(a) In this part of the question, the student is required to calculate the acceleration of the car as its speed increases. From the graph, the section where the car’s speed was increasing is as shown below:

Hence,

acceleration of the car = v-u / time taken

= (25 -20 )/4

= 1.25 m/s^{2}

(b) In part b of the question, the student is required to describe what the shaded area of the graph represents. As this is a speed-time graph, the area under the graph represents the distance travelled. Hence the suggested answer should be as follows:

The shaded area of the graph represents the distance travelled by the car for the duration from time=10s to time=16s.

(c) (i) The student is required to explain what is meant by *average speed* of the car. The answer is:

The average speed for the car is measured by the total distance divided by the total time.

(c) (ii) In this part of the question, the student is required to suggest a possible reason why the average seed of the car between time=6s and time=16s is larger than that between time=0 and time=20s.

The suggested answer is as follows:

From time t=6s and time t=16s, the duration is 10s. From time t=0 to t=20s, the duration is 20s, i.e.

the length of time from t=6s to t=16s. Since the formula of average speed is:twiceAverage speed = Total Distance/ Total time taken

For the average speed for the period t=6s to t=16s to be more than the period t=0 to t=20s, the total distance travelled during the period t=6s to t=16s must be

more thanhalfthe distance travelled during the entire period of t=0 to t=20s/In a speed-time graph, the total distance is found by the area under the graph. As observed from the given speed-time graph, the

area under the graphfrom t=6s to t=16s isunder the graph from t=0 to t=20s. Hence, the average speed which from t=6s to t=16x is larger than the average speed for t=0 to t=20s.more than half the area