A cart moves along a straight track. Its measured velocity–time coordinates are (0 s, 0 m s⁻¹), (2 s, 4 m s⁻¹), (5 s, 4 m s⁻¹), and (7 s, −2 m s⁻¹); adjacent points are joined by straight lines. Determine the acceleration on each interval, the displacement and total distance from 0 to 7 s, and the time at which the cart reverses. Sketch how the position graph must curve and identify an experimental limitation of estimating velocity by frame-to-frame video differences.

Question

Accepted Answer

Accelerations: +2.0, 0 and −3.0 m s⁻². Displacement = 18.0 m; distance = 19.3 m; reversal at 6.33 s. Position rises with changing gradient, becomes linear, reaches a maximum at reversal, then falls. Acceleration is the velocity–time gradient. The three gradients are (4−0)/(2−0) = 2.0 m s⁻², zero, and (−2−4)/(7−5) = −3.0 m s⁻². Displacement is signed area. The first triangle contributes 4 m, the rectangle contributes 12 m, and the final trapezium contributes (4 + −2)(2)/2 = 2 m, giving 18 m. On the final line v = 4 − 3(t−5). Setting v = 0 gives t = 6.33 s. Splitting that interval at the crossing gives 2.67 m forward and 0.67 m backward. Total distance is therefore 4 + 12 + 2.67 + 0.67 = 19.3 m. Distance exceeds the magnitude of displacement because the cart travels backward after the turning instant. The position graph's gradient equals velocity: it steepens initially, is straight from 2–5 s, flattens to a maximum at 6.33 s, then slopes downward. Finite video frame spacing averages velocity and can blur rapid acceleration; scale calibration and parallax also matter.

Reconstruct a journey from a velocity graph

What a complete response should do

Measured velocity of the cart

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Measured velocity of the cart — supplied values
Time / s	Velocity / m s⁻¹
0	0
2	4
5	4
7	-2