a<sub>2</sub> respectively. Then '$$\upsilon $$' is equal to : | [{"identifier": "A", "content": "$${{2{a_1}{a_2}} \\over {{a_1} + {a_2}}}t$$"}, {"identifier": "B", "content": "$$\\sqrt {2{a_1}{a_2}} t$$"}, {"identifier": "C", "content": "$$\\sqrt {{a_1}{a_2}} t$$"}, {"identifier": "D", "content": "$${{{a_1} + {a_2}} \\over 2}t$$"}] | ["C"] | null | For both car initial speed ($$\mu $$) = 0
<br><br>Let the acceleration of car A and car B is $$a$$<sub>1</sub> and $$a$$<sub>2</sub> respectively.
<br><br>Also let the time taken to reach the finishing point for car A is t<sub>1</sub> and for car B is t<sub>2</sub>.
<br><br>Let at finishing point speed of car A is $$... | mcq | jee-main-2019-online-9th-january-evening-slot | 12,094 |
U3xcTPNDrF9q7GuLFq5g4 | physics | motion-in-a-straight-line | constant-acceleration-motion | A particle moves from the point $$\left( {2.0\widehat i + 4.0\widehat j} \right)$$ m, at t = 0, with an initial velocity $$\left( {5.0\widehat i + 4.0\widehat j} \right)$$ ms<sup>$$-$$1</sup>. It is acted upon by a constant force which produces a constant acceleration $$\left( {4.0\widehat i + 4.0\widehat j} \right)$$ ... | [{"identifier": "A", "content": "15 m"}, {"identifier": "B", "content": "$$20\\sqrt 2 $$ m"}, {"identifier": "C", "content": "$$10\\sqrt 2 $$ m"}, {"identifier": "D", "content": "5 m"}] | ["B"] | null | $$\overrightarrow S = \left( {5\widehat i + 4} \right)2 + {1 \over 2}\left( {4\widehat i + 4\widehat j} \right)4$$
<br><br>$$ = 10\widehat i + 8\widehat j + 8\widehat i + 8\widehat j$$
<br><br>$$\overrightarrow {{r_f}} - \overrightarrow {{r_i}} = 18\widehat i + 16\widehat j$$
<br><br>$$\overrightarrow {{r_f}} = 20\... | mcq | jee-main-2019-online-11th-january-evening-slot | 12,095 |
1lBrfl6CGumvivDoYF1klukhpye | physics | motion-in-a-straight-line | constant-acceleration-motion | A scooter accelerates from rest for time t<sub>1</sub> at constant rate a<sub>1</sub> and then retards at constant rate a<sub>2</sub> for time t<sub>2</sub> and comes to rest. The correct value of $${{{t_1}} \over {{t_2}}}$$ wil be : | [{"identifier": "A", "content": "$${{{a_1} + {a_2}} \\over {{a_2}}}$$"}, {"identifier": "B", "content": "$${{{a_1} + {a_2}} \\over {{a_1}}}$$"}, {"identifier": "C", "content": "$${{{a_2}} \\over {{a_1}}}$$"}, {"identifier": "D", "content": "$${{{a_1}} \\over {{a_2}}}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265962/exam_images/aa6fqz7vcklexdjoopmy.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Physics - Motion in a Straight Line Question 69 English Explanation">
... | mcq | jee-main-2021-online-26th-february-evening-slot | 12,097 |
lyhEKI3Puvr1IK36cr1kmj1t82f | physics | motion-in-a-straight-line | constant-acceleration-motion | A car accelerates from rest at a constant rate $$\alpha$$ for some time after which it decelerates at a constant rate $$\beta$$ to come to rest. If the total time elapsed is t seconds, the total distance travelled is : | [{"identifier": "A", "content": "$${{4\\alpha \\beta } \\over {(\\alpha + \\beta )}}{t^2}$$"}, {"identifier": "B", "content": "$${{2\\alpha \\beta } \\over {(\\alpha + \\beta )}}{t^2}$$"}, {"identifier": "C", "content": "$${{\\alpha \\beta } \\over {2(\\alpha + \\beta )}}{t^2}$$"}, {"identifier": "D", "content": "$$... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263313/exam_images/ctgpgvnqaxdsgp9gnqoz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Morning Shift Physics - Motion in a Straight Line Question 67 English Explanation">
<br... | mcq | jee-main-2021-online-17th-march-morning-shift | 12,098 |
5JczKa1YBwDaPgOgOP1kmkrcrvd | physics | motion-in-a-straight-line | constant-acceleration-motion | The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by : | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266070/exam_images/nrybq2z0bup6lxkqt4ue.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2021 (Online) 18th March Morning Shift Physics - Motion in a Straight Li... | ["B"] | null | Acceleration(a) is constant
<br><br>$$ \therefore $$ v $$ \propto $$ t (straight line graph)
<br><br>and x $$ \propto $$ t<sup>2</sup>
(parabolic graph) | mcq | jee-main-2021-online-18th-march-morning-shift | 12,099 |
1l54uibr0 | physics | motion-in-a-straight-line | constant-acceleration-motion | <p>A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10m in t s, the distance travelled by the toy in the next t s will be :</p> | [{"identifier": "A", "content": "10 m"}, {"identifier": "B", "content": "20 m"}, {"identifier": "C", "content": "30 m"}, {"identifier": "D", "content": "40 m"}] | ["C"] | null | <p>A small toy begins to move from a standstill with a constant acceleration. It covers a distance of 10 meters in t seconds. We need to determine the distance the toy will travel in the subsequent t seconds.</p>
<p>First, we know that the initial distance traveled is given by the formula for constant acceleration sta... | mcq | jee-main-2022-online-29th-june-evening-shift | 12,100 |
1l6dzqwfz | physics | motion-in-a-straight-line | constant-acceleration-motion | <p>A car is moving with speed of $$150 \mathrm{~km} / \mathrm{h}$$ and after applying the break it will move $$27 \mathrm{~m}$$ before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ___________ m distance.</p> | [] | null | 3 | <p>$${F_R}\,d = {1 \over 2}m{v^2}$$</p>
<p>$${{{d_2}} \over {{d_1}}} = {\left( {{{{v_2}} \over {{v_1}}}} \right)^2} = {\left( {{1 \over 3}} \right)^2}$$</p>
<p>$${d_2} = {d_1} \times {1 \over 9} = 3m$$</p> | integer | jee-main-2022-online-25th-july-morning-shift | 12,101 |
1l6kmlo5h | physics | motion-in-a-straight-line | constant-acceleration-motion | <p>The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is :</p> | [{"identifier": "A", "content": "2.0"}, {"identifier": "B", "content": "1.0"}, {"identifier": "C", "content": "0.5"}, {"identifier": "D", "content": "1.5"}] | ["C"] | null | <p>S = 4 cm</p>
<p>$$v{'_4} = {v \over 3}$$, a = constant</p>
<p>$${v_{4 + x}} = 0$$</p>
<p>$$\left( {{v^2} - {{{v^2}} \over a}} \right) = 2a(4)$$</p>
<p>$$({v^2} - 0) = 2a(4 + x)$$</p>
<p>$${4 \over {4 + x}} = {8 \over 9}$$</p>
<p>$$ \Rightarrow x = 0.5$$ m</p> | mcq | jee-main-2022-online-27th-july-evening-shift | 12,102 |
1ldnzu2i0 | physics | motion-in-a-straight-line | constant-acceleration-motion | <p>For a train engine moving with speed of $$20 \mathrm{~ms}^{-1}$$, the driver must apply brakes at a distance of 500 $$\mathrm{m}$$ before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $$\sqrt{x} \math... | [] | null | 200 | By using $3^{\text {rd }}$ equation of motion
<br/><br/>$$
\begin{aligned}
& v^2=u^2+2 a s \\\\
& (0)^2=u^2+2 a s \\\\
& u^2=-2 a s \\\\
& S=\frac{u^2}{2 a}-\frac{(20)^2}{2 \times a}=500 \\\\
& \text { acceleration of the train, } a=-\frac{400}{1000}=-0.4 \mathrm{~m} / \mathrm{sec}
\end{aligned}
$$
<br/><br/>Now, if th... | integer | jee-main-2023-online-1st-february-evening-shift | 12,103 |
1lh2zn7yx | physics | motion-in-a-straight-line | constant-acceleration-motion | <p>A particle starts with an initial velocity of $$10.0 \mathrm{~ms}^{-1}$$ along $$x$$-direction and accelerates uniformly at the rate of $$2.0 \mathrm{~ms}^{-2}$$. The time taken by the particle to reach the velocity of $$60.0 \mathrm{~ms}^{-1}$$ is __________.</p> | [{"identifier": "A", "content": "30s"}, {"identifier": "B", "content": "6s"}, {"identifier": "C", "content": "3s"}, {"identifier": "D", "content": "25s"}] | ["D"] | null | <p>To find the time taken by the particle to reach the velocity of $$60.0 \mathrm{~ms}^{-1}$$, we can use the formula:</p>
<p>$$
v = u + at
$$</p>
<p>Where:
$$v$$ is the final velocity,
$$u$$ is the initial velocity,
$$a$$ is the acceleration, and
$$t$$ is the time taken.</p>
<p>Plugging in the given values:</p>
<p>$$
... | mcq | jee-main-2023-online-6th-april-evening-shift | 12,104 |
jaoe38c1lsf1tqy7 | physics | motion-in-a-straight-line | constant-acceleration-motion | <p>A body starts moving from rest with constant acceleration covers displacement $$S_1$$ in first $$(p-1)$$ seconds and $$\mathrm{S}_2$$ in first $$p$$ seconds. The displacement $$\mathrm{S}_1+\mathrm{S}_2$$ will be made in time :</p> | [{"identifier": "A", "content": "$$(2 p+1) s$$\n"}, {"identifier": "B", "content": "$$(2 p-1) s$$\n"}, {"identifier": "C", "content": "$$\\left(2 p^2-2 p+1\\right) s$$\n"}, {"identifier": "D", "content": "$$\\sqrt{\\left(2 p^2-2 p+1\\right)} s$$"}] | ["D"] | null | <p>Let's denote the constant acceleration with which the body moves as $$a$$. We know that the displacement $$S$$ covered by a body starting from rest under a constant acceleration $$a$$ in time $$t$$ is given by the equation of motion: $$S = \frac{1}{2} a t^2$$.</p>
<p>Considering the displacement $$\mathrm{S}_1$$ in... | mcq | jee-main-2024-online-29th-january-morning-shift | 12,105 |
1lsgxiwe5 | physics | motion-in-a-straight-line | constant-acceleration-motion | <p>The displacement and the increase in the velocity of a moving particle in the time interval of $$t$$ to $$(t+1) \mathrm{s}$$ are $$125 \mathrm{~m}$$ and $$50 \mathrm{~m} / \mathrm{s}$$, respectively. The distance travelled by the particle in $$(\mathrm{t}+2)^{\mathrm{th}} \mathrm{s}$$ is _________ m.</p> | [] | null | 175 | <p>The displacement and the increase in the velocity of a moving particle from time $$t$$ to $$(t + 1) \mathrm{s}$$ are $$125 \mathrm{~m}$$ and $$50 \mathrm{~m} / \mathrm{s}$$, respectively. The distance traveled by the particle in $$(\mathrm{t} + 2)^{\mathrm{th}} \mathrm{s}$$ is calculated as follows:
<p>Given that t... | integer | jee-main-2024-online-30th-january-morning-shift | 12,106 |
luxwdjas | physics | motion-in-a-straight-line | constant-acceleration-motion | <p>Two cars are travelling towards each other at speed of $$20 \mathrm{~m} \mathrm{~s}^{-1}$$ each. When the cars are $$300 \mathrm{~m}$$ apart, both the drivers apply brakes and the cars retard at the rate of $$2 \mathrm{~m} \mathrm{~s}^{-2}$$. The distance between them when they come to rest is :</p> | [{"identifier": "A", "content": "25 m"}, {"identifier": "B", "content": "100 m"}, {"identifier": "C", "content": "50 m"}, {"identifier": "D", "content": "200 m"}] | ["B"] | null | <p>Let's analyze the given information before determining the distance between the two cars when they come to rest. Each car is traveling towards the other at a speed of $$20 \, \text{m s}^{-1}$$ and they both start braking when they are $$300 \, \text{m}$$ apart. The deceleration (negative acceleration) of each car is... | mcq | jee-main-2024-online-9th-april-evening-shift | 12,107 |
APlPd3SjJvDrtvev | physics | motion-in-a-straight-line | graph-based-problem | A body is at rest at $$x=0.$$ At $$t=0,$$ it starts moving in the positive $$x$$-direction with a constant acceleration. At the same instant another body passes through $$x=0$$ moving in the positive $$x$$ direction with a constant speed. The position of the first body is given by $${x_1}\left( t \right)$$ after time $... | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264714/exam_images/gzqf0luasxfctyuu2cjb.webp\" loading=\"lazy\" alt=\"AIEEE 2008 Physics - Motion in a Straight Line Question 93 English Option 1\"> "}, {"identifier": "B", "content": "<img class=... | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263490/exam_images/b0nmzkbuamx9kfmuqjc0.webp" loading="lazy" alt="AIEEE 2008 Physics - Motion in a Straight Line Question 93 English Explanation">
<br><br><b>For the body starting from rest</b>
<br><br>$${x_1} = 0 + {1 \over 2}a{t... | mcq | aieee-2008 | 12,111 |
vhabexZH1WQvUdxX | physics | motion-in-a-straight-line | graph-based-problem | Consider a rubber ball freely falling from a height $$h=4.9$$ $$m$$ onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic.
<br/><br/>Then the velocity as a function of time and the height as a function of time will be : | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263548/exam_images/qahcebniqd3idscacibg.webp\" loading=\"lazy\" alt=\"AIEEE 2009 Physics - Motion in a Straight Line Question 92 English Option 1\"> "}, {"identifier": "B", "content": "<img src=\"... | ["B"] | null | <b>For downward motion :</b>
<br/><br/> $$v=-gt$$
<br><br>The velocity of the rubber ball increases in downward direction and we get a straight line between $$v$$ and $$t$$ with a negative slope.
<br><br>Also applying $$y - {y_0} = ut + {1 \over 2}a{t^2}$$
<br><br>We get $$y - h = - {1 \over 2}g{t^2} \Rightarrow y = ... | mcq | aieee-2009 | 12,112 |
pIoNx5jZrKM2ulVn | physics | motion-in-a-straight-line | graph-based-problem | A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs
time? | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ym679k/b64b04e1-2781-4b32-934e-b8ead9c2dc90/7cffbc80-3284-11ed-8893-19b23ee4c66d/file-1l7ym679l.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l7ym679k/b64b04e1-2781-4b32-934e-b8ead9c2dc90/7cf... | ["D"] | null | Motion of the particle is shown below<br><br>
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265752/exam_images/mgoei8a6p69pysxxgycv.webp" loading="lazy" alt="JEE Main 2017 (Offline) Physics - Motion in a Straight Line Question 94 English Explanation">
<br><br>Initially speed of... | mcq | jee-main-2017-offline | 12,113 |
Ivw9b9de0zttW79dNoHtB | physics | motion-in-a-straight-line | graph-based-problem | Which graph corresponds to an object moving with a constant negative
acceleration and a positive velocity ?
| [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l83nomz1/fa176d71-e87b-48bf-9625-cb66c114583b/5212cfd0-354a-11ed-8c5c-f3f0850e6b94/file-1l83nomz2.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l83nomz1/fa176d71-e87b-48bf-9625-cb66c114583b/521... | ["C"] | null | Given that,
<br><br>acceleration (a) = $$-$$ C (constant)
<br><br>$$\therefore\,\,\,$$ $${{dv} \over {dt}}$$ = $$-$$ c
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{dv} \over {dx}}$$ . $${{dx} \over {dt}}$$ = $$-$$ c
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\upsilon $$ $${{dv} \over {dx}}$$ = $$-$$ c
<br><br>$$ \Rightarrow $$$... | mcq | jee-main-2017-online-8th-april-morning-slot | 12,114 |
lQ2mcgZS4Esf8yPLKEADs | physics | motion-in-a-straight-line | graph-based-problem | The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in $$15$$ $$s$$ and (ii) the time at which the car will catch up with the scooter are, respectively.
<br/><br/><img src="data:image/png;base64,UklGRqYaAABXRUJQVlA4IJoaAAA... | [{"identifier": "A", "content": "$$112.5$$ $$m$$ and $$22.5$$ $$s$$"}, {"identifier": "B", "content": "$$337.5$$ $$m$$ and $$25$$ $$s$$ "}, {"identifier": "C", "content": "$$112.5$$ $$m$$ and $$15$$ $$s$$ "}, {"identifier": "D", "content": "$$225.5$$ $$m$$ and $$10$$ $$s$$ "}] | ["A"] | null | Till 15 sec car has accelerative motion and scooter has constant velocity in entire motion.
<br><br>Total Distance travelled by the car in 15 sec, = $${1 \over 2}$$ $$ \times $$ $${{45} \over {15}}$$ $$ \times $$ (15)<sup>2</sup> = $${{675} \over 2}$$ m
<br><br>Distance travelled by scooter in 15 sec.
<br><br>= V $$ ... | mcq | jee-main-2018-online-15th-april-morning-slot | 12,115 |
yjzA2KJcnwMolnku | physics | motion-in-a-straight-line | graph-based-problem | All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up. | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://gateclass.cdn.examgoal.net/SLJCGRTwWe0VZYNpK/7izXT6VjcbhVAKMWBg4cH7s8GyBpD/5w90ktu4kCcg38PR8emGQL/uploadfile.jpg\" loading=\"lazy\" alt=\"JEE Main 2018 (Offline) Physics - Motion in a Straight Line Question 95 English Option 1\">"}, {"identifi... | ["C"] | null | In option (A) you can see velocity versus time graph is a straight line with negative slope, so the acceleration is negative and constant. This motion is look like this.
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265752/exam_images/mgoei8a6p69pysxxgycv.webp" loading="lazy" a... | mcq | jee-main-2018-offline | 12,116 |
n2joOC3LkuMcqiNnqHV8E | physics | motion-in-a-straight-line | graph-based-problem | A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s ?
<br/><br/><img src="data:image/png;base64,UklGRmAgAABXRUJQVlA4IFQgAADw4ACdASoAA3UBP4G01WS2LKunIhFa+sAwCWlu/ADXnRRHZ19/... | [{"identifier": "A", "content": "3 m"}, {"identifier": "B", "content": "9 m"}, {"identifier": "C", "content": "10 m"}, {"identifier": "D", "content": "6 m"}] | ["B"] | null | S = Area under graph
<br><br>$${1 \over 2}$$ $$ \times $$ 2$$ \times $$2 + 2 $$ \times $$ 2 + 3 $$ \times $$ 1 = 9 m | mcq | jee-main-2019-online-10th-january-evening-slot | 12,117 |
VvKPSqXtnMymUVtfDSmv4 | physics | motion-in-a-straight-line | graph-based-problem | A particle starts from origin O from rest and
moves with a uniform acceleration along the
positive x-axis. Identify all figures that
correctly represent the motion qualitatively.
(a = acceleration, v = velocity,
x = displacement, t = time)
<img src="data:image/png;base64,UklGRvIDAABXRUJQVlA4IOYDAACQRQCdASrsAhABP4HA3mY2... | [{"identifier": "A", "content": "(B), (C)"}, {"identifier": "B", "content": "(A)"}, {"identifier": "C", "content": "(A), (B), (C)"}, {"identifier": "D", "content": "(A), (B), (D)"}] | ["D"] | null | Given initial velocity u = 0 and acceleration
is constant<br>
At time t<br>
v = 0 + at<br>
$$ \Rightarrow $$ v = at<br><br>
Also $$x = 0(t) + {1 \over 2}a{t^2}$$<br>
$$ \Rightarrow $$ x = $${1 \over 2}a{t^2}$$<br><br>
Graph (A), (B) and (D) are correct
| mcq | jee-main-2019-online-8th-april-evening-slot | 12,118 |
LM7XqJdILqzykfSgXNjgy2xukfl4nvq9 | physics | motion-in-a-straight-line | graph-based-problem | The velocity (v) and time (t) graph of a body in
a straight line motion is shown in the figure.
The point S is at 4.333 seconds. The total
distance covered by the body in 6 s is :
<img src="data:image/png;base64,UklGRsgRAABXRUJQVlA4ILwRAADQbACdASr0AbwAPm00l0ikIqIhI7I6IIANiWlu5VARQgRvze/IH9A/GTwJ/tn9l/ZX93fYXwZ+EP139mP7... | [{"identifier": "A", "content": "12 m"}, {"identifier": "B", "content": "11 m"}, {"identifier": "C", "content": "$${{49} \\over 4}$$ m"}, {"identifier": "D", "content": "$${{37} \\over 3}$$ m"}] | ["D"] | null | 4.333 sec = $${{13} \over 3}$$ sec
<br><br>Distance = area under the v-t graph
<br><br>= Area of Parallelogram + Area of triangle
<br><br>= $${1 \over 2}\left( 4 \right)\left( {{{13} \over 3} + 1} \right)$$ + $${1 \over 2}\left( {6 - {{13} \over 3}} \right) \times 2$$
<br><br>= $${{37} \over 3}$$ m | mcq | jee-main-2020-online-5th-september-evening-slot | 12,119 |
04Yjag2oAmcOdIfR8Ljgy2xukfaxvhsl | physics | motion-in-a-straight-line | graph-based-problem | The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by
the particle during the time interval t = 0 to t = 5 s will be________.
<img src="data:image/png;base64,UklGRuQQAABXRUJQVlA4INgQAACwAAGdASoAA74CP4HA22W2MK2nIXIY8sAwCWlu/HA3Rp6nZ1+/sb/xLWWz37ecRbD39flstDL3Ff/3quIAz8n5Pyfk... | [] | null | 20 | Distance travelled = Area under speed – time graph
<br><br>= $${1 \over 2} \times 8 \times 5$$ = 20 m | integer | jee-main-2020-online-4th-september-evening-slot | 12,120 |
fvuBRB3TQDoNhV5X1Rjgy2xukeu432t1 | physics | motion-in-a-straight-line | graph-based-problem | A Tennis ball is released from a height h and after freely falling on a wooden floor
it rebounds and reaches height $${h \over 2}$$. The
velocity versus height of the ball during its motion may be represented graphically by : <br/>(graph are drawn schematically and on not to scale)
| [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264310/exam_images/nuabnredurp5srwhdjaq.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 4th September Morning Slot Physics - Motion in a Straight ... | ["C"] | null | V, h curve will be parabolic.
<br><br>Downward velocity is negative and upward is positive
<br><br>When ball is coming down graph will be in IV quadrant and when going up graph will be in I quadrant.
<br><br>At H = h, v = 0
<br><br>at h = 0, v = $$\sqrt {2gh} $$
<br><br>also a = –g, throughout this motion. | mcq | jee-main-2020-online-4th-september-morning-slot | 12,121 |
iwOfMk3zYso6QPykaT1klrhy7on | physics | motion-in-a-straight-line | graph-based-problem | If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?<br/><br/><img src="data:image/png;base64,UklGRsQLAABXRUJQVlA4ILgLAABwqgCdASoAA74BP4HA2mU2MK0nIfMJcsAwCWlu+F+6p34MujOzr7/Xr/Z+rUwE5q7h8R5Cwfvev0p+yMdLZSHtt0UBu3qng47wcd4OO8HHeDjvBx3g400lYq2Zm/ykBHOeHLDcQRC... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l9fnvueo/11a08451-19aa-4638-ac16-a4e6fe10970e/b9a6fb00-4fb0-11ed-88d9-abb495ae8a3e/file-1l9fnvuep.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l9fnvueo/11a08451-19aa-4638-ac16-a4e6fe10970e/b9a... | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxei1rup/1ff42322-cfb9-4031-a388-2c3012567ac7/868fd400-617a-11ec-8d53-3f87ff4a83c4/file-1kxei1ruq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxei1rup/1ff42322-cfb9-4031-a388-2c3012567ac7/868fd400-617a-11ec-8d53-3f87ff4a83c4/fi... | mcq | jee-main-2021-online-24th-february-morning-slot | 12,122 |
DjhGBA1mxMMWPJs61N1kmhnczqj | physics | motion-in-a-straight-line | graph-based-problem | The velocity-displacement graph describing the motion of bicycle is shown in the figure.<br/><br/><img src="data:image/png;base64,UklGRowKAABXRUJQVlA4IIAKAACwPgCdASoZAcsAPm02mEikIyghIjUKIQANiWlu/HyYbOtQ0v0K/k3Zh/V/6J+yf8z8hv0393/J/ctv4ryS/Uf8V/EPKX9I/5t5A8Aj0x/Yf5R/QPLv/UPsY8YMAf0n/1X8n/sviifr/8g/aL+i/AH069gD9Ef8z/If7L... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267469/exam_images/hx7o8xve1quzgelwhbjj.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion in a Straight Li... | ["A"] | null | We know that, $$a = v{{dv} \over {dx}}$$<br><br>as slope is constant, so a $$\propto$$ v (from x = 0 to 200 m)
<br><br>& slope = 0 so a = 0 (from x = 200 to 400 m)<br><br>$$ \therefore $$ <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266096/exam_images/hja6qnoqxdfhdrniwxbu.webp" style="max-width... | mcq | jee-main-2021-online-16th-march-morning-shift | 12,123 |
89pWYEPOYNDl13C3Ru1kmlvvc6z | physics | motion-in-a-straight-line | graph-based-problem | The velocity $$-$$ displacement graph of a particle is shown in the figure.<br/><br/><img src="data:image/png;base64,UklGRvoDAABXRUJQVlA4IO4DAABwIACdASq1AKEAPm02mUgkIyKhJZkJQIANiWlu4XHhG/Of8gdQjldsQDVaPff5b7rtlXug6zvk7/ZvztEaX+l/l/7Ae1//XfxT+nftx7XPmv/X/zD4Cf1B/238u/rvuO+vj0P/1JHSIBnkpuYV58VDYrsPIGnCcT7ePrw8nzk9OE4n2+1... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264350/exam_images/vhzvovygbhuq9thx1crb.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2021 (Online) 18th March Evening Shift Physics - Motion in a Straight Li... | ["A"] | null | <p>The slope of the given v versus x graph is $$m = - {{{v_0}} \over {{x_0}}}$$ and intercept is c = + v<sub>0</sub>. Hence, v varies with x as</p>
<p>$$v = - \left( {{{{v_0}} \over {{x_0}}}} \right)x + {v_0}$$ ...... (1)</p>
<p>where v<sub>0</sub> and x<sub>0</sub> are constants of motion. Differentiating with respe... | mcq | jee-main-2021-online-18th-march-evening-shift | 12,124 |
1l6jetom7 | physics | motion-in-a-straight-line | graph-based-problem | <p>A bullet is shot vertically downwards with an initial velocity of $$100 \mathrm{~m} / \mathrm{s}$$ from a certain height. Within 10 s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time $$\mathrm{t}=20 \mathrm{~s}$$ will be... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6ku0wa1/1d0257ae-cae3-4d48-9538-d623b13af7ca/02075590-1724-11ed-b0f6-478ddda54bc0/file-1l6ku0wa2.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l6ku0wa1/1d0257ae-cae3-4d48-9538-d623b13af7ca/020... | ["A"] | null | <p>$$|{v_{10}}| = (100 + 10 \times 10)$$ m/s</p>
<p>$${v_{10}} = - 200$$ m/s and $${v_0} = - 100$$ m/s</p>
<p>from 10s to 20s velocity remains zero</p>
<p>$$\Rightarrow$$ from t = 0s to 10s velocity increases in magnitude linearly.</p>
<p>$$\Rightarrow$$ graph given in option A fits correctly.</p> | mcq | jee-main-2022-online-27th-july-morning-shift | 12,126 |
1ldr2duen | physics | motion-in-a-straight-line | graph-based-problem | <p>Match Column-I with Column-II :</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:s... | [{"identifier": "A", "content": "A- II, B-III, C-IV, D-I"}, {"identifier": "B", "content": "A- II, B-IV, C-III, D-I"}, {"identifier": "C", "content": "A- I, B-III, C-IV, D-II"}, {"identifier": "D", "content": "A- I, B-II, C-III, D-IV"}] | ["B"] | null | $$
\begin{aligned}
& \frac{\mathrm{dx}}{\mathrm{dt}}=\text { slope } \geq 0 \text { always increasing } \\\\
& (\mathrm{A} \text { - II) } \\\\
& \frac{\mathrm{dx}}{\mathrm{dt}}<0 \text {; and at } \mathrm{t} \rightarrow \infty \frac{\mathrm{dx}}{\mathrm{dt}} \rightarrow 0 \\\\
& \text { (B - IV) } \\\\
& \frac{\mathrm... | mcq | jee-main-2023-online-30th-january-morning-shift | 12,127 |
1ldwqjkl1 | physics | motion-in-a-straight-line | graph-based-problem | <p>The velocity time graph of a body moving in a straight line is shown in the figure.</p>
<p><img src="data:image/png;base64,UklGRh4LAABXRUJQVlA4IBILAABQogCdASoAA6YBP4HA2WU2L7+nIXJJs/AwCWlu4W2ikmNwvV6EtDbQztrxPEBbw9VJf/fqKL4M+DJj8aL7mnZZASdkXE5PvnY/fOx++dj987H752P3zp769xhp1AnzgvDHkz7IWhpUhxKzgUK6nr7vvHdw2mj987H752P3zs... | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "1 : 3"}, {"identifier": "C", "content": "1 : 4"}, {"identifier": "D", "content": "1 : 2"}] | ["B"] | null | From $v$-t graph
<br/><br/>
Displacement $=$ Area under curve considering sign also.
<br/><br/>
Distance $=$ Area under curve considering only magnitude
<br/><br/>
Displacement $= \Sigma$area $=8$$ \times $$2-4$$ \times $$2+4$$ \times $$4-2$$ \times $$4=16 \mathrm{~m}$<br/><br/>
Distance $=\Sigma \mid$ area $\mid=48 \m... | mcq | jee-main-2023-online-24th-january-evening-shift | 12,128 |
1lguy491s | physics | motion-in-a-straight-line | graph-based-problem | <p>From the $$\mathrm{v}-t$$ graph shown, the ratio of distance to displacement in $$25 \mathrm{~s}$$ of motion is:</p>
<p><img src="data:image/png;base64,UklGRroOAABXRUJQVlA4IK4OAAAw8gCdASoAA3kCP4HA22U2MK2nIfM5csAwCWlu/AkY3FsnZ19/zn/O/771q7uXuizjyF95Gqlf//1SmDd6xu9YgOiplrXCdkGKSDFJBikgxSQYpIMUkAE5pRqNZqm4gEt4HlAonNKNR... | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\\frac{3}{5}$$"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$\\frac{5}{3}$$"}] | ["D"] | null | Area under the graph from $t=0$ to $t=20 \mathrm{sec}=200 \mathrm{~m}$
<br/><br/>Area under the graph from $\mathrm{t}=20$ to $t=25 \mathrm{sec}=50 \mathrm{~m}$
<br/><br/>So distance covered $=(200+50) \mathrm{m}=250 \mathrm{~m}$
<br/><br/>Displacement $=(200-50) \mathrm{m}=150 \mathrm{~m}$
<br/><br/>$$ \therefore $$ $... | mcq | jee-main-2023-online-11th-april-morning-shift | 12,129 |
1lgyezarn | physics | motion-in-a-straight-line | graph-based-problem | <p>The position-time graphs for two students A and B returning from the school to their homes are shown in figure.</p>
<p><img src="data:image/png;base64,UklGRhAOAABXRUJQVlA4IAQOAAAwAgGdASoAA/kCP4HA3GW2Ma2nIVSo6sAwCWlu4WrjvmNwvV6t9Qd8r4bZFO2PeFedLDw+ZTu+fCX/7oRptpkZcEaiBmMMYYwxhjDGGGEXvy0WSuhkYZcN3RJiS7waWMn+TdBecvOXnL... | [{"identifier": "A", "content": "(A), (C) and (D) only"}, {"identifier": "B", "content": "(A) and (E) only"}, {"identifier": "C", "content": "(A), (C) and (E) only"}, {"identifier": "D", "content": "(B) and (E) only"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1loi4pv2l/578973da-c4ce-49f8-81c6-95faf4862d2f/5e7d13d0-7a03-11ee-bbdd-cfc3c972af88/file-6y3zli1loi4pv2m.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1loi4pv2l/578973da-c4ce-49f8-81c6-95faf4862d2f/5e7d13d0-7a03-11ee-bb... | mcq | jee-main-2023-online-10th-april-morning-shift | 12,130 |
iftt7Ukg32iOqpz9 | physics | motion-in-a-straight-line | motion-under-gravity | From a building two balls A and B are thrown such that A is thrown upwards and B downwards ( both vertically with the same speed ). If v<sub>A</sub> and v<sub>B</sub> are their respective velocities on reaching the ground, then | [{"identifier": "A", "content": "$${v_B} > {v_A}$$"}, {"identifier": "B", "content": "$${v_A} = {v_B}$$"}, {"identifier": "C", "content": "$${v_A} > {v_B}$$"}, {"identifier": "D", "content": "their velocities depend on their masses."}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264694/exam_images/j9guueau1dq0e9ehxdpw.webp" loading="lazy" alt="AIEEE 2002 Physics - Motion in a Straight Line Question 108 English Explanation">
<br>Assume the initial velocity of each particle is = u
<br><br>And height of buildi... | mcq | aieee-2002 | 12,131 |
5G6RV43sxKpCukvA | physics | motion-in-a-straight-line | motion-under-gravity | A ball is released from the top of a tower of height h meters. It takes T seconds to reach the
ground. What is the position of the ball in $${T \over 3}$$ seconds? | [{"identifier": "A", "content": "$${{8h} \\over 9}$$ meters from the ground "}, {"identifier": "B", "content": "$${{7h} \\over 9}$$ meters from the ground"}, {"identifier": "C", "content": "$${h \\over 9}$$ meters from the ground"}, {"identifier": "D", "content": "$${{7h} \\over {18}}$$ meters from the ground"}] | ["A"] | null | We know that equation of motion, $$s = ut + {1 \over 2}g{t^2},\,\,$$
<br><br>Initial speed of ball is zero and it take T second to reach the ground.
<br><br>$$\therefore$$ $$h = {1 \over 2}g{T^2}$$
<br><br>After $$T/3$$ second, vertical distance moved by the ball
<br><br>$$h' = {1 \over 2}g{\left( {{T \over 3}} \righ... | mcq | aieee-2004 | 12,132 |
s5yFHVnS9AwaD1zV | physics | motion-in-a-straight-line | motion-under-gravity | A parachutist after bailing out falls $$50$$ $$m$$ without friction. When parachute opens, it decelerates at $$2\,\,m/{s^2}.$$ He reaches the ground with a speed of $$3$$ $$m/s$$. At what height, did he bail out? | [{"identifier": "A", "content": "$$182$$ $$m$$ "}, {"identifier": "B", "content": "$$91$$ $$m$$ "}, {"identifier": "C", "content": "$$111$$ $$m$$ "}, {"identifier": "D", "content": "$$293$$ $$m$$ "}] | ["D"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/CKCw0BjOojeoZGf9E/SNtPbAVf9jNN3BpW4kIBi3UchrOMz/GcjWmEtlgVbShsKG8jmyQc/image.svg" loading="lazy" alt="AIEEE 2005 Physics - Motion in a Straight Line Question 96 English Explanation">
<br>The velocity of parachutist when parachute opens at 50 m is
<br><b... | mcq | aieee-2005 | 12,133 |
Zf24hR1sRpoTHEVm | physics | motion-in-a-straight-line | motion-under-gravity | From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the
particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation
between H, u and n is: | [{"identifier": "A", "content": "2gH = n<sup>2</sup>u<sup>2</sup>"}, {"identifier": "B", "content": "gH = (n - 2)<sup>2</sup>u<sup>2</sup>"}, {"identifier": "C", "content": "2gH = nu<sup>2</sup>(n - 2)"}, {"identifier": "D", "content": "gH = (n - 2)u<sup>2</sup>"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265375/exam_images/mmse7optyoxdv204h5st.webp" loading="lazy" alt="JEE Main 2014 (Offline) Physics - Motion in a Straight Line Question 98 English Explanation">
<br><br>Time taken to reach highest point is $$t = {u \over g}$$
<br><b... | mcq | jee-main-2014-offline | 12,134 |
9Q4t9C2UiF8BVKwVtBjgy2xukff4b06s | physics | motion-in-a-straight-line | motion-under-gravity | A helicopter rises from rest on the ground
vertically upwards with a constant acceleration
g. A food packet is dropped from the helicopter
when it is at a height h. The time taken by the
packet to reach the ground is close to :
<br/>[g is the
acceleration due to gravity] | [{"identifier": "A", "content": "t = 3.4$$\\sqrt {\\left( {{h \\over g}} \\right)} $$"}, {"identifier": "B", "content": "t = 1.8$$\\sqrt {\\left( {{h \\over g}} \\right)} $$"}, {"identifier": "C", "content": "t = $$\\sqrt {{{2h} \\over {3g}}} $$"}, {"identifier": "D", "content": "t = $${2 \\over 3}\\sqrt {\\left( {{h \... | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263843/exam_images/gpar6yukgqk4kfm3joui.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264255/exam_images/ihhwugnqdd3flxtflj9d.webp"><source media="(max-wid... | mcq | jee-main-2020-online-5th-september-morning-slot | 12,135 |
P4WZ7GFDfyZUjOf5HX1klt3asx3 | physics | motion-in-a-straight-line | motion-under-gravity | A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is : | [{"identifier": "A", "content": "50 m"}, {"identifier": "B", "content": "25 m"}, {"identifier": "C", "content": "45 m"}, {"identifier": "D", "content": "35 m"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263522/exam_images/ivzxtmw8szmnmx1lphq8.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263733/exam_images/j4fydkfk35hyfgjcliz9.webp"><source media="(max-wid... | mcq | jee-main-2021-online-25th-february-evening-slot | 12,136 |
gqd82wyBUQe34jbXuu1kmkb1992 | physics | motion-in-a-straight-line | motion-under-gravity | A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to $${{81} \over {100}}$$ of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms<sup>$$-$$2</sup>) | [{"identifier": "A", "content": "2.50 ms<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "3.0 ms<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "2.0 ms<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "3.50 ms<sup>$$-$$1</sup>"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265265/exam_images/ge7t57h4spae8iy585vg.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Physics - Motion in a Straight Line Question 66 English Explanation">
<br... | mcq | jee-main-2021-online-17th-march-evening-shift | 12,137 |
1krywvo88 | physics | motion-in-a-straight-line | motion-under-gravity | A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching $${h \over 3}$$ in both the directions. | [{"identifier": "A", "content": "$${{\\sqrt 2 - 1} \\over {\\sqrt 2 + 1}}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 - \\sqrt 2 } \\over {\\sqrt 3 + \\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 - 1} \\over {\\sqrt 3 + 1}}$$"}] | ["C"] | null | $$u = \sqrt {2gh} $$<br><br>Now, <br><br>$$S = {h \over 3}$$<br><br>a = $$-$$g<br><br>$$S = ut + {1 \over 2}a{t^2}$$<br><br>$${h \over 3} = \sqrt {2gh} t + {1 \over 2}( - g){t^2}$$<br><br>$${t^2}\left( {{g \over 2}} \right) - \sqrt {2gh} t + {h \over 3} = 0$$<br><br>From quadratic equation<br><br>$${t_1},{t_2} = {{\sqr... | mcq | jee-main-2021-online-27th-july-morning-shift | 12,140 |
1ktfjn3tk | physics | motion-in-a-straight-line | motion-under-gravity | Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor. | [{"identifier": "A", "content": "4.18 m"}, {"identifier": "B", "content": "2.94 m"}, {"identifier": "C", "content": "2.45 m"}, {"identifier": "D", "content": "7.35 m"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266536/exam_images/n6z06ntq0mibulyfw1u4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Physics - Motion in a Straight Line Question 54 English Explanation"><br... | mcq | jee-main-2021-online-27th-august-evening-shift | 12,142 |
1l546s71c | physics | motion-in-a-straight-line | motion-under-gravity | <p>Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms<sup>$$-$$1</sup>. [use g = 10 ms<sup>$$-$$2</sup... | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "30"}] | ["D"] | null | As the meeting point lies $100 \mathrm{~m}$ above ground, displacement of ball will be $80 \mathrm{~m}$.
<br/><br/>For ball $A$
<br/><br/>$$
\begin{aligned}
&u=0, \mathrm{~S}=80 \mathrm{~m}, a =+\mathrm{g}=+10 \mathrm{~m} / \mathrm{s}^2 \text {, time }=t_1 \\\\
&\Rightarrow S =u t+\frac{1}{2} a t^2 \\\\
&\Rightarrow 80... | mcq | jee-main-2022-online-29th-june-morning-shift | 12,143 |
1l58cfair | physics | motion-in-a-straight-line | motion-under-gravity | <p>A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ________ m. [Use g = 10 m/s<sup>2</sup>]</p> | [] | null | 5 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5jhsn20/d0c7769a-768d-4895-90ef-061ec8187df5/2eda5380-029b-11ed-a9b8-43edceee002f/file-1l5jhsn21.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5jhsn20/d0c7769a-768d-4895-90ef-061ec8187df5/2eda5380-029b-11ed-a9b8-43edceee002f... | integer | jee-main-2022-online-26th-june-morning-shift | 12,144 |
1l58id0re | physics | motion-in-a-straight-line | motion-under-gravity | <p>A ball is projected vertically upward with an initial velocity of 50 ms<sup>$$-$$1</sup> at t = 0s. At t = 2s, another ball is projected vertically upward with same velocity. At t = __________ s, second ball will meet the first ball (g = 10 ms<sup>$$-$$2</sup>).</p> | [] | null | 6 | <p>At t = 2 s, v<sub>1</sub> = 50 $$-$$ 2 $$\times$$ 10 = 30 m/s</p>
<p>v<sub>2</sub> = v<sub>2</sub></p>
<p>$$\therefore$$ a<sub>rel</sub> = g $$-$$ g = 0</p>
<p>$$S = {{{u^2} - {v^2}} \over {2g}} = {{{{50}^2} - {{30}^2}} \over {2 \times 10}} = {{1600} \over {20}} = 80$$ m</p>
<p>$$\therefore$$ v<sub>rel</sub> = 50 $$... | integer | jee-main-2022-online-26th-june-evening-shift | 12,145 |
1l5c4n19y | physics | motion-in-a-straight-line | motion-under-gravity | <p>From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____________ s.</p> | [] | null | 3 | <p>Based on the situation</p>
<p>$$h = - u{t_1} + {1 \over 2}gt_1^2$$ $$\to$$ throwing up ....... (i)</p>
<p>$$h = u{t_2} + {1 \over 2}gt_2^2$$ $$\to$$ throwing up ....... (ii)</p>
<p>$$h = {1 \over 2}g{t^2}$$ $$\to$$ dropping .......... (iii)</p>
<p>and $$0 = u({t_1} - {t_2}) - {1 \over 2}g{({t_1} - {t_2})^2}$$ ........ | integer | jee-main-2022-online-24th-june-morning-shift | 12,146 |
1l6nt5naz | physics | motion-in-a-straight-line | motion-under-gravity | <p>A ball is thrown vertically upwards with a velocity of $$19.6 \mathrm{~ms}^{-1}$$ from the top of a tower. The ball strikes the ground after $$6 \mathrm{~s}$$. The height from the ground up to which the ball can rise will be $$\left(\frac{k}{5}\right) \mathrm{m}$$. The value of $$\mathrm{k}$$ is __________. (use $$\... | [] | null | 392 | <p>v = 19.6 m/s</p>
<p>t = 6s</p>
<p>Time taken in upward motion above tower = 2s</p>
<p>$$\Rightarrow$$ Time taken from top most point to ground = 4s</p>
<p>$$ \Rightarrow \sqrt {{{2h} \over g}} = 4$$</p>
<p>$$h = {{16 \times 9.8} \over 2} = 8 \times 9.8$$</p>
<p>$$ \Rightarrow k = 8 \times 9.8 \times 5 = 392$$</p> | integer | jee-main-2022-online-28th-july-evening-shift | 12,148 |
1l6p3zz09 | physics | motion-in-a-straight-line | motion-under-gravity | <p>A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $$\frac{h}{3}$$ while going up and coming down respectively.</p> | [{"identifier": "A", "content": "$$\\frac{\\sqrt{2}-1}{\\sqrt{2}+1}$$"}, {"identifier": "B", "content": "$$\\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}$$"}, {"identifier": "C", "content": "$$\\frac{\\sqrt{3}-1}{\\sqrt{3}+1}$$"}, {"identifier": "D", "content": "$$\\frac{1}{3}$$"}] | ["B"] | null | <p>A ball is thrown vertically upward with a certain velocity, reaching a maximum height $ h $. We need to find the ratio of the times it is at height $ \frac{h}{3} $ while ascending and descending, respectively.</p>
<p>The initial velocity of the ball $ v $ can be given by:</p>
<p>$ v = \sqrt{2gh} $</p>
<p>When the... | mcq | jee-main-2022-online-29th-july-morning-shift | 12,149 |
1l6rgv1ad | physics | motion-in-a-straight-line | motion-under-gravity | <p>A juggler throws balls vertically upwards with same initial velocity in air. When the first ball reaches its highest position, he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is</p> | [{"identifier": "A", "content": "g/2n"}, {"identifier": "B", "content": "g/n"}, {"identifier": "C", "content": "2gn"}, {"identifier": "D", "content": "g/2n<sup>2</sup>"}] | ["D"] | null | <p>The juggler throws n balls per second.</p>
<p>$$\therefore$$ Interval between balls = $${1 \over n}$$ seconds</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9a8md/9bc69ea1-21e6-4bfc-9ab1-131394dad5ec/699d9550-2752-11ed-a077-1f1e3989e798/file-1l7e9a8me.png?format=png" data-orsrc="https://a... | mcq | jee-main-2022-online-29th-july-evening-shift | 12,150 |
1l6rhc7vi | physics | motion-in-a-straight-line | motion-under-gravity | <p>A ball is released from a height h. If $$t_{1}$$ and $$t_{2}$$ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between $$t_{1}$$ and $$t_{2}$$.</p> | [{"identifier": "A", "content": "$$t_{1}=(\\sqrt{2}) t_{2}$$"}, {"identifier": "B", "content": "$$t_{1}=(\\sqrt{2}-1) t_{2}$$"}, {"identifier": "C", "content": "$$t_{2}=(\\sqrt{2}+1) t_{1}$$"}, {"identifier": "D", "content": "$$t_{2}=(\\sqrt{2}-1) t_{1}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9ofih/3261b804-62c4-4278-80a6-960f468fdcec/f4447790-2753-11ed-a077-1f1e3989e798/file-1l7e9ofii.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7e9ofih/3261b804-62c4-4278-80a6-960f468fdcec/f4447790-2753-11ed-a077-1f1e3989e798... | mcq | jee-main-2022-online-29th-july-evening-shift | 12,151 |
1lgrikj40 | physics | motion-in-a-straight-line | motion-under-gravity | <p>A ball is thrown vertically upward with an initial velocity of $$150 \mathrm{~m} / \mathrm{s}$$. The ratio of velocity after $$3 \mathrm{~s}$$ and $$5 \mathrm{~s}$$ is $$\frac{x+1}{x}$$. The value of $$x$$ is ___________.</p>
<p>$$\left\{\right.$$ take, $$\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right\}$$</p> | [{"identifier": "A", "content": "$$-5$$"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["C"] | null | To solve this problem, we can use the following equation of motion for the vertical velocity at any given time $$t$$:
<br/><br/>
$$v = u - gt$$
<br/><br/>
Where:<br/><br/>
- $$v$$ is the final velocity at time $$t$$<br/><br/>
- $$u$$ is the initial velocity (150 m/s)<br/><br/>
- $$g$$ is the acceleration due to gravity... | mcq | jee-main-2023-online-12th-april-morning-shift | 12,152 |
lvb2956u | physics | motion-in-a-straight-line | motion-under-gravity | <p>A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $$t_1$$. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $$t_2$$. Time required to reach the ground, if it is dropped from the top of the tower, is :</p> | [{"identifier": "A", "content": "$$\\sqrt{\\mathrm{t}_1+\\mathrm{t}_2}$$\n"}, {"identifier": "B", "content": "$$\\sqrt{\\mathrm{t}_1 \\mathrm{t}_2}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{\\mathrm{t}_1}{\\mathrm{t}_2}}$$\n"}, {"identifier": "D", "content": "$$\\sqrt{\\mathrm{t}_1-\\mathrm{t}_2}$$"}] | ["B"] | null | <p>To solve this problem, we'll use the equations of motion under constant acceleration due to gravity. Let's define:</p>
<p><p>$ h $: Height of the tower</p></p>
<p><p>$ u $: Initial speed of projection (same magnitude in both cases)</p></p>
<p><p>$ g $: Acceleration due to gravity (positive downward)</p></p>
<p><p>$... | mcq | jee-main-2024-online-6th-april-evening-shift | 12,154 |
m31XpxSsMEgpqHZR | physics | motion-in-a-straight-line | relative-motion-in-one-dimension | Two stones are thrown up simultaneously from the edge of a cliff $$240$$ $$m$$ high with initial speed of $$10$$ $$m/s$$ and $$40$$ $$m/s$$ respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ?
<br/><br/>(Assume stones do not r... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91dw8zy/227e1c58-4098-46f6-a69e-17118c030811/b08a4ae0-47d6-11ed-9a0e-8f7a98995d79/file-1l91dw8zz.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l91dw8zy/227e1c58-4098-46f6-a69e-17118c030811/b08... | ["A"] | null | Using $$h = ut + {1 \over 2}g{t^2}$$
<br><br>$${y_1} = 10t - 5{t^2};\,\,{y_2} = 40t - 5{t^2}$$
<br><br>$$\therefore$$ $$\,\,\,\,{y_2} - {y_1} = 30t\,\,\,\,$$ for $$\,\,\,\,t \le 8s.$$
<br><br>Curve will be straight line when $$t \le 8s.$$
<br><br>when stone 1 reaches the ground then $$\,\,\,{y_1} = - 240m,\,\, and\,\,... | mcq | jee-main-2015-offline | 12,155 |
zrxJySXVzbi7z7PxEaecR | physics | motion-in-a-straight-line | relative-motion-in-one-dimension | A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s<sup>2</sup> and the car has acceleration 4 m/s<sup>2</sup> . The car will catch up with the bus after a time of :
| [{"identifier": "A", "content": "$$\\sqrt {110} \\,s$$ "}, {"identifier": "B", "content": "$$\\sqrt {120} \\,s$$"}, {"identifier": "C", "content": "$$10\\,\\,\\sqrt 2 \\,s$$ "}, {"identifier": "D", "content": "15 s"}] | ["C"] | null | Acceleration of Car, a<sub>C</sub> = 4 m/s<sup>2</sup>
<br><br>Acceleration of bus, a<sub>B</sub> = 2 m/s<sup>2</sup>
<br><br>Initial distance between them, S = 200 m
<br><br>Acceleration of Car with respect to bus,
<br><br>a<sub>CB</sub> = a<sub>C</sub> $$-$$ a<sub>B</sub> = 4 $$-$$ 2 = 2 m/s<sup>2</sup>
<br><br>As ... | mcq | jee-main-2017-online-9th-april-morning-slot | 12,156 |
UtGRbCl8eCFVeewELsQAz | physics | motion-in-a-straight-line | relative-motion-in-one-dimension | A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when : (i) they are moving in the same direction , and (ii) in the opposite direction is : | [{"identifier": "A", "content": "$${{25} \\over {11}}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$${{11} \\over 5}$$"}, {"identifier": "D", "content": "$${5 \\over 2}$$"}] | ["C"] | null | The total distance to be travelled by the train is
60 + 120 = 180 m.
<br><br>When the trains are moving in the same direction, relative
velocity is<br> v<sub>1</sub>
– v<sub>2</sub>
= 80 – 30 = 50 km hr<sup>–1</sup>
<br><br>So time taken to cross each other,
<br><br>t<sub>1</sub> = $${{180} \over {50 \times {{{{10}^3... | mcq | jee-main-2019-online-12th-january-morning-slot | 12,157 |
FCDZHjWtZcXdqQfoSijgy2xukev1krmx | physics | motion-in-a-straight-line | relative-motion-in-one-dimension | Train A and train B are running on parallel
tracks in the opposite directions with speeds of
36 km/hour and 72 km/hour, respectively. A
person is walking in train A in the direction
opposite to its motion with a speed of 1.8 km/
hour. Speed (in ms<sup>–1</sup>) of this person as
observed from train B will be close to :... | [{"identifier": "A", "content": "30.5 ms<sup>\u20131</sup>"}, {"identifier": "B", "content": "29.5 ms<sup>\u20131</sup>"}, {"identifier": "C", "content": "31.5 ms<sup>\u20131</sup>"}, {"identifier": "D", "content": "28.5 ms<sup>\u20131</sup>"}] | ["B"] | null | Velocity of man with respect to ground
<br><br>$${\overrightarrow V _{m/g}}$$ = $${\overrightarrow V _{m/A}}$$ + $${\overrightarrow V _{A}}$$
<br><br>= -1.8 + 36
<br><br>Velocity of man w.r.t. B
<br><br>$${\overrightarrow V _{m/B}}$$ = $${\overrightarrow V _{m}}$$ - $${\overrightarrow V _{B}}$$
<br><br>= –1.8 + 36 – (–... | mcq | jee-main-2020-online-2nd-september-morning-slot | 12,158 |
1krqarmk6 | physics | motion-in-a-straight-line | relative-motion-in-one-dimension | A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t<sub>1</sub>. If he remains stationary on a moving escalator then the escalator takes him up in time t<sub>2</sub>. The time taken by him to walk up on the moving escalator will be : | [{"identifier": "A", "content": "$${{{t_1}{t_2}} \\over {{t_2} - {t_1}}}$$"}, {"identifier": "B", "content": "$${{{t_1} + {t_2}} \\over 2}$$"}, {"identifier": "C", "content": "$${{{t_1}{t_2}} \\over {{t_2} + {t_1}}}$$"}, {"identifier": "D", "content": "$${t_2} - {t_1}$$"}] | ["C"] | null | L = Length of escalator<br><br>$${V_{b/esc}} = {L \over {{t_1}}}$$<br><br>When only escalator is moving.<br><br>$${V_{esc}} = {L \over {{t_2}}}$$<br><br>when both are moving<br><br>$${V_{b/g}} = {V_{b/esc}} + {V_{esc}}$$<br><br>$${V_{b/g}} = {L \over {{t_1}}} + {L \over {{t_2}}} \Rightarrow \left[ {t = {L \over {{V_{b/... | mcq | jee-main-2021-online-20th-july-evening-shift | 12,159 |
1l59lv0ce | physics | motion-in-a-straight-line | relative-motion-in-one-dimension | <p>Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by $${X_P}(t) = \alpha t + \beta {t^2}$$ and $${X_Q}(t) = ft - {t^2}$$. At what time, both the buses have same velocity?</p> | [{"identifier": "A", "content": "$${{\\alpha - f} \\over {1 + \\beta }}$$"}, {"identifier": "B", "content": "$${{\\alpha + f} \\over {2(\\beta - 1)}}$$"}, {"identifier": "C", "content": "$${{\\alpha + f} \\over {2(1 + \\beta )}}$$"}, {"identifier": "D", "content": "$${{f - \\alpha } \\over {2(1 + \\beta )}}$$"}] | ["D"] | null | <p>$${X_P} = \alpha t + \beta {t^2}$$</p>
<p>$${X_Q} = ft - {t^2}$$</p>
<p>$$\therefore$$ $${V_P} = \alpha + 2\beta t$$</p>
<p>$${V_Q} = f - 2t$$</p>
<p>$$\because$$ $${V_P} = {V_Q}$$</p>
<p>$$ \Rightarrow \alpha + 2\beta t = f - 2t$$</p>
<p>$$ \Rightarrow t = {{f - \alpha } \over {2(1 + \beta )}}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift | 12,160 |
lsamvrk0 | physics | motion-in-a-straight-line | relative-motion-in-one-dimension | <p>Train A is moving along two parallel rail tracks towards north with speed $$72 \mathrm{~km} / \mathrm{h}$$ and train B is moving towards south with speed $$108 \mathrm{~km} / \mathrm{h}$$. Velocity of train B with respect to A and velocity of ground with respect to B are (in $$\mathrm{ms}^{-1}$$):</p> | [{"identifier": "A", "content": "-50 and -30"}, {"identifier": "B", "content": "-50 and 30"}, {"identifier": "C", "content": "-30 and 50"}, {"identifier": "D", "content": "50 and -30"}] | ["B"] | null | <p>To find the velocity of Train B with respect to Train A, we have to subtract the velocity of Train A from the velocity of Train B, keeping in mind that they are moving in opposite directions. Since they are moving in opposite directions, the relative velocity is calculated by adding their magnitudes when converting ... | mcq | jee-main-2024-online-1st-february-evening-shift | 12,163 |
G62rUQ4WvQn4wwUv | physics | motion-in-a-straight-line | variable-acceleration-motion | The relation between time t and distance x is t = ax<sup>2</sup> + bx where a and b are constants.
The acceleration is | [{"identifier": "A", "content": "2bv<sup>3</sup>"}, {"identifier": "B", "content": "-2abv<sup>2</sup>"}, {"identifier": "C", "content": "2av<sup>2</sup>"}, {"identifier": "D", "content": "-2av<sup>3</sup>"}] | ["D"] | null | <p>The relationship between time $ t $ and distance $ x $ is defined as $ t = ax^2 + bx $, where $ a $ and $ b $ are constants.</p>
<p>To find the acceleration, follow these steps:</p>
<p>First, differentiate the equation with respect to time $ t $:</p>
<p>$ \frac{d}{dt}(t) = a \frac{d}{dt}(x^2) + b \frac{dx}{dt} $<... | mcq | aieee-2005 | 12,164 |
uY6K2wAVhD70ftdr | physics | motion-in-a-straight-line | variable-acceleration-motion | A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a velocity
'v' that varies as $$v = \alpha \sqrt x $$. The displacement of the particle varies with time as | [{"identifier": "A", "content": "t<sup>2</sup>"}, {"identifier": "B", "content": "t"}, {"identifier": "C", "content": "t<sup>1/2</sup>"}, {"identifier": "D", "content": "t<sup>3</sup>"}] | ["A"] | null | <p>Given the velocity function: $$v = \alpha \sqrt{x}$$</p>
<p>We know that velocity $v$ is the rate of change of displacement with respect to time:</p>
<p>$$\therefore \frac{dx}{dt} = \alpha \sqrt{x}$$</p>
<p>Rearranging and separating variables, we get:</p>
<p>$$\Rightarrow \frac{dx}{\sqrt{x}} = \alpha \, dt$$</p... | mcq | aieee-2006 | 12,165 |
VmhYRHaJiN9y2G0KU3fD6 | physics | motion-in-a-straight-line | variable-acceleration-motion | The position of a particle as a function of time
t, is given by<br/>x(t) = at + bt<sup>2</sup> – ct<sup>3</sup><br/>
where a, b and c are constants. When the
particle attains zero acceleration, then its
velocity will be : | [{"identifier": "A", "content": "$$a + {{{b^2}} \\over {c}}$$"}, {"identifier": "B", "content": "$$a + {{{b^2}} \\over {4c}}$$"}, {"identifier": "C", "content": "$$a + {{{b^2}} \\over {3c}}$$"}, {"identifier": "D", "content": "$$a + {{{b^2}} \\over {2c}}$$"}] | ["C"] | null | x = at + bt<sup>2</sup> – ct<sup>3</sup><br><br>
$$V = {{dx} \over {dt}} = a + 2bt - 3c{t^2}$$<br><br>
$$a = {{dv} \over {dt}} = 2b - 6ct$$<br><br>
Put acceleration = 0<br><br>
$$ \Rightarrow t = {b \over {3c}}$$<br><br>
Now V at t = $${b \over {3c}}$$<br><br>
$$V = a + {{{b^2}} \over {3c}}$$
| mcq | jee-main-2019-online-9th-april-evening-slot | 12,168 |
xilgUXOlQLZMwr8DE13rsa0w2w9jx6le17d | physics | motion-in-a-straight-line | variable-acceleration-motion | A particle is moving with speed v = b$$\sqrt x $$ along positive x-axis. Calculate the speed of the particle at
time t = $$\tau $$(assume that the particle is at origin t = 0)
| [{"identifier": "A", "content": "$${{{b^2}\\tau } \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${{b^2}\\tau }$$"}, {"identifier": "C", "content": "$${{{b^2}\\tau } \\over 2}$$"}, {"identifier": "D", "content": "$${{{b^2}\\tau } \\over 4}$$"}] | ["C"] | null | v = b$$\sqrt x $$
<br>$$ \Rightarrow $$ $${{dx} \over {dt}}$$ = b$$\sqrt x $$
<br>$$ \Rightarrow $$$$\int\limits_0^x {{{dx} \over {\sqrt x }}} = \int\limits_0^t {bdt} $$
<br>$$ \Rightarrow $$$$\left[ {{{{x^{ - {1 \over 2} + 1}}} \over { - {1 \over 2} + 1}}} \right]_0^x$$ = $$b\left[ t \right]_0^t$$
<br>$$ \Rightarrow... | mcq | jee-main-2019-online-12th-april-evening-slot | 12,169 |
mmq3v8rGIK3qZlQNSb7k9k2k5inzg9b | physics | motion-in-a-straight-line | variable-acceleration-motion | The distance x covered by a particle in one
dimensional motion varies with time t as
<br/>x<sup>2</sup> = at<sup>2</sup> + 2bt + c. If the acceleration of the
particle depends on x as x<sup>–n</sup>, where n is an
integer, the value of n is __________ | [] | null | 3 | x<sup>2</sup> = at<sup>2</sup> + 2bt + c .....(1)
<br><br>$$ \Rightarrow $$ 2xv = 2at + 2b .....(2)
<br><br>$$ \Rightarrow $$ v = $${{at + b} \over x}$$
<br><br>differentiating (2) w.r.t. time
<br><br>$$ \Rightarrow $$ xa' + v<sup>2</sup> = a
<br><br>Here a' is acceleration.
<br><br>$$ \Rightarrow $$ a'x = a - v<sup>2<... | integer | jee-main-2020-online-9th-january-morning-slot | 12,170 |
xuIndchfXveJFsmLrS1kmkarpkg | physics | motion-in-a-straight-line | variable-acceleration-motion | The velocity of a particle is v = v<sub>0</sub> + gt + Ft<sup>2</sup>. Its position is x = 0 at t = 0; then its displacement after time (t = 1) is : | [{"identifier": "A", "content": "v<sub>0</sub> + g + f"}, {"identifier": "B", "content": "v<sub>0</sub> + $${g \\over 2}$$ + $${F \\over 3}$$"}, {"identifier": "C", "content": "v<sub>0</sub> + 2g + 3F"}, {"identifier": "D", "content": "v<sub>0</sub> + $${g \\over 2}$$ + F"}] | ["B"] | null | <p>The velocity of a particle is given by $ v = v_0 + gt + Ft^2 $. Its position is $ x = 0 $ at $ t = 0 $. To find its displacement after time $ t = 1 $, follow these steps:</p>
<p>Given:</p>
<p>$ v = v_0 + gt + Ft^2 $</p>
<p>We know that:</p>
<p>$ \frac{dx}{dt} = v_0 + gt + Ft^2 $</p>
<p>To find the displacement,... | mcq | jee-main-2021-online-17th-march-evening-shift | 12,171 |
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