Section A — Recall Questions 1–5
1.
State what is meant by a travel graph (distance-time graph).
Answer
A travel graph (distance-time graph) is a graph that shows the distance travelled from a starting point, plotted against time.
2.
Write the formula used to find the speed from a distance-time graph, in terms of the change in distance and the change in time.
Answer
$$\text{speed} = \frac{\text{change in distance}}{\text{change in time}}$$
3.
State what a horizontal (flat) section of a distance-time graph tells you about the object’s motion.
Answer
It means the object is stationary (not moving).
4.
State what it means when the line on a distance-time graph slopes back down towards the time axis.
Answer
It means the object is returning towards the starting point.
5.
State what is meant by a conversion graph.
Answer
A conversion graph is a straight-line graph used to change a value from one unit or currency into another.
Section B — Application Questions 6–10
6.
A car travels 150 km in 3 hours at a constant speed. Calculate its speed in km/h.
Answer
- Formula: $\text{speed} = \dfrac{\text{distance}}{\text{time}}$
- Given: distance $= 150\text{ km}$, time $= 3\text{ h}$
- Substitute: $150 \div 3$
- Calculate: $150 \div 3 = 50$
50 km/h
7.
The graph below shows a bus’s journey from the depot to town and back. Use the graph to find how far the bus was from the depot at $t = 75$ minutes.
Answer
Reading the graph at t = 75 minutes: draw a line up to the graph, then across to the distance axis
- Find 75 minutes on the time axis
- Draw a line up to the graph line, then across to the distance axis
12 km
8.
A ferry crosses a channel of 18 km in 45 minutes at a constant speed. Calculate its speed in km/h.
Answer
- Formula: $\text{speed} = \dfrac{\text{distance}}{\text{time}}$
- Given: distance $= 18\text{ km}$, time $= 45$ minutes
- Unit conversion: $45$ minutes $= \dfrac{45}{60}\text{ h} = \dfrac{3}{4}\text{ h}$
- Substitute: $18 \div \dfrac{3}{4} = 18 \times \dfrac{4}{3}$
- Calculate (cross-cancel 18 and 3 by 3): $\dfrac{\cancelto{6}{18}}{1} \times \dfrac{4}{\cancelto{1}{3}} = 6 \times 4 = 24$
24 km/h
9.
The graph below is a conversion graph between British pounds (£) and euros (€). Use the graph to convert £24 into euros.
Answer
Reading the graph: £24 up to the line, then across to the euros axis, gives €30
- Find £24 on the pounds axis
- Draw a line up to the graph line, then across to the euros axis
€30
10.
Two sections of a travel graph have gradients of 20 km/h and 35 km/h. State which section represents the faster speed, and explain your answer using the word gradient.
Answer
The section with a gradient of 35 km/h represents the faster speed, because a steeper gradient means a greater speed.
Section C — Challenge Questions 11–15
11.
The graph shows a hiker’s walk from a camp to a lake and back.
(a)
State what happened between hour 2 and hour 3.(b)
Calculate the hiker’s speed on the walk to the lake (0 to 2 hours).(c)
Calculate the hiker’s speed on the return journey (3 to 7 hours), and explain why this speed is different from the outward journey.A hiker’s walk from camp to a lake and back
Answer
(a)
The hiker was stationary (resting at the lake) — the distance from camp stayed at 8 km.
(b) Outward speed
- Formula: $\text{speed} = \dfrac{\text{distance}}{\text{time}}$
- Given: distance $= 8\text{ km}$, time $= 2\text{ h}$
- Substitute and calculate: $8 \div 2 = 4$
4 km/h
(c) Return speed
- Given: distance $= 8\text{ km}$, time $= 7 – 3 = 4\text{ h}$
- Substitute and calculate: $8 \div 4 = 2$
2 km/h
Explanation: the return speed (2 km/h) is slower than the outward speed (4 km/h) because the return section of the graph is less steep — a smaller gradient means a slower speed.
12.
The table shows a student’s distance from home during a bike ride.
| Time (minutes) | 0 | 15 | 30 | 45 | 60 |
|---|---|---|---|---|---|
| Distance from home (km) | 0 | 4 | 4 | 4 | 10 |
(a)
Draw the distance-time graph for this data on the axes provided.(b)
Calculate the speed for the first 15 minutes.(c)
Describe what is happening between $t=15$ minutes and $t=45$ minutes.Answer
(a) Completed graph
The completed distance-time graph, plotted from the data table
(b) Speed for the first 15 minutes
- Formula: $\text{speed} = \dfrac{\text{distance}}{\text{time}}$
- Given: distance $= 4\text{ km}$, time $= 15$ minutes
- Unit conversion: $15$ minutes $= \dfrac{15}{60}\text{ h} = 0.25\text{ h}$
- Substitute and calculate: $4 \div 0.25 = 16$
16 km/h
(c) Between t = 15 and t = 45 minutes
The student is stationary — the distance from home stays at 4 km, so they are not moving during this time.
13.
The graph below is a conversion graph between US dollars (\$) and British pounds (£).
(a)
Use the graph to convert \$50 into pounds.(b)
Use the graph to convert £60 into dollars.(c)
Explain why the same graph can be used to convert in both directions.Reading the graph both ways: \$50 up to £40 (dashed), and £60 across then down to \$75 (dashed)
Answer
(a) \$50 to pounds
- Find \$50 on the dollars axis
- Draw a line up to the graph line, then across to the pounds axis
£40
(b) £60 to dollars
- Find £60 on the pounds axis
- Draw a line across to the graph line, then down to the dollars axis
\$75
(c) Explanation
Because a conversion graph is a straight line, the gradient (rate) is the same everywhere on the line. This means you can start from either axis and use the same line to convert in either direction.
14.
A delivery driver leaves the depot and drives 30 km to a warehouse at a steady speed, taking 30 minutes. He unloads for 20 minutes. He then drives back to the depot at a steady speed, taking 40 minutes.
(a)
Draw the distance-time graph for this journey on the axes provided.(b)
Calculate his speed on the outward journey.(c)
Calculate his speed on the return journey.Answer
(a) Completed graph
The completed distance-time graph for the delivery driver’s journey
(b) Outward speed
- Formula: $\text{speed} = \dfrac{\text{distance}}{\text{time}}$
- Given: distance $= 30\text{ km}$, time $= 30$ minutes $= 0.5\text{ h}$
- Substitute and calculate: $30 \div 0.5 = 60$
60 km/h
(c) Return speed
- Given: distance $= 30\text{ km}$, time $= 40$ minutes
- Unit conversion: $40$ minutes $= \dfrac{40}{60}\text{ h} = \dfrac{2}{3}\text{ h}$
- Substitute: $30 \div \dfrac{2}{3} = 30 \times \dfrac{3}{2}$
- Calculate (cross-cancel 30 and 2 by 2): $\dfrac{\cancelto{15}{30}}{1} \times \dfrac{3}{\cancelto{1}{2}} = 15 \times 3 = 45$
45 km/h
15.
The graph shows three sections, P, Q and R, of a delivery van’s journey.
(a)
State which section shows the van travelling fastest, giving a reason based on the graph.(b)
State which section shows the van stationary.(c)
Calculate the speed for each of sections P, Q and R, to confirm your answer to part (a).Three sections, P, Q and R, of a delivery van’s journey
Answer
(a)
Section R — it has the steepest line (steepest gradient) on the graph, which means the greatest speed.
(b)
Section Q — the line is flat (horizontal), so the distance is not changing.
(c) Speed of each section
- Section P: distance $= 20\text{ km}$, time $= 20$ minutes $= \dfrac{1}{3}\text{ h}$. Speed $= 20 \div \dfrac{1}{3} = 20 \times 3 = 60$ km/h
- Section Q: distance $= 0\text{ km}$, time $= 10$ minutes. Speed $= 0$ km/h
- Section R: distance $= 60 – 20 = 40\text{ km}$, time $= 30$ minutes $= 0.5\text{ h}$. Speed $= 40 \div 0.5 = 80$ km/h
P = 60 km/h, Q = 0 km/h, R = 80 km/h
Check: R has the greatest speed (80 km/h), confirming the answer to part (a).
