Question

A body moves in a straight line according to the law of motion s=t^3-4t^2-3t.Find its acceleration when its velocity is zero.
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Answer 1

ANSWER :

ANSWER REFER TO THE ATTACHMENT PROVIDED.

Answer 2

(a)  The displacement of a particle at time t is given s = ut + 1/2at2

At time (t - 1), the displacement of a particle is given by

S' = u (t-1) + 1/2a(t-1)2

So, Displacement in the last 1 second is,

St = S - S'

= ut + 1/2 at2 – [u(t-1)+1/2 a(t-1)2 ]

= ut + 1/2at2 - ut + u - 1/2a(t - 1)2

= 1/2at2 + u - 1/2 a (t+1-2t) =  1/2at2 + u - 1/2at2 - a/2 + at

S = u + a/2(2t - 1)

Solution:

(a) x(t) = 3t - 4t2 + t3

=> x(2) = 3 x 2 - 4 x (2)2 + (2)3 = 6 - 4 x 4 + 8 = -2m.

(b) x(o) = 0

X(4) = 3 x 4 - 4 x (4)2 + (4)3 = 12 m.

Displacement = x(4) - x(0) = 12 m.

(c) < v > = X(4)X(2)/(4-2) = (12-(-2))/2 m/s = 7 m/s

(d) dx/dt = 3 - 8t + 3t2

v(2) (dx/dt)2 = 3 - 8 x 2 + 3 x (2)2 = -1m/s

(b) Putting the values of u = 2 m/s, a = 1 m/s2 and t = 5 sec, we get

S = 2 + 1/2(2 x 5 - 1) = 2 + 1/2 x 9

= 2 + 4.5 = 6.5 m