Motion in a Straight Line
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Question : 22
Total: 27
Figure gives a speed-time graph of a particle in one dimensional 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:
(i) The magnitude of the average acceleration is given by
=
i.e. average acceleration in a small interval of time is equal to the slope of(v-t) graph in that time interval.
As the slope of (v-t) graph is maximum in the interval 2 as compared tointervals 1 and 3, hence the magnitude of average acceleration is greatestin interval 2 .
(ii) The average speed is greatest in the interval 3 as peak D is at maximumon speed axis.
(iii) v > 0 i.e. positive in all the three intervals.
(iv) The slope is positive in intervals 1 and 3, so ‘a’ i.e. acceleration ispositive in these intervals while the slope is negative in interval 2, soacceleration is negative in it. So, a > 0 i.e. positive in intervals 1 and 3 anda < 0 i.e. negative in interval 2.
(v) As slope is zero at points A, B, C and D, so the acceleration is zero atall the four points.
i.e. average acceleration in a small interval of time is equal to the slope of(v-t) graph in that time interval.
As the slope of (v-t) graph is maximum in the interval 2 as compared tointervals 1 and 3, hence the magnitude of average acceleration is greatestin interval 2 .
(ii) The average speed is greatest in the interval 3 as peak D is at maximumon speed axis.
(iii) v > 0 i.e. positive in all the three intervals.
(iv) The slope is positive in intervals 1 and 3, so ‘a’ i.e. acceleration ispositive in these intervals while the slope is negative in interval 2, soacceleration is negative in it. So, a > 0 i.e. positive in intervals 1 and 3 anda < 0 i.e. negative in interval 2.
(v) As slope is zero at points A, B, C and D, so the acceleration is zero atall the four points.
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