Question

A particle moving along a straight line has a velocity $v \,m \,s^{-1}$, when it cleared a distance of $x\, m$. These two are connected by the relation $v=\sqrt{49+x}$. When its velocity is $1 \,m \,s^{-1}$, its acceleration is

• A. $2\,m\,s^{-2}$
• B. $7\,m\,s^{-2}$
• C. $1\,m\,s^{-2}$
• D. $0.5\,m\,s^{-2}$

Answer 1

Answer: (D)

Given : $v=\sqrt{49+x}$
Squaring both sides, we get, $v^{2} = 49 + x$
Differentiating both sides w.r.t. $t$, we get ; $2v \frac{dv}{dt}=\frac{dx}{dt}$
$2v \frac{dv}{dt}=v$ ;
$\frac{dv}{dt}=\frac{1}{2}=0.5\quad\left(\because v=\frac{dx}{dt}\right)$
Acceleration, $a=\frac{dv}{dt}=0.5\,m\,s^{-2}$