Thu May 31, 2012 By: Shourya Mukherjee

A car starts from rest and moves with uniform acceleration of x m/s^2 along a straight line. It then retards uniformly at a rate y m/s^2 and stops. If ‘t’ is the time elapsed since it starts moving and stops, find the average speed of the car. Please explain clearly.

Expert Reply
Mon June 04, 2012
v = u + at1 where t1 is the time at which it accelearte with x m/s2
 v = 0 + (x) t1
 S1 = ut +1/2 at12
S1 = 0 + 1/2 x (t1)2
Now v = u +a t2 where t2 is the time at which it retarded with ym/s2
0 = v - y t2
S2 = ut2 +1/2 at22
as u=v in this case=xt1
S2 = vt2 -1/2 y t22
as the final vel after retardation:0=initial vel before retardation-retardationxtime=xt1-yt2   from first law of motion
xt1 = yt2
So t1/t2 = y/x
Average speed = total distance / total time 
total time taken = (t1 + t2)
Total distance = S1 + S2
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