Asked by ronit
a car starting from rest acquired a speed of 25ms-¹ in 10sec,after which it maintain this speed for 10sec. find (a)the acceleration (b) distance traveled during acceleration (c)total distance traveled?
Answers
Answered by
Steve
acceleration is (25m/s)/(10s) = 2.5 m/s^2
distance during acceleration: s = 1/2 at^2 = 125m
so, total distance is 125 + 10*25 = 375 m
distance during acceleration: s = 1/2 at^2 = 125m
so, total distance is 125 + 10*25 = 375 m
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