Question

A point moves in a straight line so that its displacement $x$ metre at time $t$ second is given by $x^2=1+t^2$. Its acceleration in $m{s}^{-2}$ at time $t$ second is

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

Answer 1

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

Given $x^2=1+t^2$

Differentiating the above equation we get

$2xdx=2tdt$

$x\frac{dx}{dt}=t$------(1)

Now taking differentiation for acceleration we get,

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

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

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

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