3.3 Safety in public transport — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

A car of mass 800 kg accelerates at $2.5\,\text{m/s}^2$ . Calculate the resultant force.
A cyclist travels 150 m in 20 s. Calculate her average speed.

Answers

$F = ma = 800 \times 2.5 = 2000\,\text{N} = 2.0 \times 10^3\,\text{N}$ .
$v = \dfrac{s}{t} = \dfrac{150}{20} = 7.5\,\text{m/s}$ .

If question 1 went wrong, look again at Section 3.2: multiply the mass by the acceleration. If question 2 went wrong, divide the distance by the time.

Practise with help

1. A car travels at 18 m/s. The driver has a reaction time of 0.50 s. Calculate the thinking distance.

Hint

During the reaction time the car keeps moving at its original speed.

Next step

Multiply the speed by the reaction time: $s = vt$ .

Full working

\begin{aligned} s &= v t \\ &= 18 \times 0.50 \\ &= 9.0\,\text{m} \end{aligned}

2. A car travels at 20 m/s. The driver's reaction time is 0.60 s and the braking distance is 35 m. Calculate the thinking distance and the stopping distance.

Hint

The stopping distance is the thinking distance plus the braking distance. Find the thinking distance first.

Next step

Work out $20 \times 0.60$ for the thinking distance, then add the 35 m braking distance.

Full working

Thinking distance:

s = v t = 20 \times 0.60 = 12\,\text{m}

Stopping distance:

\text{stopping distance} = 12 + 35 = 47\,\text{m}

3. A car travels at 12 m/s and stops in a distance of 18 m. Calculate the deceleration.

Hint

The car ends at rest, so the final velocity is zero.

Next step

Write $v^2 - u^2 = 2as$ , then put in $u = 12$ , $v = 0$ and $s = 18$ .

Full working

\begin{aligned} v^2 - u^2 &= 2 a s \\ 0^2 - 12^2 &= 2 \times a \times 18 \\ -144 &= 36 a \\ a &= \frac{-144}{36} \\ &= -4.0\,\text{m/s}^2 \end{aligned}

The deceleration is $4.0\,\text{m/s}^2$ .

4. A car of mass 900 kg travels at 18 m/s and stops in a distance of 27 m. Calculate the deceleration and the braking force.

Hint

Find the deceleration first, then use $F = ma$ with the mass in kilograms.

Next step

Use $v^2 - u^2 = 2as$ with $u = 18$ , $v = 0$ and $s = 27$ . Then multiply the mass by the size of the deceleration.

Full working

\begin{aligned} v^2 - u^2 &= 2 a s \\ 0^2 - 18^2 &= 2 \times a \times 27 \\ -324 &= 54 a \\ a &= -6.0\,\text{m/s}^2 \end{aligned}

The deceleration is $6.0\,\text{m/s}^2$ . The braking force is

F = m a = 900 \times 6.0 = 5400\,\text{N} = 5.4 \times 10^3\,\text{N}

5. Explain why a wet road increases the braking distance of a car.

Hint

Think about the grip between the tyres and the road, and about the braking force the tyres can transmit.

Next step

Say what happens to the deceleration when the grip is reduced, and what a smaller deceleration means for the distance travelled while stopping.

Full working

On a wet road there is less grip between the tyres and the road, so the tyres can transmit a smaller braking force. A smaller braking force gives a smaller deceleration. The car takes longer to stop, and it travels further while it is stopping, so the braking distance is longer. This answer is a model answer for you to compare with your own wording.

6. Explain why a large deceleration is dangerous for the brakes and for the control of a vehicle.

Hint

Think about where the kinetic energy goes, and about what happens if the braking force is larger than the grip of the tyres.

Next step

Name the store that receives the energy, and say what happens to the wheels if they lock.

Full working

A large deceleration transfers a large amount of kinetic energy to the thermal store of the brakes in a short time, so the brakes may overheat and work less well. If the braking force is too large for the grip of the tyres, the wheels lock and the vehicle skids. A skidding vehicle cannot be steered, so the driver loses control. This answer is a model answer for you to compare with your own wording.