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

Explanation

(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)`