KCSE 2025 Maths P2 Q9 — Rate of Change from a Displacement-Time Graph
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The Question
“The graph shows the displacement S metres of a moving particle from a point O after time t seconds, for t between 0 and 1. The curve starts at 6 m, rises to a peak of about 7.4 m near t = 0.3 s, then falls back to zero at t = 1 s. Use the graph to determine the rate of change of displacement S at time t = 0.5 seconds.”
Recognise the rate of change as a gradient
The rate of change of displacement with respect to time is the velocity of the particle. On a displacement-time graph the velocity at any instant is the gradient of the curve at that point, which you read off by drawing the tangent line that just touches the curve there.
Draw the tangent at t = 0.5 s
Locate the point on the curve at t = 0.5 seconds. From the graph the displacement there is about 6.75 m. Place a sharp point at that spot and carefully draw a straight line that touches the curve at only that point, leaning the ruler so it matches the direction of the curve, so the line is a true tangent.
Read two points off the tangent
Pick two convenient points on the tangent line and read their change in time along the bottom and their change in displacement up the side. On this tangent, moving across a run of 0.2 seconds the displacement falls by 1.2 metres, so the rise is negative because the line slopes downwards.
Compute the gradient
Divide the rise by the run to get the gradient of the tangent. The negative sign is kept because the displacement is decreasing, which tells you the particle is now moving back towards the point O it started from.
Final Result
The rate of change of displacement at t = 0.5 seconds is about -6 m/s. The negative sign shows the particle is moving back towards O. Because the value is read from a tangent drawn by eye, any answer from roughly -5 to -7 m/s is accepted.
Why this method works
This works because the gradient of a displacement-time graph measures how fast displacement changes per unit time, which is exactly the definition of velocity, the rate of change of displacement. At a single instant the curve is not straight, so its steepness is captured by the tangent line that matches the curve's direction at that point; the tangent's rise over run gives the instantaneous gradient. A negative gradient means displacement is falling, so the particle is heading back towards its starting point.
The curve is past its peak at t = 0.5 s and heading down to zero at t = 1 s, so the displacement must be decreasing there, which agrees with the negative gradient of -6 m/s.