Question

A particle is moving in the $x-y$ plane. At certain instant of time, the components of its velocity and acceleration are as follows: $v_x=3{\text{ms}}^{-1},v_y=4{\text{ms}}^{-1},a_x=2{\text{ms}}^{-2}$ and $a_y=1{\text{ms}}^{-2}$. The rate of change of speed at this moment is

• A. $\sqrt{10}{\text{ms}}^{-2}$
• B. $4{\text{ms}}^{-2}$
• C. $\sqrt{5}{\text{ms}}^{-2}$
• D. $2{\text{ms}}^{-2}$

Answer 1

The correct option is D $2{\text{ms}}^{-2}$

At a certain instant, $v=\sqrt{v_x^2+v_y^2}$
Rate of change of velocity is given by
$\frac{dv}{dt}=\frac{2v_x\frac{d{v}_x}{dt}+2v_y\frac{d{v}_y}{dt}}{2\sqrt{v_x^2+v_y^2}}$
$=\frac{v_x{a}_x+v_y{a}_y}{\sqrt{v_x^2+v_y^2}}(\because \frac{d{v}_x}{dt}=a_x\text{and}\frac{d{v}_y}{dt}=a_y)$
$=\frac{3\times 2+4\times 1}{\sqrt{3^2+4^2}}$
$=2{\text{m/s}}^2$