GCE “O” Level November 2015 Paper 2 Section B Question 10(a)(b)(c)

This is a suggested solution to the GCE O Level Physics examination question in 2015 November, Paper 2, Section B, question 10.
For this question, the velocities of a skier sliding down a slope is marked on a given graph paper for time t=0, t=2, t=4 and t=8.
Given graph- N2015-p2-B-10
(a) The student is required to draw a line in the given graph to complete the velocity-time graph from t=0 to t=10s. There are few important things to note here:
    1. The graph must be drawn to the point where t=10s as this is stated in the question even though the “x” marked in the given graph stopped at t=8. Some books provided the solution with graph ending at t=8. This is not correct.
    2. It is given in the question that the acceleration between A (t=0) and C (t=4.0s) is uniform. This means that the graph must be a straight line between A and C whereas the graph between C and D (t=8s) should not be straight because the acceleration is not uniform.
    3. It is also stated in the question that after D (t=8s), the velocity is constant. Constant velocity means the velocity remains unchanged after D, and hence should be a horizontal straight line.

The graph should therefore look like this:

Suggested solution N2015-P2-B-a

(b) The second part of the question require the student to calculate the distance between A and B as well as between B and C.
(i)  The distance between point A (t=0) and point B (t=2s) is given by the area under the graph from time t=0 to t=2s. Hence,
Distance between point A and B =( 1/2) x (2 s)x  (3 m/s) = 3 m
(ii)  The distance between point B (t=2s) and point C (t=4s) is given by the area under the graph from time t=2 to t=4s. Hence,
Distance between point A and B =( 1/2) x (2 s)x  (3 m/s + 6 m/s) = 9 m
(c)  (i) In the third part of the question, it is given that the distance between point A and D is 44m and that the displacement of the skier at point A is zero. The student is required to draw a displacement-time graph for the motion. By using the data computed in (b) (i) and (b) (ii) above, we can plot the displacement-time graph using the following values:
t=0, displacement = 0 (given)
t=2s, displacement = 3m (computed in b(i))
t=4s, displacement = 9m (computed in b(ii))
t=8s, displacement = 44m (given)
Hence, the graph should look like this:
(c) (ii) In this part of the question, the student is required to state how the displacement-time graph shows the velocity is constant. The suggested answer should be:
In  a displacement-time graph, the gradient of the graph at any point gives the instantaneous velocity at that point. Hence, when the velocity is constant,the graph will be a straight line. Hence, in the above displacement-time graph, the graph is a straight line after t=8s (point D).
Note that this is a suggested solution and not a solution provided by Cambridge. It is a suggested solution to help students and we hold not responsibilities for any error. If you spot any error, kindly drop us a gentle word so we may correct it and it will benefit other students.
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