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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 |
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Answer» (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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