A car accelerates from rest at a constant rate of 3ms-2 for sometime. Then it retards at constant rate of 6ms-2 and comes to rest. If the total time for which it remains in motion is 3 seconds, what is the total distance travelled?

Asked by  | 29th May, 2013, 07:13: PM

Expert Answer:

u = 0 m/s
a1 = 3m/s2 
Let the velocity after first segment of acceleration be u1
a2 = -6m/s2
v = 0
 
Total time = 3 sec
Let the time taken for first segment be t and for the second segment would then be (3-t) sec
 
v = u +at 
u1 = 0+a1t = 3t
 
Further for the second segment
v = u +at 
0 = 3t -6(3-t)
3t = 18-6t
9t = 18
 t = 2 sec
u1  = 6 m/s
 
Hence the distance travelled in the first segment 
s1 = ut+1/2a1t^2 = 0+1/2*3*(2)^2 = 6 m 
s2 = u1(3-t)+1/2a2(3-t)^2 = 6*1+1/2*(-6)*(1)^2 = 3m
 
Hence total distance travelled = 9m 
 
 

Answered by  | 30th May, 2013, 02:52: AM

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