Chapter 3: Review

The chapter problem, solved

The bus and its passengers have a mass of 12000 kg. The bus travels at 12 m/s. The driver's reaction time is 0.60 s. Take the dry braking distance as 9.0 m and the wet braking distance as 14 m.

1. The acceleration from rest to 12 m/s in 8.0 s.

\begin{aligned} a &= \frac{\Delta v}{t} \\ &= \frac{12 - 0}{8.0} \\ &= 1.5\,\text{m/s}^2 \end{aligned}

2. The resultant force that gives this acceleration.

\begin{aligned} F &= m a \\ &= 12000 \times 1.5 \\ &= 18000\,\text{N} \\ &= 1.8 \times 10^4\,\text{N} \end{aligned}

3. The thinking distance.

\begin{aligned} s &= v t \\ &= 12 \times 0.60 \\ &= 7.2\,\text{m} \end{aligned}

4. The stopping distance on a dry road.

\text{stopping distance} = 7.2 + 9.0 = 16.2\,\text{m}

The hazard is 20 m ahead. The bus stops after 16.2 m, so it stops before the hazard with 3.8 m to spare.

5. The deceleration while braking on a dry road. The bus slows from 12 m/s to rest in 9.0 m.

\begin{aligned} v^2 - u^2 &= 2 a s \\ 0^2 - 12^2 &= 2 \times a \times 9.0 \\ -144 &= 18 a \\ a &= -8.0\,\text{m/s}^2 \end{aligned}

A deceleration of $8.0\,\text{m/s}^2$ is large. It is close to the acceleration of a falling object, which is why the brakes become hot and why the passengers need restraints.

6. The stopping distance on a wet road. The braking distance is 14 m.

\text{stopping distance} = 7.2 + 14 = 21.2\,\text{m}

The hazard is 20 m ahead, so on a wet road the bus does not stop in time. It is still moving when it reaches the hazard.

7. The braking force on a dry road.

\begin{aligned} F &= m a \\ &= 12000 \times 8.0 \\ &= 96000\,\text{N} \\ &= 9.6 \times 10^4\,\text{N} \end{aligned}

This is the force that produces the large deceleration of $8.0\,\text{m/s}^2$ .

8. The collision with the stationary van. The bus has a momentum of $12000 \times 12 = 144000\,\text{kg m/s}$ . After the collision the bus and the van move together.

\begin{aligned} 144000 &= (12000 + 8000) \times v \\ 144000 &= 20000 v \\ v &= 7.2\,\text{m/s} \end{aligned}

The two vehicles move off together at 7.2 m/s.

The judgement that the bus is safe on a dry road but not on a wet road is a reasoned answer. It follows from the numbers above. It is not the kind of claim a computer can check, so compare your reasoning with the evidence in the working.

Mixed practice — choose the method first

For each question, say which equation or idea you will use before you calculate. Keep your units with your numbers.

In the chapter problem the bus accelerates from rest to 12 m/s in 8.0 s. Calculate its acceleration.
The bus and its passengers have a mass of 12000 kg and accelerate at 1.5 m/s². Calculate the resultant force on the bus.
The bus travels at 12 m/s and the driver reacts in 0.60 s. Calculate the thinking distance.
On a dry road the thinking distance is 7.2 m and the braking distance is 9.0 m. Calculate the stopping distance.
On a dry road the bus slows from 12 m/s to rest in a braking distance of 9.0 m. Calculate the deceleration.
H The bus has a mass of 12000 kg and decelerates at 8.0 m/s². Calculate the braking force.
On a wet road the thinking distance is 7.2 m and the braking distance is 14 m. Calculate the stopping distance, and say whether the bus stops before a hazard 20 m ahead.
H The bus has a mass of 12000 kg and travels at 12 m/s. Calculate its momentum.
H The bus of mass 12000 kg travelling at 12 m/s hits the stationary van of mass 8000 kg. The two move off together. Calculate their common velocity.
A car of mass 1200 kg travels at 20 m/s. Calculate the energy in its kinetic store.
A train travels 3600 m in 120 s. Calculate its average speed.
A force of 250 N moves a crate 12 m along the line of the force. Calculate the work done.
A passenger has a mass of 80 kg. Take $g = 9.8\,\text{N/kg}$ . Calculate her weight.
H A ball of mass 0.50 kg moves at 12 m/s. Calculate its momentum.
A car brakes to a stop. Say which equation you would use to find the deceleration, and which equation you would then use to find the braking force. Name the quantities you need for each.
Explain why an object falling through a fluid reaches a terminal velocity.

Answers and concise solutions

1. Acceleration of the bus

Answer: $a = \dfrac{\Delta v}{t} = \dfrac{12 - 0}{8.0} = 1.5\,\text{m/s}^2$ .

2. Resultant force on the bus

Answer: $F = ma = 12000 \times 1.5 = 18000\,\text{N} = 1.8 \times 10^4\,\text{N}$ .

3. Thinking distance of the bus

Answer: $s = vt = 12 \times 0.60 = 7.2\,\text{m}$ .

4. Stopping distance on a dry road

Answer: Stopping distance $= 7.2 + 9.0 = 16.2\,\text{m} = 1.6 \times 10^1\,\text{m}$ . The hazard is 20 m ahead, so the bus stops in time.

5. Deceleration of the bus

Answer: $v^2 - u^2 = 2as$ with $v = 0$ , so $a = \dfrac{12^2}{2 \times 9.0} = 8.0\,\text{m/s}^2$ . The deceleration is $8.0\,\text{m/s}^2$ .

6. Braking force on the bus

Answer: $F = ma = 12000 \times 8.0 = 96000\,\text{N} = 9.6 \times 10^4\,\text{N}$ . A deceleration of $8.0\,\text{m/s}^2$ is large.

7. Stopping distance on a wet road

Answer: Stopping distance $= 7.2 + 14 = 21.2\,\text{m} = 21\,\text{m}$ to 2 significant figures. This is more than 20 m, so on a wet road the bus does not stop in time.

8. Momentum of the bus

Answer: $p = mv = 12000 \times 12 = 144000\,\text{kg m/s} = 1.4 \times 10^5\,\text{kg m/s}$ .

9. Common velocity of the bus and van

Answer: Total momentum before $= 12000 \times 12 = 144000\,\text{kg m/s}$ . Total momentum after $= 20000 \times v$ , so $v = \dfrac{144000}{20000} = 7.2\,\text{m/s}$ .

10. Kinetic energy of the car

Answer: $E_k = \tfrac{1}{2}mv^2 = 0.5 \times 1200 \times 20^2 = 240000\,\text{J} = 2.4 \times 10^5\,\text{J}$ .

11. Average speed of the train

Answer: $v = \dfrac{s}{t} = \dfrac{3600}{120} = 30\,\text{m/s} = 3.0 \times 10^1\,\text{m/s}$ .

12. Work done on the crate

Answer: $W = Fs = 250 \times 12 = 3000\,\text{J} = 3.0 \times 10^3\,\text{J}$ .

13. Weight of the passenger

Answer: $W = mg = 80 \times 9.8 = 784\,\text{N} = 7.8 \times 10^2\,\text{N}$ .

14. Momentum of the ball

Answer: $p = mv = 0.50 \times 12 = 6.0\,\text{kg m/s}$ .

15. Choosing the equation for each step

For the deceleration, use $v^2 - u^2 = 2as$ with the final velocity zero. I need the initial velocity and the braking distance.
For the braking force, use $F = ma$ . I need the mass of the car and the deceleration I have just found.

16. Reaching terminal velocity

At first the weight is larger than the drag, so the object accelerates. As the object gets faster, the drag force increases.
Eventually the drag force equals the weight, so the resultant force is zero.
A zero resultant force means the velocity stops changing, so the object falls at a steady terminal velocity.

What you can now do

tell distance from displacement, and speed from velocity, and recall typical speeds;
calculate speed, distance, time and acceleration;
read speed and acceleration from motion graphs, and find the distance from the area under a velocity-time graph (Higher tier);
use $v^2 - u^2 = 2as$ and describe terminal velocity;
apply Newton's three laws of motion and use $F = ma$ ;
work out thinking distance, braking distance and stopping distance, and explain what affects each one;
explain the dangers of large decelerations and estimate the force involved (Higher tier);
calculate momentum, use conservation of momentum, and explain how safety features reduce the force in a crash (Higher tier).

Your relevance reflection

Return to the prompt from the chapter opener: where in your own week do you notice speed, or the time something takes to stop? Write one more example now that you have finished the chapter. Has the chapter changed what you notice about a journey, a brake or a safety feature?