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Question
The given figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?
Solution
- The average acceleration's magnitude is determined by
= `"Change in speed"/"Time interval"`
The average acceleration over a brief period of time matches the v-t graph's slope during that specific interval. The v-t graph's slope is maximum during interval 2 when compared with intervals 1 and 3, therefore the greatest average acceleration in magnitude is observed in interval 2. - Interval 3 has the highest average speed since peak D reaches the maximum on the speed axis.
- v > 0 is positive throughout all three time intervals.
- The graph's slope is positive in intervals 1 and 3, which means a > 0; thus, acceleration is positive in these intervals. Conversely, the slope is negative in interval 2, indicating a negative acceleration. Hence, a > 0 signifies a positive acceleration in intervals 1 and 3, and a < 0 indicates a negative acceleration in interval 2.
- Since the slope is zero at points A, B, C, and D, the acceleration at these points is also zero.
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