Question

A motor car starting from rest moves with uniform acceleration and attains a velocity of 8 m/s ‘ in 8 s. It then moves with uniform velocity and is finally brought to rest in 32 m under uniform retardation. The total distance covered by the car is 464 m. Find the value of (i)acceleration,(ii) retardation and(iii) the total time is taken.

Answer 1

Answer:

Acceleration = Change in velocity/time = (8-0)/8 = 1m/s^2

Now, for the retardation part, u = 8 m/s; v = 0m/s, distance travelled(s) = 32m

Using third law of motion

v^2 = u^2+ 2as

0 = 64 + 2d(32)

d = -1 m/s^2

Hence, the time taken during retardation

v = u + dt

0 = 8 -1t

t = 8 sec

During Uniform acceleration phase, the distance travelled

s= ut + 1/2at^2 = 1/2 *1(8)^2 = 32m

Now during uniform velocity phase, the velocity of the body will be 8m/s

distance travelled = total distance - distance travelled during acceleration and deceleration = 464 - (32+32) = 400 m

time taken = distance/speed = 400/8 = 50 m.

Hence, total time taken = 8+50+8 = 66 sec

Hope it helps u......✌