| | "text": "Physics Motion Graphs - StickMan Physics\nSkip to content\nStickMan Physics\nAnimated Physics Lessons\nMenu\nHome\nStickman Physics Music: Blending Science with Sound\nUnit 1: One Dimensional Motion: Physics Introduction\nUnit 2: Two Dimensional Motion: Projectile and Non-Projectile\nUnit 3: Newton’s Laws of Motion and Force\nUnit 4: Universal Gravitation and Circular Motion\nUnit 5: Work, Power, Mechanical Energy, and Simple Machines\nUnit 6: Momentum Impulse and Conservation of Momentum\nUnit 7: Electrostatics\nUnit 8: Current and Circuits\nUnit 9: Magnets and Magnetism\nUnit 10: Waves\nUnit 11: Electromagnetic Waves\nUnit 12: Nuclear Physics\nTable of Contents\nPractice\nEquation Sheet\nDigital Learning\nContact\nMenu\nPhysics Motion Graphs\nPhysics Motion Graphs\nPhysics motion graphs include position time graphs (also called displacement time graphs) and velocity time graphs. It is important to look at the Y axis and determine which one you have before analyzing the data. Position time graphs show if a person is moving forward or backwards as the line goes up or down. Velocity time graphs show if the person is going faster or slower when the line moves up or down.\nWatch our position time graph lesson here or on our YouTube channel\nPosition Time Graphs\nPosition time graphs show how displacement (another term for position) changes in the Y axis relative to time in the X axis. Since slope equals rise over run and since rise is change in displacement and run is the change in time the slope of a PT graph equals the velocity. See more about calculating velocity in a prior lesson here.\nBe aware that position and displacement are used here interchangeably as X variables. Position means where you are located and displacement means how far you are from a prior position with a direction from start to finish.\nTotal displacement equals final position minus initial position (x = xf – xi)\nx : displacement\nxi : initial position\nxf : final position\nWhat the Slope of a PT Graph Represents\nA flat line in a PT graph shows no slope, meaning no motion, and therefore at rest (v = 0).\nA positive slope in a PT graph is constant positive velocity (+v).\nA negative slope in a PT graph is constant backward motion (-v).\nA curved line represents an acceleration or deceleration (See the examples below)\nPT Graph at Rest\nA flat line (no slope) on a position time graph. Think of a number line on the ground as seen in the video.\nThe person starts at the 12m position and stays there the whole time.\nPT Graph Constant Forward Velocity\nA straight line means constant \"something\". In the previous example the stickman was constantly at 12 meters position, so at rest. In this example the line is straight with a positive slope.\nHere this stickman starts at the 0 m position and travels to 24 meters (x = xf – xi = 24 - 0) at a constant rate in 6 seconds. Since v = x/t, the velocity here is +24m/6s or 6 m/s forward which is also equal to the slope of the line.\nObserve videos, tables, and graphs of various constant motion cars on this page\nPT Graph Constant Backward Velocity\nIn this example the line is straight so we have constant velocity but the slope is negative which represents backwards motion. The stickman in the example starts at the 16 meter position and travels to the 2 meter position.\nx = xf – xi = 2 - 16 = -14\nSo -14 meters in 6 seconds.\nv=d/t\nv= -14m/6s = -2 1/3 m/s\nThis stickman travels at a velocity of 2 1/3 m/s backwards since it was negative (also the slope of the line).\nPT Graph Acceleration Moving Forward\nThis stickman in this example starts at rest vi = 0 m/s and accelerates from the 0 position to the 24 meter position with a line going up (forward motion) and a slope constantly curving more upward (accelerating here).\nPT Graph Deceleration Moving Forward\nThe stickman in this example is always going forward as seen by the line always going up until the end. The slope here is getting less and less as the stickman slows down until the end where it is flat and he is stopped.\nSolving Problems with Position Time Graphs\nThe two things you can read off a position time graph are positions (or the displacement) and time. Anything else you will have to solve for. You can find velocity using a position time graph by taking the initial and final positions and the time it took to do this. Use the equations below to do this.\nv = Δx/t\nv = (xf – xi)/t\nTo determine the velocity any where before 5 seconds in the PT graph to the right you can use the initial and final position for that interval.\nxf = 8 m/s , xi = 0 m/s , t = 5s\nv = 8/5 or + 1.6\nThe velocity anywhere in the 0-5 second interval is 1.6 m/s forward\nDistance Verses Displacement on a Position Time Graph\nRemember that going up on a PT graph means going forward and down means backward.\nLets analyze this motion above. This object went:\nA constant velocity forward 8 meters in 5 seconds\nWas at rest between 5 and 15 seconds\nWent constant velocity backwards 4 meters between 15 and 30 seconds.\nTo determine the vector displacement subtract where the individual ended from where they began.\nTo determine the scalar distance using this graph add any time you are going forward or backwards.\nThe displacement from 0-30 seconds is (+8) + (-4) = +4, or 4 meters forward.\nThe distance from 0-30 seconds is 8 + 4 = 12, or 12 meters.\nVelocity Time Graphs\nVelocity time graphs show how velocity changes in the Y axis relative to time in the X axis. Slope equals rise over run and since rise is change in velocity and run is the change in time the slope of a VT graph equals the acceleration.\nWhat a velocity time does not show is the actual position of the object, just how the position is changing. You can determine the change in displacement using information from the graphs and our acceleration or 1D motion equations as seen on our equation sheet.\nWhat the Slope of a VT Graph Represents\nA positive slope in a VT graph is acceleration (+a).\nA negative slope in a VT graph is deceleration (-a).\nA flat line in a VT graph shows no slope, meaning no acceleration, and therefore constant motion (a = 0).\nWatch the lesson on velocity time graphs directly on our YouTube channel here\nVT Graph Rest\nThe only place on a velocity time graph that you see rest is at 0 m/s. This is the only point on a velocity time graph the stickman is traveling 0 meters every second\nVT Graph Constant Positive Velocity\nA velocity time graphs slope shows how many meters per second an individual is moving per second. A flat line on a VT graph shows an individual moving a constant m/s. If the line is above the 0 m/s point the person is moving forward.\nVT Graph Constant Negative Velocity\nA velocity time graphs slope shows how many meters per second an individual is moving per second. A flat line on a VT graph shows an individual moving a constant m/s. If the line is below the 0 m/s point the person is moving backwards.\nVT Graph Acceleration In Forward Motion\nA velocity time graphs slope shows how many meters per second an individual is moving per second. A curved line on a VT graph shows an individual moving a increasing or decreasing their velocity (meters traveled per second). If the line is above the 0 m/s point the person is moving forward. The curve up above the 0 m/s velocity is accelerating or increasing velocity since the velocity is positive and the change in velocity or acceleration is also positive.\nVT Graph Deceleration In Forward Motion\nA velocity time graphs slope shows how many meters per second an individual is moving per second. A curved line on a VT graph shows an individual moving a increasing or decreasing their velocity (meters traveled per second). If the line is above the 0 m/s point the person is moving forward. The curve down above the 0 m/s velocity is decelerating or decreasing velocity since the velocity is positive and the change in velocity or acceleration is negative.\nSolving Problems with Velocity Time Graphs\nThe two things you can read off a velocity time graph are velocities and time. Anything else you will have to solve for. You can find acceleration in any straight lined interval by taking the initial and final velocity and the time. For displacement use the equation below or just calculate the area under the slope.\na = Δv/t\na = (vf – vi)/t\nx = ((vi + vf)/2)t\nNotice that the shape of the graph to the right is the same as the one used in the PT graph before but means something different on a VT graph. Lets analyze the motion.\nThe object started from rest and had a constant acceleration until 8 m/s for the first 5 seconds\nFrom 5 seconds to 15 seconds we had a constant velocity of 8 m/s\nThe object decelerated from 8 m/s to 4 meters per second between 15 and 30 seconds\nThe object never had a negative velocity (never below 0 m/s) so was always traveling forward\nThe acceleration during the first interval (0-5s) above was:\na = (vf – vi)/t\na = (8 - 0)/5 = 1.6 m/s2\nThe displacement during the first interval (0-5s) was:\nx = ((vi + vf)/2)t\nx = ((0 + 8)/2)(5) = 20 meters forward\nExample PT and VT Graph Problems\n1. Which displacement time graph on the right shows the same motion as the velocity time graph below?\nSee Solution\n(Graph A)\nFirst (0-1.5 seconds)\nV/T: the object travels 4 m/s for 1.5 seconds\nP/T: the object is at 6 meters after 1.5 minutes at the above velocity\nSecond (1.5-3.0 seconds)\nV/T: the object travels 0 m/s (not moving)\nP/T: the object stays at the 6 meter mark\nThird (3.0-4.0 seconds)\nV/T: the object travels 2 m/s (not moving)\nP/T: the object moves 2 more meters and is at 8 meters\nUse the displacement time graph to the right for questions 2-6\n2. How long did it take the object to the 6 m position?\nSee Solution\n(Click on the image to see larger)\n3. What is the objects displacement at 2 seconds?\nSee Solution\n(Click on the image to see larger)\n4. What is the objects velocity at 2 seconds?\nSee Solution\n(Click on the image to see larger)\n5. What is the objects velocity at 1 second?\nSee Solution\n(Click on the image to see larger)\n6. What is the objects velocity during the 3-4 second interval?\nSee Solution\n(Click on the image to see larger)\nUse the VT graph to the right to answer questions 7-10\n7. What is the velocity of the object in the graphs at 7.5 seconds?\nSee Solution\n(Click on the image to see larger)\n8. What is the acceleration of the object in the graphs at 5 seconds?\nSee Solution\n(Click on the image to see larger)\n9. What is the displacement of the object in the graph from 0 to 7.5 seconds?\nSee Solution\n(Click on the image to see larger)\n10. Is the object in the graph ever at rest and when?\nSee Solution\n(Click on the image to see larger)\nMotion Graphs Quiz Do you know how to tell the difference between what is happening on a position time or velocity time graph.Try our quiz and find out\n1 / 11\nWhat does the slope of a position time graph determine?\ndisplacement\nvelocity\nacceleration\nThe rate at which the position or displacement is changing. This object goes from 16 meters to 8 meters or is going backwards for the 30 seconds. The straight line means the type of motion is constant.Constant backward velocity\n2 / 11\nWhat does the slope of a velocity time graph determine?\ndisplacement\nvelocity\nacceleration\nAcceleration which is the rate at which the (Y axis) velocity on a V/T graph is changing.In the graph provided the object is going 8 meters every seconds (so is moving) but no slope or acceleration.\n3 / 11\nWhat is the displacement between 5 and 15 seconds\n0 m forward\n8 m forward\n40 m forward\n80 m forward\n4 / 11\nWhat type of motion is seen in this graph?\nconstant forward velocity\nconstant backwards velocity\nrest\nacceleration\ndeceleration\nWas moving at a constant rate of -4 m/s velocity for 30 seconds or had a constant backward velocity of 4 meters per second.Slope on a V/T graph is equal to acceleration. There was no acceleration because the slope was flat, but there was a velocity the whole time.\n5 / 11\nWhat type of motion is seen in this graph?\nconstant forward velocity\nconstant backwards velocity\nrest\nacceleration\ndeceleration\nWent from 16 meters to 8 meters (backwards 8 meters) in 30 seconds. The slope on a position or displacement time graph is equal to velocityConstant velocity backwards\n6 / 11\nWhat type of motion is seen in this graph?\nconstant forward velocity\nconstant backwards velocity\nrest\nacceleration\ndeceleration\nA straight line means something is constant. A straight sloped up line on a position time graph meant the position is moving at a constant rate or constant velocity.\n7 / 11\nWhat type of motion is seen in this graph?\nconstant forward velocity\nconstant backwards velocity\nrest\nacceleration\ndeceleration\nOn a V/T graphs the slope is acceleration. The slope here is sloped so there is acceleration. Another way to think about it is by what is happening in the graph. The object went 0 m/s to 12 m/s during the 30 seconds. They sped up during the entire time.\n8 / 11\nWhat type of motion is seen in this graph?\nconstant forward velocity\nconstant backwards velocity\nrest\nacceleration\ndeceleration\nOn a V/T graphs the slope is acceleration. The slope here is flat so no acceleration but it does have a constant velocity of 8 m/s the entire time. Another way to think about it is by what is happening in the graph. The object went 8 m/s and kept going 8 m/s for 30 seconds. Therefore moves 8 meters more every second for 30 seconds.\n9 / 11\nWhat type of motion is seen in this graph?\nconstant forward velocity\nconstant backwards velocity\nrest\nacceleration\ndeceleration\nOn a P/T graphs the slope is velocity. The slope here is flat so no velocity. Another way to think about it is by what is happening in the graph. The object went from 16 meters to 16 meters in 30 seconds. Therefore is in the same place and did not move.\n10 / 11\nWhat is the velocity of the object in this graph between 15 and 30 seconds\n0.18 m/s\n0.27 m/s\n0.67 m/s\n3.75 m/s\n600 m/s\n11 / 11\nWhat is the acceleration of this object over the 30 second time period?\n0.2 m/s/s\n0.4 m/s/s\n2.5 m/s/s\n360 m/s/s\nSlope of a velocity time graph is acceleration. Either find the slope or turn it into an equations picking givens off the graph (either way you are doing the same thing)\nYour score is\nLinkedIn\nFacebook\nTwitter\n0%\nLinks\nOn to Acceleration Due to Gravity\nBack to the Unit Main Menu\nBack to the Stickman Physics Home Page\nFind many of your animation resources in one place at the StickMan Physics Gallery\nShare\nPost\n1\nPin\nCopy\nShare\nShare\nMail\nShare\nSearch for:\nStickMan Physics Home\nStickman Physics Music Page\nUnit 1: One Dimensional Motion\nUnit 2: 2D Motion\nUnit 3: Newton’s Laws and Force\nUnit 4: Universal Gravitation and Circular Motion\nUnit 5: Work, Power, Mechanical Advantage, and Simple Machines\nUnit 6: Momentum, Impulse, and Conservation of Momentum\nUnit 7: Electrostatics\nUnit 8: Current and Circuits\nUnit 9: Magnetism and Electromagnetism\nUnit 10: Intro to Waves\nUnit 11: Electromagnetic Waves\nUnit 12: Nuclear Physics\nAP Physics 1 Pages (Deeper Dive into Concepts)\nDIY Creations for Fun and Physics\nTeachers: Do you want lessons and handouts already put together? Find resources at TeachersPayTeachers.\n©2025 StickMan Physics\n| Built using WordPress and Responsive Blogily theme by Superb\nMenu\nHome\nStickman Physics Music: Blending Science with Sound\nUnit 1: One Dimensional Motion: Physics Introduction\nUnit 2: Two Dimensional Motion: Projectile and Non-Projectile\nUnit 3: Newton’s Laws of Motion and Force\nUnit 4: Universal Gravitation and Circular Motion\nUnit 5: Work, Power, Mechanical Energy, and Simple Machines\nUnit 6: Momentum Impulse and Conservation of Momentum\nUnit 7: Electrostatics\nUnit 8: Current and Circuits\nUnit 9: Magnets and Magnetism\nUnit 10: Waves\nUnit 11: Electromagnetic Waves\nUnit 12: Nuclear Physics\nTable of Contents\nPractice\nEquation Sheet\nDigital Learning\nContact\nTerms and Conditions - Privacy Policy\nNotifications\nPin",
|
