question_answer 22)

Figure 3.92 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown, (a) In which interval is the average acceleration greatest in magnitude ? (b) In which interval is the average speed greatest ? (c) Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals, (d) What are the accelerations at the points A, B, C and D ?     Fig. 3.92  

Answer:

(a) As the change in speed is greatest in interval 2, so magnitude of average acceleration is greatest in interval 2. (b) Obviously from the graph, average speed is greatest in interval 3. (c) is positive in all three intervals, a is positive in the intervals 1 and 3 as speed is increasing in these intervals. But a is negative in interval 2 as speed is decreasing in this interval. (d) At the points A, B, C and D, the v-t graph is parallel to the time-axis or has a zero slope, so a = 0 at all these points.