Question

If the time and displacement of particle along the positive X-axis are related as $t=(x^2-1{)}^{\frac{1}{2}}$, then find the acceleration in terms of x.

• A. $\frac{1}{x^4}$
• B. $\frac{1}{x^7}$
• C. $\frac{1}{x}$
• D. $\frac{1}{x^3}$

Answer 1

Answer: (D)

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

The relation between time and the displacement is given by the equation,

$t=(x^2-1{)}^{1/2}$

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

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

Differentiate equation (1) with respect to time,

$2t=2x\frac{dx}{dt}$

Velocity, $v=\frac{dx}{dt}$

$2t=2xv$

$v=\frac{t}{x}$. . . . . . .(3)

Again differentiate the above equation,

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

Acceleration, $a=\frac{d^{2}x}{d{t}^2}$

$1=xa+v^2$

$xa=1-v^2$

$xa=1-\frac{t^2}{x^2}$ (from equation 3)

$xa=\frac{x^2-t^2}{x^2}=\frac{1}{x^2}$ (from equation 2)

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

The correct option is D.