How to solve this variable acceleration Q, parts a to c?

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1 Answer
Ananda Dasgupta
Jun 8, 2018

a) t=0, t=1/2 and t =1.
b) 33/16
c) see below

Explanation:

The expression for the position can be rewritten in the form :
x = 1/2 t^2(t-1)^2

a) The velocity is

v = dx/dt = t(t-1)^2+t^2(t-1)
quad = t(t-1)(2t-1)

Thus the particle is at rest (v=0) at the times t=0, t=1/2 and t =1.

b) Since the velocity, which in this case is a continuous function, vanishes at t=0, t=1/2 and t = 1, it must change sign at this points. So, to calculate the distance traveled, we must consider each of these time intervals separately. Now

  • displacement between t=0 and t = 1/2 :
    x(1/2)-x(0) = 1/2times (1/2)^2(1/2-1)^2-0
    qquad = 1/32
  • displacement between t=1/2 and t = 1 :
    x(1)-x(1/2) = 0-1/2times (1/2)^2(1/2-1)^2
    qquad = -1/32
  • displacement between t=1 and t = 2 :
    x(2)-x(1) = 1/2times (2)^2(2-1)^2-0
    qquad = 2

The distances traveled in these three intervals are 1/32, 1/32 and 2 respectively.

Thus the total distance traveled is 1/32+1/32+2 = 33/16

c) It is easy to see that 1/2t^2(t-1)^2 can never be negative.

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