Draw the acceleration-time graph from a velocity-time graph
On a velocity-time graph, what does the slope of the line at a moment give?
A car rolls past a stop line at 2 m/s and climbs steadily to 10 m/s over 4 s. You've already drawn a velocity-time graph from an acceleration record. Now run the translation the other way, starting from the car's velocity record.
The car's motion told twice on one time axis: its velocity-time graph on top, one straight segment, and beneath it the acceleration-time graph you are about to draw.
Take the velocity-time graph one straight segment at a time. Each segment has one slope: rise over run, sign included. For the car's segment:
rise = 10 − 2 = 8 m/s
run = 4 s
slope = 8 ÷ 4 = 2 m/s²
That slope is the car's acceleration, because the slope of a velocity-time graph at a moment is the acceleration at that moment, sign included.
On the acceleration-time graph, draw a flat level at the slope's value, lasting exactly as long as the segment. For the car: a level at 2 m/s² from 0 s to 4 s.
A flat velocity-time segment, the car cruising at one speed, has slope zero. Draw its level on the time axis, at 0 m/s².
A falling velocity-time segment, the car braking, has a negative slope. Draw its level below the time axis.
Do not copy the velocity-time graph's shape. The acceleration-time graph plots each segment's slope, so the car's straight climb becomes a flat level. The same segment-by-segment translation works for any velocity-time graph made of straight segments.
A taxi's velocity-time graph as it brakes: a straight line from 12 m/s down to 3 m/s over 3 s.
Sketch the taxi's acceleration-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 segment's rise is 3 − 12 = −9 m/s and its run is 3 s; the slope is −9 ÷ 3 = −3 m/s²; the level sits at −3 m/s², below the time axis, from 0 s to 3 s.
The figure shows a shuttle bus's velocity-time graph and four candidate acceleration-time graphs, A to D. Which panel shows the correct acceleration-time graph?
A train's velocity-time graph: a straight rise from 5 m/s to 11 m/s over the first 2 s, then a flat line at 11 m/s from 2 s to 4 s.
Sketch the train's acceleration-time graph on paper. Then continue to compare your sketch with the model answer.
The model answer. Check your sketch: the first segment's rise is 11 − 5 = 6 m/s over a run of 2 s, so its level sits at 3 m/s² from 0 s to 2 s; the second segment is flat, slope zero, so its level sits on the time axis from 2 s to 4 s.
A motorbike's velocity-time graph: one straight segment.
On paper, draw the motorbike's acceleration-time graph for the stretch shown. Label both axes with quantity and unit, and mark the acceleration value. Then continue to compare your drawing with the model answer.
Check your drawing: the graph is one flat level, not a ramp, because the segment has a single slope; the rise is 15 − 3 = 12 m/s over a run of 4 s, so the level sits at 12 ÷ 4 = 3 m/s²; the level runs the full 0 s to 4 s; the axes are labeled acceleration (m/s²) and time (s).
A truck's velocity-time graph: one straight falling segment.
On paper, draw the truck's acceleration-time graph for the stretch shown. Label both axes with quantity and unit, and mark the acceleration value. Then continue to compare your drawing with the model answer.
Check your drawing: the rise is 1 − 9 = −8 m/s over a run of 4 s, so the slope is −8 ÷ 4 = −2 m/s²; the slope is negative, so the level sits below the time axis; the level runs the full 0 s to 4 s; the axes are labeled acceleration (m/s²) and time (s).
Your turn
A city bus's velocity-time graph with three straight segments.
On paper, draw the bus's acceleration-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 drawing with the model answer.
Check your drawing: the slopes are (8 − 2) ÷ 3 = 2 m/s², then 0 m/s² for the flat segment, then (2 − 8) ÷ 2 = −3 m/s²; each slope becomes one flat level lasting exactly as long as its segment; a straight ramp on the velocity-time graph never becomes a ramp here, because its slope is constant.
The figure shows a harbor boat's velocity-time graph and four candidate acceleration-time graphs, A to D. Which panel shows the correct acceleration-time graph?
A river boat's velocity-time graph with three straight segments.
On paper, draw the boat's acceleration-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 drawing with the model answer.
Check your drawing: the slopes are (4 − 10) ÷ 2 = −3 m/s², then 0 m/s², then (12 − 4) ÷ 4 = 2 m/s²; a falling segment's level sits below the axis, a flat segment's level sits on it, and a rising segment's level sits above it; the level heights are the slopes, never the velocity values themselves.