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A particle moves on straight line accord...

A particle moves on straight line according to the velocity-time graph shown in fig. Calculate -

(i) Total distance covered
(ii) Average speed
(iii) In which part of the graph the acceleration is maximum and also find its value.
(iv) Retardation

Text Solution

Verified by Experts

(i) Total distance covered = Area under v - t curve
` " "= (1)/(2) (2xx2) + 2(4-2) + (1)/(2) (10+2) xx 1`
`(1)/(2) xx 5 xx 10 = 37` m
(ii) Average speed = `("Total distance")/("Total time") = (37)/(10) = 3.7 m//s`
(iii) Acceleration = Slope of v-t curve
So maximum acceleration will be in the part where the slope will be maximum i.e. BC
`a_(max) = a_(BC) = (10-2)/(1) = 8m//s^(2)`
(iv) Retardation = Slope of CD (As it is negative)
`" " = |(0-10)/(5)| = |-2| = 2m//s^(2)`
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