A car starts from rest and accelerates uniformly for10 s to a velocity

1.	A car starts from rest and accelerates uniformly for10 s to a velocity of 8 ms-1.It then runs at a constantvelocity and is finally brought to res in 64 m with aconstant retardation. The total distance covered bythe car is 584 m. Find the value of accelerationretardation and total time taken.
Answer» 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/s² Distance traveled during this time is, S = 0 + ½ at² => S = 0.5 X 0.8 X 10² => 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, v²= u²- 2aS => 0 = 8²- 2(a)(64) => a = 0.5 m/s² Retardation = 0.5 m/s² t= (8)/(0.5) = 16 s Total time taken is= 10 s + 60 s + 16 s =86 s

A car starts from rest and accelerates uniformly for10 s to a velocity of 8 ms-1.It then runs at a constantvelocity and is finally brought to res in 64 m with aconstant retardation. The total distance covered bythe car is 584 m. Find the value of accelerationretardation and total time taken.

Answer»

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/s²

Distance traveled during this time is,

S = 0 + ½ at²

=> S = 0.5 X 0.8 X 10²

=> 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,

v²= u²- 2aS

=> 0 = 8²- 2(a)(64)

=> a = 0.5 m/s²

Retardation = 0.5 m/s²

t= (8)/(0.5) = 16 s

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