Question

A point moves in a straight line so that its displacement x metre at a time t second is such that t =(x²-1)^{1/2 .}Its acceleration in m/s² at time t second is

1/x

1/x³

-t/x²

t/x²

Answer 1

We are given that the displacement of the point is given by the following relation

X2 = 1 + T2 (1)

we know that the acceleration is the double derivative of the displacement. Hence it will be given by d2X / dT2

from equation (1), we get

X = (1 + T2) 1/2

differentiating with respect to T, we get

dX / dT = 1/2 ( 1 + T2) -1/2 x 2T

= T ( 1 + T2) -1/2

So

d2X / dT2 = 1/ ( 1 + T2)1/2 - T2 / (1+T2) 3/2

But

( 1+ T2)1/2 = X [ from (1)]

Thus

d2X / dT2 = [1/ X ] - [(X2 -1) / X3]

= [X2 -X2 +1] / X3

= 1 / X3

Hence the answer is (1) 1 / X3