Interpreting a Velocity–Time Graph with a Changing Force
A 0.50 kg model car moves along a straight track. Its velocity–time graph is described in words: from t = 0 to 2.0 s the velocity rises steadily from 0 to 3.0 m/s; from 2.0 s to 5.0 s it stays at 3.0 m/s; from 5.0 s to 6.0 s it falls steadily to 0. A student claims the car 'experiences no force' between 2.0 s and 5.0 s because the velocity is constant, and that the largest force acts during the 5.0–6.0 s interval because 'the change is fastest there'. The track is horizontal and friction is stated to be negligible during the first interval only.
1. Calculate the acceleration in each interval and the net force acting during each interval.
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Interval 1 (0–2.0 s): a = (3.0 − 0) / 2.0 = 1.5 m/s². Net force F = ma = 0.50 × 1.5 = 0.75 N.
Interval 2 (2.0–5.0 s): a = 0. Net force = 0.
Interval 3 (5.0–6.0 s): a = (0 − 3.0) / 1.0 = −3.0 m/s². Net force = 0.50 × (−3.0) = −1.5 N, i.e. 1.5 N opposite the motion.
So the largest net force is in the third interval (1.5 N), which does support the student's second claim, but for the correct reason: the magnitude of acceleration is greatest there. The first claim is also correct in the sense that the net force is zero, but it is wrong to say 'no force' - the driving force and friction (or other resistances) may balance. Since friction is stated negligible only in the first interval, during the constant-velocity phase the driving force must equal the resistive force.
2. Evaluate the student's reasoning that 'the change is fastest' in the third interval, and explain what the graph's gradient represents.
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The student's conclusion happens to be right but the reasoning should be made precise. On a velocity–time graph, the gradient is acceleration, not force directly. Force is proportional to acceleration only for constant mass, which is stated here (0.50 kg).
The gradient in interval 3 is −3.0 m/s², compared with +1.5 m/s² in interval 1. So the magnitude of acceleration is twice as large in interval 3, and the force is twice as large.
The phrase 'the change is fastest' is vague; the correct statement is that the rate of change of velocity is greatest in magnitude during 5.0–6.0 s. A good answer distinguishes velocity (the value on the graph) from acceleration (the gradient) and from force (mass times gradient).
3. Use the graph description to find the total displacement and the average velocity over the whole 6.0 s, and comment on whether average speed and average velocity are the same here.
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Displacement is the area under the velocity–time graph.
Interval 1: triangle, ½ × 2.0 × 3.0 = 3.0 m. Interval 2: rectangle, 3.0 × 3.0 = 9.0 m. Interval 3: triangle, ½ × 1.0 × 3.0 = 1.5 m.
Total displacement = 3.0 + 9.0 + 1.5 = 13.5 m.
Average velocity = 13.5 / 6.0 = 2.25 m/s.
Because the motion is along a straight line in one direction, the distance travelled equals the magnitude of displacement (13.5 m), so average speed = average velocity = 2.25 m/s. If the car had reversed direction, the areas would have opposite signs and average speed and average velocity would differ. This is a common trap: the equality here is a consequence of one-way motion, not a general rule.