3.1 Kinematics and motion graphs — Practise

Work through these on your own. Write the positive direction at the top of each answer, and name the equation you use.

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

A car accelerates uniformly from rest at $2.5\,\text{m}\,\text{s}^{-2}$ for 6.0 s. Find its final speed and the distance it travels.
A train slows from $40\,\text{m}\,\text{s}^{-1}$ to $15\,\text{m}\,\text{s}^{-1}$ over a distance of 500 m. Find its acceleration.
A stone is dropped from rest and falls for 3.2 s. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ . Find its speed and the distance it falls.
A ball is thrown vertically upwards at $12\,\text{m}\,\text{s}^{-1}$ . Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ . Find the maximum height it reaches and the time it takes to reach it.
A velocity–time graph has the points $(0, 0)$ , $(2, 6)$ , $(4, 12)$ , $(6, 12)$ and $(8, 0)$ . The first two points are in a straight line. Find the acceleration in the first 2.0 s and the total displacement over the 8.0 s.
A ball is thrown vertically upwards at $20\,\text{m}\,\text{s}^{-1}$ . Take upwards as positive and $g = 9.81\,\text{m}\,\text{s}^{-2}$ . Find its velocity after 3.0 s.
A car accelerates uniformly from rest at $1.5\,\text{m}\,\text{s}^{-2}$ for 8.0 s, then travels at the final speed for a further 12 s. Find the total distance travelled.
A runner completes one lap of a 400 m track in 50 s and stops at the start. Explain the difference between the distance travelled, the displacement and the average speed.
A model rocket lifts off from rest and accelerates upwards at $25\,\text{m}\,\text{s}^{-2}$ for 4.0 s. The engine then cuts off and the rocket decelerates at $g = 9.81\,\text{m}\,\text{s}^{-2}$ . Find the speed at engine cut-off and the time from cut-off until the rocket reaches its highest point.
A car travels at $25\,\text{m}\,\text{s}^{-1}$ . Write this speed in $\text{km}\,\text{h}^{-1}$ .
A car accelerates uniformly from $10\,\text{m}\,\text{s}^{-1}$ to $22\,\text{m}\,\text{s}^{-1}$ over 120 m. Say which equation of motion you would choose, and why, before you calculate the acceleration.
A trolley starts from rest and travels 60 m in 8.0 s with constant acceleration. Find the acceleration to two significant figures.

Answers and concise solutions

1. The final speed and the distance

Answer: From rest, $v = u + at = 0 + 2.5 \times 6.0 = 15\,\text{m}\,\text{s}^{-1}$ . The distance is $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2} \times 2.5 \times 6.0^2 = 45\,\text{m}$ .

2. The acceleration

Answer: Use $v^2 = u^2 + 2as$ with the time absent. Rearranging gives $a = \dfrac{v^2 - u^2}{2s} = \dfrac{15^2 - 40^2}{2 \times 500} = \dfrac{-1375}{1000} = -1.375\,\text{m}\,\text{s}^{-2}$ . The negative sign shows that the train is slowing down.

3. The speed and the distance

Answer: Take downwards as positive. The speed is $v = gt = 9.81 \times 3.2 = 31.4\,\text{m}\,\text{s}^{-1}$ . The distance is $s = \tfrac{1}{2}gt^2 = \tfrac{1}{2} \times 9.81 \times 3.2^2 = 50.2\,\text{m}$ .

4. The maximum height and the time

Answer: At the highest point the velocity is zero, so $0 = u^2 - 2gs$ gives $s = \dfrac{u^2}{2g} = \dfrac{144}{19.62} = 7.34\,\text{m}$ . The time is $t = \dfrac{u}{g} = \dfrac{12}{9.81} = 1.22\,\text{s}$ .

5. The acceleration and the displacement

Answer: The acceleration is the gradient in the first 2.0 s: $\dfrac{6 - 0}{2 - 0} = 3.0\,\text{m}\,\text{s}^{-2}$ . The displacement is the area under the graph, split into trapezia: $\tfrac{1}{2}(2)(0 + 6) + \tfrac{1}{2}(2)(6 + 12) + \tfrac{1}{2}(2)(12 + 12) + \tfrac{1}{2}(2)(12 + 0) = 6 + 18 + 24 + 12 = 60\,\text{m}$ .

6. The velocity after 3.0 s

Answer: Using $v = u + at$ with $a = -g$ : $v = 20 - 9.81 \times 3.0 = -9.43\,\text{m}\,\text{s}^{-1}$ . The velocity is negative, so the ball is now moving downwards.

7. The total distance

Answer: In the first stage the car reaches $v = 1.5 \times 8.0 = 12\,\text{m}\,\text{s}^{-1}$ and covers $s_1 = \tfrac{1}{2} \times 1.5 \times 8.0^2 = 48\,\text{m}$ . In the second stage it covers $s_2 = 12 \times 12 = 144\,\text{m}$ . The total is $48 + 144 = 192\,\text{m}$ . One equation cannot describe both stages, because the acceleration changes.

8. Distance, displacement and average speed

The distance travelled is 400 m, because the runner has covered the whole lap.
The displacement is zero, because the runner finishes at the starting point. Displacement is a vector from the start to the finish.
The average speed is 400 / 50 = 8.0 m/s. The average velocity is zero, because the displacement is zero.

Answer: The distance travelled is 400 m. The displacement is zero, because the runner returns to the start. The average speed is 8.0 m/s; the average velocity is zero.

9. The speed at cut-off and the time to the highest point

Answer: At cut-off the speed is $v = at = 25 \times 4.0 = 100\,\text{m}\,\text{s}^{-1}$ . After cut-off the only force is gravity, so the rocket slows at $g$ . The time to the highest point is $t = \dfrac{v}{g} = \dfrac{100}{9.81} = 10.2\,\text{s}$ .

10. The speed in kilometres per hour

Answer: $25\,\text{m}\,\text{s}^{-1} = 25 \times \dfrac{3600}{1000} = 90\,\text{km}\,\text{h}^{-1}$ . Multiplying metres per second by 3.6 gives kilometres per hour.

11. The equation and the reason

The acceleration is wanted, and the time is not given and is not wanted.
Use v^2 = u^2 + 2as, because it is the equation that leaves out the time.
Each of the four equations leaves out one quantity, so choose the one that omits the quantity you do not have.

Answer: The acceleration is wanted and the time is not given, so use $v^2 = u^2 + 2as$ . Rearranging gives $a = \dfrac{22^2 - 10^2}{2 \times 120} = \dfrac{384}{240} = 1.6\,\text{m}\,\text{s}^{-2}$ .

12. The acceleration to two significant figures

Answer: Using $s = ut + \tfrac{1}{2}at^2$ from rest: $60 = \tfrac{1}{2}a(8.0)^2$ , so $a = \dfrac{120}{64} = 1.9\,\text{m}\,\text{s}^{-2}$ to two significant figures.