Question

A point moves in a straight line so that its displacement is x m at time t sec, given by $x^2=t^2+1$. Its acceleration in $m/s^2$ at time t sec is:

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

Answer 1

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

Given,

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

Velocity, $v=\frac{dx}{dt}$. . . . . . . . . . .. . . . . . . . (2)

Acceleration, $a=\frac{d^{2}x}{d{t}^2}$. . . . . . . . . . . . .(3)

Differentiate equation (1) with respect to time $t$,

$2x\frac{dx}{dt}=2t+0=2t$. . . . . . . . .(4)

$2xv=2t$

$xv=t$

$v=\frac{t}{x}$. . . . . . . . . .(5)

Again differentiate the equation (4) with respect to time $t$

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

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

$xa+v^2=1$ (from equation 1 and 2)

$a=\frac{1-v^2}{x}$

$a=\frac{1}{x}-\frac{v^2}{x}$

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

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

$a=\frac{1}{x}-\frac{x^2-1}{x^3}$ (From equation 1 $,t^2=x^2-1$)

$a=\frac{x^2-x^2+1}{x^3}$

$a=\frac{1}{x^3}$

The correct option is D.