15.5 Calculus in kinematics — Practise

Work through these on your own. Decide whether you need a derivative or an integral, and apply the initial condition before you call the answer finished. When a particle turns around, split the interval before you find a distance.

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

The displacement of a particle is $s = t^2 + 3t$ metres. Find its velocity and its acceleration.
The displacement of a particle is $s = 2t^3 - 3t^2$ metres. Find its velocity and its acceleration.
A particle has velocity $v = 4t + 1$ m/s. Its displacement is $0$ m when $t = 0$ . Find its displacement as a function of time.
A particle has constant acceleration $a = 10$ m/s $^2$ . Its velocity is $3$ m/s when $t = 0$ . Find its velocity as a function of time.
A particle has constant acceleration $a = 10$ m/s $^2$ . Its velocity is $3$ m/s and its displacement is $0$ m when $t = 0$ . Find its displacement as a function of time.
The displacement of a particle is $s = t^3 - 3t^2$ metres. Find the times when the particle is at rest.
The displacement of a particle is $s = 6t - t^2$ metres. Find the time when the particle is at rest.
The displacement of a particle is $s = 6t - t^2$ metres for $0 \le t \le 6$ . Find the displacement over the whole interval.
The displacement of a particle is $s = 6t - t^2$ metres for $0 \le t \le 6$ . The particle is at rest at $t = 3$ . Find the total distance travelled.
A particle has acceleration $a = 6t$ m/s $^2$ . Its velocity is $4$ m/s when $t = 0$ . Find its velocity as a function of time.
A particle has velocity $v = 3t^2 + 4$ m/s and displacement $0$ m when $t = 0$ . Find its displacement as a function of time.
A particle starts with velocity $5$ m/s and moves with constant acceleration $4$ m/s $^2$ . Use calculus to find its displacement after $3$ s.

Answers and concise solutions

1. The velocity and the acceleration

Answer: Differentiate once: $v = \dfrac{ds}{dt} = 2t + 3$ m/s. Differentiate again: $a = \dfrac{dv}{dt} = 2$ m/s $^2$ , which is constant.

2. The velocity and the acceleration

Answer: Differentiate once: $v = 6t^2 - 6t$ m/s. Differentiate again: $a = 12t - 6$ m/s $^2$ . The acceleration depends on $t$ , so the constant-acceleration formulae do not apply.

3. The displacement

Answer: Integrate: $s = \int (4t + 1)\,dt = 2t^2 + t + C$ . The condition $s = 0$ at $t = 0$ gives $C = 0$ , so $s = 2t^2 + t$ m.

4. The velocity

Answer: Integrate: $v = \int 10\,dt = 10t + C$ . The condition $v = 3$ at $t = 0$ gives $C = 3$ , so $v = 10t + 3$ m/s.

5. The displacement

Answer: The velocity is $v = 10t + 3$ m/s from the starting condition. Integrate: $s = \int (10t + 3)\,dt = 5t^2 + 3t + C$ . With $s = 0$ at $t = 0$ , $C = 0$ , so $s = 5t^2 + 3t$ m.

6. The times at rest

Answer: The velocity is $v = 3t^2 - 6t = 3t(t - 2)$ . Setting $v = 0$ gives $t = 0$ or $t = 2$ , so the particle is at rest at $0$ s and $2$ s.

7. The time at rest

Answer: The velocity is $v = 6 - 2t$ . Setting $v = 0$ gives $t = 3$ s. After that time the velocity is negative, so the particle turns around.

8. The displacement

Answer: At $t = 6$ , $s = 36 - 36 = 0$ m. At $t = 0$ , $s = 0$ m. The displacement is $0 - 0 = 0$ m, so the particle returns to its starting point.

9. The total distance

Answer: At $t = 3$ , $s = 18 - 9 = 9$ m. The particle moves from $0$ m to $9$ m, then back to $0$ m, so the distance travelled is $9 + 9 = 18$ m.

10. The velocity

Answer: Integrate: $v = \int 6t\,dt = 3t^2 + C$ . The condition $v = 4$ at $t = 0$ gives $C = 4$ , so $v = 3t^2 + 4$ m/s.

11. The displacement

Answer: Integrate term by term: $s = \int (3t^2 + 4)\,dt = t^3 + 4t + C$ . With $s = 0$ at $t = 0$ , $C = 0$ , so $s = t^3 + 4t$ m.

12. The displacement

Answer: Integrate the acceleration: $v = 4t + 5$ m/s. Integrate the velocity: $s = 2t^2 + 5t$ m, because $s = 0$ at $t = 0$ . At $t = 3$ , $s = 18 + 15 = 33$ m. The constant-acceleration formula $s = ut + \frac{1}{2}at^2$ gives the same value.