# 2.2 Motion (II)

F.4 Physics · Topic II Force and Motion · Lesson 2

Reading motion graphs, the suvat equations, stopping distance, free fall, and how motion is recorded.

## Figure

A stone thrown vertically upwards, upward positive: the three graphs of one motion. At the top v = 0, but a = −g throughout.

## Key points

- **s-t graph**: **slope = velocity** (at an instant: the slope there). Horizontal line = **at rest**; straight line = **constant velocity**; curve = **velocity changing**. The area under it means **nothing**.
- **v-t graph**: **slope = acceleration**; **area = displacement** (area below the t-axis is **negative**). Horizontal line = constant velocity.
- **a-t graph**: **area = change in velocity**. Chain: s-t → (slope) v-t → (slope) a-t, and back by areas.
- **suvat ／ 勻加速運動方程**: $v = u + at$, $s = \frac{u+v}{2}t$, $s = ut + \frac{1}{2}at^{2}$, $v^{2} = u^{2} + 2as$. Only for a **straight line** with **constant** $a$; any 3 of $s, u, v, a, t$ give the other two.
- **Stopping distance** = **thinking distance ／ 思考距離** ($s_1 = ut_1$: constant speed during the reaction time) + **braking distance ／ 剎車距離** ($s_2 = u^{2}/2a$). Doubling $u$ **doubles** $s_1$ but makes $s_2$ **4 times** as long.
- **Free fall ／ 自由下落**: gravity only; $a = g = 9.81$ m s⁻² (about 10) **vertically downwards**, the same for all masses. At the highest point $v = 0$ but $a$ is still $g$. Pick one positive direction: upward positive means $a = -g$.
- **Bouncing ball**: $a = g$ downward on the way down and up; a large **upward** acceleration during the short impact; with energy loss each bounce is lower. **Smooth inclined plane**: $a = g\sin\theta$ down the slope.
- **Recording motion**: **ticker-tape timer ／ 打點計時器** 50 Hz, so 1 tick = **0.02 s** (equal gaps = uniform velocity, growing gaps = acceleration); **stroboscope** images $T = 1/f$ apart; **motion sensor** sends ultrasound and times the echoes, motion **away** from it positive.

## Must know

| Graph | Slope gives | Area under the graph gives | A horizontal line means |
|---|---|---|---|
| **s-t graph** | velocity | nothing (no physical meaning) | at rest |
| **v-t graph** | acceleration | displacement (area below the t-axis is negative) | constant velocity |
| **a-t graph** | – | change in velocity | constant acceleration |

Free fall with **downward positive**: $v = u + gt$, $s = ut + \frac{1}{2}gt^{2}$; with **upward positive**: $v = u - gt$, $s = ut - \frac{1}{2}gt^{2}$.

## Common mistakes

- ✗ "Area under the s-t graph = distance travelled." ✓ The area under an **s-t** graph means nothing. The area under a **v-t** graph is the displacement, and area below the t-axis counts as **negative**.
- ✗ One suvat equation for the whole stop, or the whole bounce. ✓ suvat only while $a$ is **constant**: split the journey into parts (constant speed during the reaction time, then braking; falling, then rising).
- ✗ "At the highest point $v = 0$, so $a = 0$." ✓ $a = g$ **downwards** the whole time; only $v$ is zero, for an instant.
- ✗ "$a < 0$ means the body is slowing down." ✓ A body moving in the **negative** direction with $a < 0$ is **speeding up**. Slowing down means $a$ and $v$ have **opposite** signs.
- ✗ Upward positive, yet $s = 20 \times 3 + \frac{1}{2} \times 10 \times 3^{2} = 105$ m. ✓ Keep one positive direction: upward positive makes $a = -10$ m s⁻², so $s = 60 - 45 = 15$ m. 105 m is impossible: the stone rises only $u^{2}/2g = 20$ m.

## Quick check

1. A car travelling at 5 m s⁻¹ accelerates at 2 m s⁻². Find its displacement after 10 s, and its speed after travelling 100 m.
2. A car at 50 km h⁻¹ can be braked to a stop in 15 m. With the same deceleration, what braking distance does it need from 70 km h⁻¹?
3. A stone is thrown vertically upwards at 20 m s⁻¹ (take g = 10 m s⁻²). Find the maximum height reached and the velocity after 3 s.
4. A car decelerates uniformly from 30 m s⁻¹ to 20 m s⁻¹ over 100 m. How much further does it travel before coming to rest?
5. A ball is released from rest at the top of a smooth slope 3 m long inclined at 30° (g = 10 m s⁻²). How long does it take to reach the bottom?

## Answers

1. $s = 5 \times 10 + \frac{1}{2} \times 2 \times 10^{2} = 150$ m; $v^{2} = 5^{2} + 2 \times 2 \times 100 = 425$, $v = 20.6$ m s⁻¹.
2. Braking distance $\propto u^{2}$: $15 \times (70/50)^{2} = 29.4$ m.
3. Upward positive, $a = -10$: $H = u^{2}/2g = 20^{2}/20 = 20$ m; $v = 20 - 10 \times 3 = -10$, i.e. 10 m s⁻¹ downwards.
4. $a = (20^{2} - 30^{2})/(2 \times 100) = -2.5$ m s⁻²; further distance $= 20^{2}/(2 \times 2.5) = 80$ m.
5. $a = g\sin 30° = 5$ m s⁻²; $3 = \frac{1}{2} \times 5 \times t^{2}$, so $t = 1.10$ s.

## Did you know?

In 1971, astronaut David Scott dropped a hammer and a feather at the same time on the Moon, where there is no air. They landed together, just as Galileo had predicted.