A car starts from rest and accelerates uniformly for 10 s to a velocity of 8 m/s. It then runs at a constant velocity and is finally brought to 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.

The car starts from rest and accelerates uniformly for 10 s to a velocity of 8 m/s. The acceleration is found to be,

**a** = (v – u)/t = (8 – 0)/10 = **0.8 m/s**^{2}

Distance traveled during this time is,

S = 0 + ½ at^{2}

=> S = 0.5 X 0.8 X 10^{2}

=> S = 40 m

Suppose it travels 'x' distance with constant velocity, 8 m/s, for time 't'.

It then travels 64 m with uniform retardation and comes to rest.

Total distance traveled = 40+x+64 = 584

=> x = 480 m

So, 8 X t = 480

=> t = 60 s

Let the car travel 64 m with uniform retardation for time t^{/}.

Using,

v^{2} = u^{2} - 2aS

=> 0 = 8^{2} - 2(a)(64)

=> a = 0.5 m/s^{2}

**Retardation = 0.5 m/s ^{2}**

t^{/} = (8)/(0.5) = 16 s

**Total time taken is** = 10 s + 60 s + 16 s = **86 s**

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