Select Page

This is a suggested solution to the GCE O Level Physics examination, in year 2019, Paper 2, Section A, question 1.

The below velocity time graph of an athelete is given: (a) The student is asked to state a time during the race when the aceleration of the athelet is largest. In a velocity-time graph, acceleration is obtained by measuring the gradient of the graph. From the given graph, the section when the gradient is largest, ie when the graph is steepest, is the period from time t=0 to t=1.8s.

(b) It was stated in the question that the velocity of the athelete is always positive, but his acceleration is sometims negative. The student is required to explain this statement. The suggested answer to this explanation is :

The acceleration is measured by the gradient of the velocity-time graph. When the athlete is slowing down, i.e. reducing his velocity, the gradient of the velocity-time graph will be negative, i.e. the acceleration becomes negative. As shown in the section of the graph from time t=8s onwards, the gradient of the graph is negative, indicating that acceleration is negative although the velocity is still positive because the athlete is still running in the same direction but with reducing speed.

(c) The student is required to show how by using the above given graph to show that the race is 100m long. The distance can be computed from a velocity-time graph by calculating the area under the graph.

(d) For this final part of the question, it is required to determine the average speed of the athlete. To compute the average speed, we use the formula: From the velocity-time graph, the total time taken for the whole journey is 9.8s.

Hence, with the total distance of 100m given in part (c) and using the  above formula,

average speed of athlete = 100m/9.8s =10.2 m/s