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*t**^{2})/2

t = 10s v = 8m/s

a = v/t = 0.8 m/s^{2}

s = 40 m

2. retardation moving

**s = 64 m v = 8 m/s**

using formula: r = v^{2}/(2*s) =>

retardation r = 0.5 m/s^{2}

** 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**

t = 10s v = 8m/s

a = v/t = 0.8 m/s

s = 40 m

using formula: r = v

retardation r = 0.5 m/s

3. A car runs with uniform velocity

s = 584 - 40 - 64 = 480m

t = s/v = 60 s

So, total time

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