Kinematics Graph Explorer
Drag the corners of a velocity–time graph and watch a car follow it, with displacement–time and acceleration–time graphs drawn as it moves.
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Learning objectives
- Interpret position, displacement, velocity and acceleration using equations and graphs.
- Derive the uniformly accelerated motion equations from the definitions of velocity and acceleration.
- Derive and apply uniformly accelerated motion equations with a stated sign convention.
- Plot and interpret displacement–time and velocity–time graphs in one dimension
- Deduce rest and uniform or non-uniform velocity from a displacement–time graph
- Deduce rest, uniform velocity and uniform or non-uniform acceleration from a velocity–time graph
- Use signed area under a velocity–time graph to determine displacement
Velocity–time graph through 0.0 at 0 s, 4.0 at 2 s, 8.0 at 4 s, 8.0 at 6 s, 8.0 at 8 s, 0.0 at 10 s metres per second. At 0.0 s the car is 0.0 metres from the start, moving at 0.0 metres per second.
- Displacement, s
- 0.0 m
- Velocity, v
- 0.0 m s−1
- Acceleration, a
- 2.00 m s−2
- Distance travelled
- 0.0 m
Try this
0 of 4 doneMake the displacement–time graph one straight sloping line, then play the motion. (not done yet)
A straight s–t line means constant velocity: its gradient is the velocity and the acceleration is zero.
Make the car travel 30 m forwards in 10 s and finish at rest. (not done yet)
The area under the velocity–time graph is the displacement, whatever the shape.
Make the car speed up while its acceleration is negative. (not done yet)
Negative acceleration speeds the car up when it is already moving in the negative direction: speed rises when v and a have the same sign.
Make the car finish where it started after moving at least 10 m. (not done yet)
The area above the time axis equals the area below it, so the displacement is zero even though the distance is not.