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Answer to Question #2975 in Mechanics | Relativity for Prathamesh

Question #2975
A car starts from rest and accelerates uniformly for 10 s to a velocity of 8 m/s . It runs at a constant velocity and is finally brought at rest in 64 m with a constant retardation . the total distance covered by the car is 584 m . find the value of acceleration, retardation and total time taken.
Expert's answer
1. Using the equations of the accelerated motion we have:

v = a*t, s = (a*t2)/2
t = 10s v = 8m/s
a = v/t = 0.8 m/s2
s = 40 m

2. retardation moving

s = 64 m v = 8 m/s
using formula: r = v2/(2*s) =>
retardation r = 0.5 m/s2
v = t*r => t = v/r =16 s

3. A car runs with uniform velocity
v = 8 m/s a=0
s = 584 - 40 - 64 = 480m
t = s/v = 60 s

So, total time T = 10+16+60 = 86 s

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