15.3 Displacement–time and velocity–time graphs — Practise

Work through these on your own. Decide whether the question needs a gradient or an area, then watch the sign of every area below the time axis. A distance adds the sizes; a displacement adds the signs.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A displacement–time graph is a straight line through $(0, 0)$ and $(8, 24)$ , with $s$ in metres and $t$ in seconds. Find the velocity.
A displacement–time graph is a straight line through $(0, 10)$ and $(5, 25)$ , with $s$ in metres. Find the velocity.
A displacement–time graph is horizontal from $t = 2$ s to $t = 9$ s. Find the velocity in that interval.
A velocity–time graph is a straight line from $(0, 5)$ to $(4, 17)$ , with $v$ in m/s. Find the acceleration and the distance travelled.
A velocity–time graph is a straight line from $(0, 20)$ to $(5, 5)$ , with $v$ in m/s. Find the acceleration and the distance travelled.
A velocity–time graph is a straight line from $(0, 0)$ to $(6, 18)$ , with $v$ in m/s. Find the acceleration and the distance travelled.
A velocity–time graph is a straight line from $(0, 24)$ to $(8, 0)$ . Find the acceleration and the distance travelled.
A velocity–time graph has segments $(0, 0)$ to $(3, 6)$ , $(3, 6)$ to $(9, 6)$ , and $(9, 6)$ to $(12, 0)$ . Find the total distance travelled.
A velocity–time graph is a straight line from $(0, 8)$ to $(4, -8)$ , with $v$ in m/s. Find the displacement and the distance travelled.
A velocity–time graph is a straight line from $(0, 4)$ to $(5, 14)$ . Find the distance travelled.
A velocity–time graph rises from $(0, 0)$ to $(5, 15)$ , then stays at $15$ m/s from $(5, 15)$ to $(15, 15)$ . Find the total distance travelled.
A velocity–time graph has a straight line from $(0, 12)$ to $(6, 0)$ , then a straight line from $(6, 0)$ to $(10, -4)$ . Find the displacement and the distance travelled over the whole $10$ s.

Answers and concise solutions

1. The velocity

Answer: The velocity is the gradient: $v = \dfrac{24 - 0}{8 - 0} = 3$ m/s.

2. The velocity

Answer: Use the change in displacement over the change in time: $v = \dfrac{25 - 10}{5 - 0} = 3$ m/s. The starting value of $10$ m does not change the gradient.

3. The velocity

Answer: A horizontal line has zero gradient, so the velocity is $0$ m/s. The object is at rest.

4. The acceleration and the distance

Answer: Acceleration $= \dfrac{17 - 5}{4} = 3$ m/s $^2$ . The distance is the area of the trapezium: $\dfrac{5 + 17}{2} \times 4 = 44$ m.

5. The acceleration and the distance

Answer: Acceleration $= \dfrac{5 - 20}{5} = -3$ m/s $^2$ . The distance is the trapezium area: $\dfrac{20 + 5}{2} \times 5 = 62.5$ m.

6. The acceleration and the distance

Answer: Acceleration $= \dfrac{18}{6} = 3$ m/s $^2$ . The area is a triangle: $\frac{1}{2} \times 6 \times 18 = 54$ m.

7. The acceleration and the distance

Answer: Acceleration $= \dfrac{0 - 24}{8} = -3$ m/s $^2$ . The car brakes to rest, and the distance is the triangle $\frac{1}{2} \times 8 \times 24 = 96$ m.

8. The total distance

Answer: The three areas are a triangle $9$ , a rectangle $36$ and a triangle $9$ . The total distance is $9 + 36 + 9 = 54$ m.

9. The displacement and the distance

Answer: The line crosses the axis at $t = 2$ s. The two triangles each have area $\frac{1}{2} \times 2 \times 8 = 8$ . The displacement is $8 - 8 = 0$ m and the distance travelled is $8 + 8 = 16$ m.

10. The distance

Answer: The distance is the trapezium area: $\dfrac{4 + 14}{2} \times 5 = 9 \times 5 = 45$ m.

11. The total distance

Answer: The first stage is a triangle, $\frac{1}{2} \times 5 \times 15 = 37.5$ m. The second stage is a rectangle, $10 \times 15 = 150$ m. The total is $37.5 + 150 = 187.5$ m.

12. The displacement and the distance

Answer: The area above the axis is $\frac{1}{2} \times 6 \times 12 = 36$ . The area below is $\frac{1}{2} \times 4 \times 4 = 8$ , which counts as $-8$ for the displacement. Displacement $= 36 - 8 = 28$ m, and distance travelled $= 36 + 8 = 44$ m.