Speed/Distance time graphs

Ratio, proportion and rates of change

Questions marked Higher are aimed at the Higher tier.

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The speed-time graph shows a journey. Find the acceleration during the first $10$ seconds.

Speed-time graph for questions 1 to 5 and question 8

2 marks
For how long does the journey in question $1$ stay at a constant speed?

2 marks
Find the distance travelled while the journey in question $1$ is accelerating.

3 marks
Find the total distance travelled in the $40$ seconds shown in question $1$.

3 marks
Find the average speed for the whole $40$ seconds in question $1$.

4 marks
A second journey is a straight line on a speed-time graph from $(0, 0)$ to $(20, 10)$. Find the acceleration and the distance travelled.

4 marks
A car accelerates from rest at $2$ m/s$^{2}$ for $8$ seconds, then travels at that speed for $10$ seconds. Find the speed after $8$ seconds, the distance while accelerating, and the total distance. Higher

5 marks
After the $40$ seconds in question $1$, the traveller continues at a constant $8$ m/s for another $30$ seconds. Find the extra distance, the distance for the whole $70$ seconds, and the average speed over the $70$ seconds to $3$ significant figures. Higher

6 marks

Mark scheme

Answer. $1.2$ m/s$^{2}$
- Acceleration $= \dfrac{12}{10}$
- $1.2$ m/s$^{2}$
Answer. $20$ seconds
- The horizontal section runs from $10$ s to $30$ s.
- $20$ seconds
Answer. $60$ m
- The area is a triangle: $\dfrac{1}{2} \times 10 \times 12$.
- $60$ m
Answer. $360$ m
- Accelerating: $60$ m. Constant speed: $20 \times 12 = 240$ m. Slowing: $60$ m.
- Total $= 360$ m.
Answer. $9$ m/s
- Average speed $= \dfrac{\text{total distance}}{\text{total time}}$
- $\dfrac{360}{40} = 9$ m/s
Answer. $0.5$ m/s$^{2}$ and $100$ m
- Acceleration $= \dfrac{10}{20} = 0.5$ m/s$^{2}$.
- Distance $= \dfrac{1}{2} \times 20 \times 10 = 100$ m.
Answer. $16$ m/s, $64$ m, and $224$ m
- Speed $= 2 \times 8 = 16$ m/s.
- Accelerating distance $= \dfrac{1}{2} \times 8 \times 16 = 64$ m.
- Constant-speed distance $= 16 \times 10 = 160$ m, so the total is $224$ m.
Answer. Extra $240$ m, total $600$ m, average speed $8.57$ m/s
- Extra distance $= 8 \times 30 = 240$ m.
- Total distance $= 360 + 240 = 600$ m.
- Average speed $= \dfrac{600}{70} = 8.571\ldots = 8.57$ m/s to $3$ s.f.