Question

A particle moves rectilinearly. its displacement x at time t is given by $x^2=t^2+1$. Its acceleration at time t is proportional to

• A. $\frac{1}{x^3}$
• B. $\frac{1}{x}-\frac{1}{x^2}$
• C. - $\frac{t}{x^2}$
• D. $\frac{1}{x}-\frac{t^2}{x^3}$

Answer 1

The correct option is A $\frac{1}{x^3}$

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

$x^2=1+t^2$....(I)

We know that the acceleration is the double derivative of the displacement

Hence, it will be given by $a=w\frac{d^{2}x}{d{t}^2}$

From equation (I), we get

$x=(1+t^2{)}^{\frac{1}{2}}$

Differentiating with respect to T, we get

$\frac{dx}{dt}=t(1+t^2{)}^{\frac{-1}{2}}$

$\frac{d^{2}x}{d{t}^2}=\frac{1}{(1+t^2{)}^{\frac{1}{2}}}-\frac{t^2}{(1+t^2{)}^{\frac{3}{2}}}$

From equation (I)

$(1+t^2{)}^{\frac{1}{2}}=x$

The acceleration is

$\frac{d^{2}x}{d{t}^2}=\frac{1}{x}-\frac{(x^2-1)}{x^3}$

$\frac{d^{2}x}{d{t}^2}=\frac{(x^2-x^2+1)}{x^3}$

$\frac{d^{2}x}{d{t}^2}=\frac{1}{x^3}$

The acceleration at time t is $\frac{1}{x^3}$

Hence, Option A is correct.