A-level Physics (Advancing Physics)/Graphs/Worked Solutions

1. In the following distance-time graph, what is the velocity 4 seconds after the beginning of the object's journey?

At this point, the graph is a straight line. So:

$v=\frac{ds}{dt}=\frac{10-0}{6-0}=\frac{5}{3}\approx 1.67{\text{ms}}^{-1}$

2. What is the velocity at 12 seconds?

Approximately, there is a rough straight line at this point. The object is travelling towards home 2 ms⁻¹ each second (look at the previous second, for example). So, its velocity is -2 ms⁻¹.

3. In the following velocity-time graph, how far does the object travel between 7 and 9 seconds?

Distance travelled is equal to the area under the graph. Between 7 and 9 seconds, this is the shaded area of the graph. So, calculate the area of the triangle:

$A=\frac{bh}{2}=\frac{2\times 3}{2}=3\text{m}$

4. What is the object's acceleration at 8 seconds?

There is a straight line between 7 and 9 seconds which we can use to answer this question. The acceleration is equal to the gradient of the graph, so:

$a=\frac{3}{2}=1.5{\text{ms}}^{-2}$

5. A car travels at 10ms⁻¹ for 5 minutes in a straight line, and then returns to its original location over the next 4 minutes, travelling at a constant velocity. Draw a distance-time graph showing the distance the car has travelled from its original location.

10ms⁻¹ = 600 metres / minute.

6. Draw the velocity-time graph for the above situation.

The velocity from 0-5s is 10ms⁻¹. The velocity from 5-9s is:

$v=\frac{3\times 10^3}{4}=750\text{ metres/minute}=12.5{\text{ms}}^{-1}$

7. The velocity of a ball is related to the time since it was thrown by the equation $v=30-9.8t$. How far has the ball travelled after 2 seconds?

$s=\underset{t_1}{\overset{t_2}{\int }}f(t)dt=\underset{0}{\overset{2}{\int }}30-9.8tdt=[30t-4.9t^2{]}_0^2=30(2)-4.9(2^2)-30(0)+4.9(0^2)=60-19.6=40.4\text{ m}$