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CBSE Class 9 Answered

The body starts from rest and moves with uniform acceleration and attains a velocity of 8 m/s in 8s. 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 464m. Find the value of acceleration and retardation and the total time taken.
Asked by Aryan Suresh | 22 Jun, 2013, 02:35: PM
answered-by-expert Expert 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
Answered by | 23 Jun, 2013, 08:22: AM

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