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

Question

Aicharya · Accepted Answer

Given $x^2=1+t^2$
Differentiating the above equation we get
$2 xdx=2tdt$
$x \frac{dx}{dt}=t$ ...(1)
Now taking differentiation for acceleration we get,
$x \frac{d^2 x}{d t^2}+\left(\frac{d x}{d t}\right)^2=1$
$x \frac{d^2 x}{d t^2}=1-\left(\frac{t}{x}\right)^2$
$x \frac{d^2 x}{d^2}=\frac{x-\frac{t^2}{x}}{x}=\frac{(x^2 - t^2)}{x^3}=\frac{x^2-t^2}{x^3}$
$\frac{d^2 x}{d t^2}=\frac{x^2-\left(x^2+1\right)}{x^3}=\frac{1}{x^3}$ ^

A point moves in a straight line so that its displacement \(x \mathrm{~m}\) at time \(t \mathrm{~s}\) is…

Solution