Sketch the velocity-time graph from a position-time graph
What does the slope of a position-time graph at a moment give?
A runner jogs along a straight path at a steady 3 m/s for 4 s, starting 2 m past the timing post. Then she stops and stands still for 2 s. Her trip has two records: where she was, and how fast she was moving. The first record builds the second.
The runner's trip on one time axis: where she was on top, and beneath it, how fast she was moving. The lower graph is the one you are about to sketch.
The top graph is her position-time graph: a straight climb from 2 m to 14 m over 4 s, then a flat stretch at 14 m while she stands still.
Take the position-time graph one straight segment at a time. Each segment has one slope: rise over run, sign included. For the runner's climb:
rise = 14 − 2 = 12 m
run = 4 s
slope = 12 ÷ 4 = 3 m/s
That slope is the runner's velocity, because the slope of a position-time graph at a moment is the velocity at that moment, sign included.
On the velocity-time graph beneath, draw a flat level at the slope's value, lasting exactly as long as the segment. For the runner: a level at 3 m/s from 0 s to 4 s.
The flat stretch has slope zero. The runner is not moving, so from 4 s to 6 s her velocity level drops to 0 m/s, on the time axis.
If she had turned and walked back toward the post, that falling segment's slope would be negative. Its velocity level would sit below the time axis.
Do not copy the position-time graph's shape. The velocity-time graph plots each segment's slope, so straight ramps become flat levels. The same segment-by-segment sketch works for any position-time graph made of straight segments.
A cyclist's position-time graph: a straight fall from 10 m to −2 m over 4 s.
Sketch the cyclist's velocity-time graph on paper. Mark the level's value. Then continue to compare your sketch with the model answer.
The model answer. Check your sketch: the rise is −2 − 10 = −12 m over a run of 4 s; the slope is −12 ÷ 4 = −3 m/s, so the level sits at −3 m/s, below the time axis, from 0 s to 4 s.
The figure shows a dog's position-time graph along a straight path, and four candidate velocity-time graphs, A to D. Which panel shows the correct velocity-time graph?
A remote-control car's position-time graph: from 1 m up to 7 m over 2 s, flat at 7 m for 2 s, then down to −1 m over the next 4 s.
Sketch the car's velocity-time graph on paper. Then continue to compare your sketch with the model answer.
The model answer. Check your sketch: the first segment's slope is (7 − 1) ÷ 2 = 3 m/s; the flat segment gives 0 m/s; the last segment's slope is (−1 − 7) ÷ 4 = −2 m/s, a level below the axis from 4 s to 8 s.
A mail carrier's position-time graph: a climb, then a flat stretch.
On paper, sketch the mail carrier's velocity-time graph for the full stretch shown. Label both axes with quantity and unit, and mark each level's value. Then continue to compare your sketch with the model answer.
Check your sketch: the first level sits at 2 m/s, because the segment rises 8 m over 4 s; the level is flat, not a ramp, because one straight segment has one slope; from 4 s to 6 s the level sits on the axis at 0 m/s, because the position holds still; each level lasts exactly as long as its segment.
A hiker's position-time graph, measured from the trailhead: one straight falling segment.
On paper, sketch the hiker's velocity-time graph for the stretch shown. Label both axes with quantity and unit, and mark the level's value. Then continue to compare your sketch with the model answer.
Check your sketch: the rise is 0 − 12 = −12 m over a run of 3 s, so the level sits at −4 m/s; the level is below the time axis, because the position falls, so the hiker moves in the negative direction; the level is flat and runs the full 0 s to 3 s.
Your turn
A warehouse robot's position-time graph along a straight track, with three straight segments.
On paper, sketch the robot's velocity-time graph for the full stretch shown. Label both axes with quantity and unit, and mark each level's value. Then continue to compare your sketch with the model answer.
Check your sketch: the slopes are (9 − 1) ÷ 2 = 4 m/s, then 0 m/s, then (3 − 9) ÷ 3 = −2 m/s; each slope is that stretch's velocity; a rising segment's level sits above the axis, a flat segment's level sits on it, and a falling segment's level sits below it.
The figure shows a rover's position-time graph, which falls through negative positions and then holds flat, and four candidate velocity-time graphs, A to D. Which panel shows the correct velocity-time graph?
A museum tour cart's position-time graph, with three straight segments.
On paper, sketch the cart's velocity-time graph for the full stretch shown. Label both axes with quantity and unit, and mark each level's value. Then continue to compare your sketch with the model answer.
Check your sketch: the slopes are 9 ÷ 3 = 3 m/s, then 0 m/s, then (3 − 9) ÷ 2 = −3 m/s; each straight segment's slope becomes one flat velocity level over its own stretch; the return trip's level sits below the axis, because its slope is negative.