Chapter 15: Review

The chapter problem, solved

A free kick is taken from the ground, $40$ m from the goal line. The ball leaves the ground at $20$ m/s at $45^\circ$ to the horizontal. The crossbar is $2.44$ m high. Model the ball as a particle, ignore air resistance, and take $g = 9.8$ m/s $^2$ .

Step 1. Check the units. Length is measured in metres, m; time in seconds, s; velocity in m/s; acceleration in m/s $^2$ . The mass of the ball is not needed, because the motion is fixed by the initial velocity and gravity alone.

Step 2. Resolve the initial velocity. The launch angle is $45^\circ$ , so the two components are equal:

u_x = 20\cos 45^\circ = 10\sqrt{2} \approx 14.1 \text{ m/s}, \qquad u_y = 20\sin 45^\circ = 10\sqrt{2} \approx 14.1 \text{ m/s}.

There is no horizontal acceleration, so $v_x = 10\sqrt{2}$ throughout the flight.

Step 3. Find the time to the goal line. The horizontal distance is $40$ m, and $x = u_x t$ :

40 = 10\sqrt{2} t, \qquad t = \frac{40}{10\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \approx 2.83 \text{ s}.

Step 4. Find the height at that time. Use the vertical equation with upwards positive, $a = -9.8$ :

y = u_y t - \frac{1}{2}gt^2 = 10\sqrt{2} \times 2\sqrt{2} - \frac{1}{2} \times 9.8 \times (2\sqrt{2})^2.

Now $10\sqrt{2} \times 2\sqrt{2} = 40$ , and $(2\sqrt{2})^2 = 8$ , so

y = 40 - 4.9 \times 8 = 40 - 39.2 = 0.8 \text{ m}.

Step 5. Compare with the crossbar. The ball is $0.8$ m above the ground when it reaches the goal line. The crossbar is $2.44$ m above the ground. Since

0.8 < 2.44,

the ball passes under the crossbar. It does not go over the bar.

Step 6. Check with the vertical velocity. From $v_y = u_y - gt$ ,

v_y = 10\sqrt{2} - 9.8 \times 2\sqrt{2} = (10 - 19.6)\sqrt{2} = -9.6\sqrt{2} \approx -13.6 \text{ m/s}.

The vertical velocity is negative, so the ball is falling as it crosses the line. It is on its way down and still below the bar, which agrees with Step 5.

Step 7. The rest of the flight. The greatest height is

H = \frac{u_y^2}{2g} = \frac{200}{19.6} \approx 10.2 \text{ m},

reached at $t = \dfrac{u_y}{g} = \dfrac{10\sqrt{2}}{9.8} \approx 1.44$ s. The range over level ground is

R = \frac{U^2 \sin 2\theta}{g} = \frac{400}{9.8} \approx 40.8 \text{ m}.

So the ball crosses the goal line at $0.8$ m and would land $40.8$ m from the kick, past the line.

Step 8. Check the model. The calculation treats the ball as a particle and ignores air resistance. Drag would reduce both the height and the range, so a real ball would be lower at the goal line. The conclusion that the ball passes under the bar is therefore safe for this kick. The same model would need adjusting for a strong wind or a spinning ball.

The chapter's answer. The ball reaches the goal line at $0.8$ m above the ground, below the $2.44$ m crossbar. The free kick does not go over the bar. It passes under the bar, so it can be a goal. The greatest height of the flight is about $10.2$ m, and the ball would land about $40.8$ m from where it was kicked.

Mixed practice — choose the method first

For each question, say which method you will use and why before you calculate. The items mix this chapter with earlier chapters, so the choice is not always obvious.

Convert $108$ km/h to m/s.
A force of $12$ N acts on a mass of $4$ kg. Find the acceleration.
A walker walks $700$ m east, then turns and walks $200$ m west. Taking east as positive, find the distance travelled and the displacement.
A boat moves $300$ m due north in $25$ s, then $100$ m due south in $15$ s. Find its average speed and its average velocity over the whole journey.
A velocity–time graph is a straight line from $(0, 6)$ to $(4, 18)$ , with $v$ in m/s. Find the acceleration and the displacement.
A velocity–time graph is a straight line from $(0, 10)$ to $(5, -10)$ , with $v$ in m/s. Find the displacement and the distance travelled.
A particle moves at $8$ m/s and accelerates at $3$ m/s $^2$ for $6$ s. Find its final velocity and its displacement.
A car travelling at $20$ m/s brakes at $4$ m/s $^2$ until it stops. Find the stopping distance.
A ball is thrown vertically upwards at $24.5$ m/s. Find the time to its greatest height and the greatest height. Take $g = 9.8$ m/s $^2$ .
The displacement of a particle is $s = t^3 - 6t^2 + 9t$ metres. Find its velocity and its acceleration.
A particle has acceleration $a = 12$ m/s $^2$ . Its velocity is $5$ m/s and its displacement is $0$ m when $t = 0$ . Find its velocity and its displacement as functions of time.
The displacement of a particle is $s = 8t - t^2$ metres for $0 \le t \le 8$ . Find the displacement and the distance travelled over the whole interval.
A particle is projected at $30$ m/s at $60^\circ$ to the horizontal. Find the horizontal and vertical components of its initial velocity.
A ball is projected at $30$ m/s at $30^\circ$ from level ground. Find its time of flight, its greatest height and its range. Take $g = 9.8$ m/s $^2$ .
A ball is kicked at $25$ m/s at $60^\circ$ towards a wall $25$ m away. Find the time it takes to reach the wall and its height then. Take $g = 9.8$ m/s $^2$ .

Answers and concise solutions

1. The speed in m/s

Answer: Multiply by $\dfrac{5}{18}$ : $108 \times \dfrac{5}{18} = 30$ m/s.

2. The acceleration

Answer: Rearrange $F = ma$ : $a = \dfrac{F}{m} = \dfrac{12}{4} = 3$ m/s $^2$ .

3. The distance and the displacement

Answer: Distance travelled $= 700 + 200 = 900$ m. Displacement $= (+700) + (-200) = +500$ m, which is $500$ m east.

4. The average speed and the average velocity

Answer: Total distance $= 400$ m over $40$ s, so the average speed is $10$ m/s. The displacement is $200$ m north, so the average velocity is $5$ m/s north. The speed is larger because the boat doubled back.

5. The acceleration and the displacement

Answer: The acceleration is the gradient: $\dfrac{18 - 6}{4} = 3$ m/s $^2$ . The displacement is the trapezium area: $\dfrac{6 + 18}{2} \times 4 = 48$ m.

6. The displacement and the distance

Answer: The line crosses the axis at $t = 2.5$ s. Each triangle has area $\frac{1}{2} \times 2.5 \times 10 = 12.5$ , so the displacement is $12.5 - 12.5 = 0$ m and the distance travelled is $12.5 + 12.5 = 25$ m.

7. The final velocity and the displacement

Answer: Use $v = u + at$ : $v = 8 + 3 \times 6 = 26$ m/s. Then $s = ut + \frac{1}{2}at^2 = 48 + 54 = 102$ m.

8. The stopping distance

Answer: First find the time: $0 = 20 - 4t$ , so $t = 5$ s. Then $s = ut + \frac{1}{2}at^2 = 100 - 50 = 50$ m. The formula $v^2 = u^2 + 2as$ gives the same distance directly.

9. The time and the greatest height

Answer: At the top $v = 0$ , so $t = \dfrac{u}{g} = \dfrac{24.5}{9.8} = 2.5$ s. The greatest height is $\dfrac{u^2}{2g} = \dfrac{600.25}{19.6} = 30.625$ m.

10. The velocity and the acceleration

Answer: Differentiate once: $v = 3t^2 - 12t + 9$ m/s. Differentiate again: $a = 6t - 12$ m/s $^2$ . Setting $v = 0$ gives $t = 1$ or $t = 3$ , where the particle is at rest.

11. The velocity and the displacement

Answer: Integrate the acceleration: $v = 12t + C$ . With $v = 5$ at $t = 0$ , $C = 5$ , so $v = 12t + 5$ m/s. Integrate again: $s = 6t^2 + 5t + K$ . With $s = 0$ at $t = 0$ , $K = 0$ , so $s = 6t^2 + 5t$ m.

12. The displacement and the distance

Answer: The particle is at rest at $t = 4$ s. At that time $s = 32 - 16 = 16$ m, and at $t = 8$ s $s = 64 - 64 = 0$ m. The displacement is $0$ m and the distance travelled is $16 + 16 = 32$ m.

13. The two components

Answer: Horizontal $= 30\cos 60^\circ = 15$ m/s. Vertical $= 30\sin 60^\circ = 15\sqrt{3} \approx 26.0$ m/s.

14. The time, the greatest height and the range

Answer: The vertical component is $u_y = 15$ m/s. The time of flight is $\dfrac{2u_y}{g} = \dfrac{30}{9.8} = \dfrac{150}{49} \approx 3.06$ s. The greatest height is $\dfrac{u_y^2}{2g} = \dfrac{225}{19.6} = \dfrac{1125}{98} \approx 11.5$ m. The range is $\dfrac{U^2 \sin 60^\circ}{g} = \dfrac{900 \times \frac{\sqrt{3}}{2}}{9.8} = \dfrac{2250\sqrt{3}}{49} \approx 79.5$ m.

15. The time and the height

Answer: The horizontal component is $u_x = 25\cos 60^\circ = 12.5$ m/s, so the time is $\dfrac{25}{12.5} = 2$ s. The height is $y = u_y t - \frac{1}{2}gt^2 = 25\sin 60^\circ \times 2 - 4.9 \times 4 = 25\sqrt{3} - 19.6 \approx 23.7$ m.

What you can now do

name the fundamental SI quantities and build derived units from them;
convert between units, including km/h and m/s;
use the language of position, displacement, distance, speed, velocity and acceleration;
read the gradient of a displacement–time graph and the gradient and area of a velocity–time graph;
choose and use the constant-acceleration formulae;
move between displacement, velocity and acceleration with calculus;
treat a projectile as two constant-acceleration motions sharing one time;
resolve an initial velocity, and find a time of flight, a greatest height, a range and a height at a given distance;
check a model and say what has been assumed.

Your relevance reflection

Return to the prompt in the chapter opener: something you have seen move in two directions at once. Write the two directions again, and say which one is steady and which one changes. Now add three things you could not have said before this chapter: how to resolve the starting velocity, how the two motions share one time, and how a graph or a formula gives the height at a chosen distance. Finish with one sentence about a moving object where the vertical motion decides what happens.