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

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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:

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

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:

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

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:

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

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:

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

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?

$s = \int_{t_1}^{t_2} f(t)\, dt = \int_{0}^{2} 30 - 9.8t\, dt = [30t - 4.9t^2]_{0}^{2} = 30(2) - 4.9(2^2) - 30(0) + 4.9(0^2) = 60 - 19.6 = 40.4\mbox{ m}$