(a) Fig. 1.1 is a speed–time graph for the first 5 minutes of a bus journey.
Describe the motion between:
$t = 0.90\,\text{min}$ and $t = 2.9\,\text{min}$
$t = 2.9\,\text{min}$ and $t = 3.5\,\text{min}$
$t = 3.5\,\text{min}$ and $t = 4.5\,\text{min}$
[3]
Fig. 1.1 — Speed–Time Graph: Bus Journey (first 5 minutes)
On a speed–time graph, the shape of the line tells us the motion:
A horizontal line = constant speed (no acceleration)
A downward sloping line = decelerating (slowing down)
A line at v = 0 = stationary (not moving)
① $t = 0.90\,\text{min}$ to $t = 2.9\,\text{min}$ (blue region):
The bus moves at constant speed of 10 m/s. The line is horizontal — there is no acceleration or deceleration.
② $t = 2.9\,\text{min}$ to $t = 3.5\,\text{min}$ (orange region):
The bus decelerates uniformly. The speed falls in a straight line from 10 m/s down to 0 m/s.
③ $t = 3.5\,\text{min}$ to $t = 4.5\,\text{min}$ (green strip):
The bus is stationary (not moving). Speed = 0 m/s. The line rests on the time axis.
(b) Another bus travels at a speed of 8.9 m/s. The brakes apply a constant force and the bus stops in a distance of 23 m. This bus has a mass of 18 000 kg.
(i) Calculate the kinetic energy of the bus before the brakes are applied.
[2]
The kinetic energy before braking is 712 890 J (accept 710 000 J).
(b)(ii) Calculate the force applied to stop the bus.
[3]
The braking force does work on the bus to remove all its kinetic energy. Using the work–energy theorem:
Principle: Work done by brakes = kinetic energy of bus
$W = F \times d \quad \Rightarrow \quad F \times d = E_k$
Rearranging for F:
$F = \dfrac{E_k}{d}$
Given:
$E_k$ = 712 890 J (from part i)
stopping distance (d) = 23 m
Calculation:
$F = \dfrac{712\,890}{23}$
$\boxed{F = 30\,995 \; \text{N} \approx 31\,000 \; \text{N}}$
Both the cyclist and the runner start from rest and accelerate uniformly. Their lines are straight, which means constant acceleration. However, the cyclist’s line is steeper, so the cyclist has a greater acceleration. At $t = 6\,\text{s}$, the cyclist reaches 9 m/s while the runner reaches only 6 m/s.
Both accelerate uniformly. The cyclist has a greater acceleration (1.5 m/s²) than the runner (1.0 m/s²), shown by the steeper gradient of the cyclist’s line.
(b) Describe the motion of the cyclist between time $t = 6.0\,\text{s}$ and time $t = 12.0\,\text{s}$.
[1]
The cyclist moves at constant speed of 9 m/s. The line is horizontal between $t = 6\,\text{s}$ and $t = 12\,\text{s}$, meaning there is no acceleration.
(c) Calculate the total distance travelled by the cyclist between $t = 0$ and $t = 12.0\,\text{s}$.
[4]
On a speed–time graph, area under the line = distance travelled. The cyclist’s journey has two sections:
Area under cyclist’s graph = total distance travelled
(d) After the first 6.0 seconds, the runner moves at constant speed for 4.0 seconds. He then slows down uniformly and stops in a further 2.0 seconds.
On Fig. 1.1, complete the graph for the runner’s motion.
[2]
Fig. 1.1 — Completed: Runner’s full motion
Cyclist
Runner (given)
Runner (answer — draw these on your graph)
At $t = 6\,\text{s}$, the runner is travelling at 6 m/s (read from the graph). The two lines you must draw are:
Mark 1 — Constant speed section ($t = 6\,\text{s}$ to $t = 10\,\text{s}$):
Draw a horizontal straight line from the point $(6,\;6)$ to the point $(10,\;6)$. Speed stays at 6 m/s for 4 seconds.
Mark 2 — Uniform deceleration ($t = 10\,\text{s}$ to $t = 12\,\text{s}$):
Draw a straight line sloping downward from the point $(10,\;6)$ to the point $(12,\;0)$. The runner decelerates uniformly to rest.
✓ Mark 1: horizontal line at 6 m/s from t = 6 s to t = 10 s
✓ Mark 2: straight line with negative gradient from (10, 6) down to (12, 0)
