15.5 Calculus in kinematics — Practise with help

Work each one first, closed book. Decide whether the question needs a derivative or an integral, then write down the initial condition before you integrate.

Warm-up: remember

Differentiate $y = x^3$ with respect to $x$ .
Find $\displaystyle\int 4t\,dt$ .
Work out $(-3)^2$ .

Answers

$\dfrac{dy}{dx} = 3x^2$ .
$\displaystyle\int 4t\,dt = 2t^2 + C$ .
$(-3)^2 = 9$ .

If question 2 was hard, revise integrating powers of $t$ before you continue. Every velocity on this page is an integral of an acceleration.

Practise with help

1. The displacement of a particle is $s = t^2 + 3t$ metres. Find its velocity and its acceleration.

Hint

Differentiate once for the velocity and twice for the acceleration.

Next step

$v = \dfrac{ds}{dt} = 2t + 3$ . Differentiate that for $a$ .

Full working

v = 2t + 3 \text{ m/s}, \qquad a = \frac{dv}{dt} = 2 \text{ m/s}^2.

The acceleration is constant at $2$ m/s $^2$ .

2. The displacement of a particle is $s = 2t^3 - 3t^2$ metres. Find its velocity and its acceleration.

Hint

Use the power rule. Take the coefficient down and reduce the power by one.

Next step

$v = 6t^2 - 6t$ . Differentiate again for $a$ .

Full working

v = 6t^2 - 6t \text{ m/s}, \qquad a = 12t - 6 \text{ m/s}^2.

This acceleration is not constant, so the constant-acceleration formulae do not apply here.

3. 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.

Hint

Integrate the velocity. Add a constant, then use the initial condition to fix it.

Next step

$s = \displaystyle\int (4t + 1)\,dt = 2t^2 + t + C$ . Use $s = 0$ at $t = 0$ .

Full working

s = 2t^2 + t + C.

At $t = 0$ , $s = 0$ , so $C = 0$ and

s = 2t^2 + t \text{ m}.

4. 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.

Hint

Integrate the acceleration and use the starting velocity to find the constant.

Next step

$v = \displaystyle\int 10\,dt = 10t + C$ . Use $v = 3$ at $t = 0$ .

Full working

v = 10t + C.

At $t = 0$ , $v = 3$ , so $C = 3$ and

v = 10t + 3 \text{ m/s}.

5. A particle has velocity $v = 3t^2 + 4$ m/s. Its displacement is $0$ m when $t = 0$ . Find its displacement as a function of time.

Hint

Integrate term by term: the integral of $3t^2$ is $t^3$ , and the integral of $4$ is $4t$ .

Next step

$s = t^3 + 4t + C$ . Use $s = 0$ at $t = 0$ .

Full working

s = t^3 + 4t + C.

At $t = 0$ , $s = 0$ , so $C = 0$ and

s = t^3 + 4t \text{ m}.

6. The displacement of a particle is $s = t^3 - 3t^2$ metres. Find the times when the particle is at rest.

Hint

The particle is at rest when $v = 0$ . Differentiate the displacement first.

Next step

$v = 3t^2 - 6t = 3t(t - 2)$ . Set each factor equal to zero.

Full working

v = 3t^2 - 6t = 3t(t - 2) = 0.

So $t = 0$ or $t = 2$ . The particle is at rest at $t = 0$ s and at $t = 2$ s.

7. The displacement of a particle is $s = 6t - t^2$ metres for $0 \le t \le 6$ . Find the time at rest, the displacement over the whole interval, and the distance travelled over the whole interval.

Hint

The particle is at rest when $v = 0$ . That time splits the interval into two parts. The displacement uses the signs, and the distance adds the sizes.

Next step

$v = 6 - 2t$ , so the particle is at rest at $t = 3$ . Work out $s$ at $t = 0$ , $t = 3$ and $t = 6$ .

Full working

$v = 6 - 2t$ , so $v = 0$ at $t = 3$ s.

$s(0) = 0$ , $s(3) = 18 - 9 = 9$ m and $s(6) = 36 - 36 = 0$ m.

Displacement $= 0 - 0 = 0$ m.

Distance travelled $= |9 - 0| + |0 - 9| = 9 + 9 = 18$ m.

The particle moves forward $9$ m and then backward $9$ m.