Question

A point moves in a straight line so that its displacement $x$ is given by $x^2=1+t^2$ where, $x$ is in $\text{m}$ and time $t$ is in $\text{s}$. Its acceleration in ${\text{m/s}}^2$ at time $t$ is

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

Answer 1

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

Given, $x^2=1+t^2$
Differentiating the above equation we get
$2xdx=2tdt\Rightarrow x\frac{dx}{dt}=t$.........(i)
Now, taking differentiation of above equation for acceleration we get,
$x\frac{d^{2}x}{d{t}^2}+{\left(\frac{dx}{dt}\right)}^2=1$
Or, $x\frac{d^{2}x}{d{t}^2}=1-{\left(\frac{t}{x}\right)}^2$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=\frac{x^2-t^2}{x^3}=\frac{1}{x^3}$ $(\because x^2=1+t^2\Rightarrow x^2-t^2=1)$
Hence, the acceleration is $a=\frac{d^{2}x}{d{t}^2}=\frac{1}{x^3}{\text{m/s}}^2$