Question

Position $x$ of a particle moving along a straight line with time $t$ is given by $x=\sqrt{t^2+2}$. Acceleration of the particle is given by.

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

Answer 1

Answer: (B)

$\frac{2}{x^3}$

$x=\sqrt{t^2+2}$

$\frac{dx}{dt}=\frac{1}{2\sqrt{t^2+2}}2t=\frac{t}{\sqrt{t^2+2}}$

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

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

$a=\frac{2}{x^3}$