Check an acceleration-time graph against its velocity-time graph
A student sees a graph with one straight rising line and says the object shown must be accelerating. On which kind of graph is the student right?
A train rolls through a station zone at a steady 5 m/s for 3 s. A classmate turns that motion into a pair of records, and gets one of them wrong. One check catches the mistake and names it: compare the pair one velocity-time segment at a time.
The train's velocity record: a flat line at 5 m/s for the whole 3 s.
The classmate's offered acceleration record: a steady 2 m/s² throughout.
Run the check. The flat segment's slope is 0 m/s², so the acceleration record must sit on the time axis, at 0 m/s².
The offered 2 m/s² claims the train speeds up while its velocity never moves: a nonzero level where the velocity-time line is flat.
Run the check one velocity-time segment at a time. Find each straight segment's slope, sign included. The acceleration-time graph must show a flat level equal to that slope, lasting exactly as long as the segment, because the slope of a velocity-time graph at a moment is the acceleration at that moment, sign included.
Watch for three common flaws in any offered pair: a nonzero level where the velocity-time line is flat, a level with the wrong sign, and an acceleration-time graph that copies the velocity-time graph's shape instead of plotting its slopes.
The same train braking: the velocity-time slope is −2 m/s², but the offered level is +2 m/s². A level with the wrong sign.
The offered graph copies the rising shape of the velocity-time line. The slope says it should be flat, at 2 m/s².
The figure shows a car's velocity-time graph and an offered acceleration-time graph. What is the flaw in the offered graph?
A cart's velocity-time graph, a straight fall from 8 m/s to 2 m/s over 3 s, above an offered acceleration-time graph: a flat level at +2 m/s².
What is the flaw in the offered graph? Decide before you continue.
The segment's slope is (2 − 8) ÷ 3 = −2 m/s², sign included. The offered level has the right size and the wrong sign. The velocity is falling, so the level must sit at −2 m/s², below the time axis.
The figure shows a minibus's velocity-time graph and an offered acceleration-time graph. What is the flaw in the offered graph?
A van's velocity-time graph, a straight rise from 3 m/s to 11 m/s over 4 s, above an offered acceleration-time graph: a ramp rising from 0 to 4 m/s².
What is the flaw this time? Decide before you continue.
The velocity-time segment is one straight line, so its slope is one constant value: (11 − 3) ÷ 4 = 2 m/s². The offered graph copies the rising shape instead of plotting the slope. A constant slope must appear as a flat level at 2 m/s², not a ramp.
The figure shows a scooter's velocity-time graph, which rises and then holds flat, and an offered acceleration-time graph. What is the flaw in the offered graph?
Your turn
The figure shows a delivery truck's velocity-time graph and an offered acceleration-time graph. What is the flaw in the offered graph?
The figure shows a monorail's velocity-time graph, which is flat and then falls to zero, and an offered acceleration-time graph. What is the flaw in the offered graph?
The figure shows a tour boat's velocity-time graph, which rises steeply and then rises gently, and an offered acceleration-time graph. What is the flaw in the offered graph?