15.3 Displacement–time and velocity–time graphs — Practise with help

Work each one first, closed book. Before you use a graph, decide whether the gradient or the area answers the question. Watch the sign of every area below the time axis.

Warm-up: remember

Find the gradient of the straight line through $(0, 3)$ and $(4, 11)$ .
Find the area of a trapezium with parallel sides $6$ and $10$ and height $5$ .
Work out $\frac{1}{2} \times 6 \times 8$ .

Answers

Gradient $= \dfrac{11 - 3}{4 - 0} = \dfrac{8}{4} = 2$ .
Area $= \dfrac{6 + 10}{2} \times 5 = 8 \times 5 = 40$ .
$\frac{1}{2} \times 6 \times 8 = 24$ .

If question 1 was hard, revise the gradient of a straight line before you continue. Every velocity and acceleration on this page comes from a gradient.

Practise with help

1. A displacement–time graph is a straight line through $(0, 0)$ and $(5, 15)$ , with $s$ in metres and $t$ in seconds. Find the velocity.

Hint

The velocity is the gradient of the displacement–time graph.

Next step

Divide the change in displacement by the change in time: $\dfrac{15 - 0}{5 - 0}$ .

Full working

v = \frac{15 - 0}{5 - 0} = 3 \text{ m/s}.

The line is straight, so the velocity is constant.

2. A displacement–time graph is horizontal from $t = 2$ s to $t = 9$ s at $s = 12$ m. Find the velocity in that interval.

Hint

A horizontal line has zero gradient. The displacement does not change.

Next step

The change in displacement is $0$ m over $7$ s.

Full working

v = \frac{12 - 12}{9 - 2} = 0 \text{ m/s}.

The object is at rest from $2$ s to $9$ s.

3. A velocity–time graph is a straight line from $(0, 8)$ to $(4, 20)$ , with $v$ in m/s and $t$ in s. Find the acceleration and the distance travelled.

Hint

Use the gradient for the acceleration and the area for the distance.

Next step

The gradient is $\dfrac{20 - 8}{4}$ . The area is the trapezium $\dfrac{8 + 20}{2} \times 4$ .

Full working

a = \frac{20 - 8}{4} = 3 \text{ m/s}^2, \qquad s = \frac{8 + 20}{2} \times 4 = 14 \times 4 = 56 \text{ m}.

4. A velocity–time graph is a straight line from $(0, 15)$ to $(5, 0)$ . Find the acceleration and the distance travelled.

Hint

The line falls, so the gradient is negative. The area is still a triangle, because the line stays above the axis.

Next step

The gradient is $\dfrac{0 - 15}{5}$ . The area is $\frac{1}{2} \times 5 \times 15$ .

Full working

a = \frac{0 - 15}{5} = -3 \text{ m/s}^2, \qquad s = \frac{1}{2} \times 5 \times 15 = 37.5 \text{ m}.

The negative acceleration shows the object is slowing down.

5. A velocity–time graph has three straight segments: $(0, 0)$ to $(4, 8)$ , $(4, 8)$ to $(10, 8)$ , and $(10, 8)$ to $(14, 0)$ . Find the total distance travelled and the acceleration on the last segment.

Hint

Split the graph into a triangle, a rectangle and a triangle. The acceleration on the last segment comes from its gradient.

Next step

The areas are $\frac{1}{2} \times 4 \times 8$ , $6 \times 8$ and $\frac{1}{2} \times 4 \times 8$ . The last gradient is $\dfrac{0 - 8}{14 - 10}$ .

Full working

\text{distance} = 16 + 48 + 16 = 80 \text{ m}, \qquad a = \frac{0 - 8}{4} = -2 \text{ m/s}^2.

The train speeds up, travels at a steady speed, then slows down.

6. A velocity–time graph is a straight line from $(0, 6)$ to $(2, -6)$ , with $v$ in m/s. Find the displacement and the distance travelled over the two seconds.

Hint

There are two triangles, one above the axis and one below it. Give the lower one a negative sign for the displacement.

Next step

The upper triangle has area $\frac{1}{2} \times 1 \times 6 = 3$ , and the lower triangle has the same size.

Full working

Displacement $= 3 - 3 = 0$ m, and distance travelled $= 3 + 3 = 6$ m.

The line crosses the axis at $t = 1$ s, so the object turns around and returns to its starting point.

7. A velocity–time graph is a straight line from $(0, 12)$ to $(6, 0)$ , and then a straight line from $(6, 0)$ to $(10, -4)$ . Find the displacement and the distance travelled over the whole $10$ s.

Hint

The first area is above the axis and the second is below it. Add the second with a negative sign for the displacement.

Next step

The first triangle has area $\frac{1}{2} \times 6 \times 12 = 36$ . The second has area $\frac{1}{2} \times 4 \times 4 = 8$ .

Full working

Displacement $= 36 - 8 = 28$ m. Distance travelled $= 36 + 8 = 44$ m.

The object moves forward, stops at $6$ s, then moves backwards.