Question

Equation of motion of a particle moving in straight line is given by $x=\frac{1}{2}vt$ then its acceleration is

• A. $\propto x$
• B. $\propto \frac{1}{x}$
• C. $\propto \frac{1}{v^2}$
• D. constant

Answer 1

The correct option is D constant

Given the equation of motion of a particle moving in a straight line is given by

$x=\frac{1}{2}vt$

Now differentiating both sides with respect to $t$ we get,

$\frac{dx}{dt}=\frac{1}{2}v+\frac{t}{2}\frac{dv}{dt}$

or, $v=\frac{1}{2}v+\frac{t}{2}\frac{dv}{dt}$

or, $\frac{v}{2}=\frac{t}{2}\frac{dv}{dt}$

or, $v=t\frac{dv}{dt}$

Now again differentiating both sides with respect to $t$ we get,

or, $\frac{dv}{dt}=\frac{dv}{dt}+t\frac{d^{2}v}{d{t}^2}$

or, $t\frac{d^{2}v}{d{t}^2}=0$

or, $\frac{d^{2}v}{d{t}^2}=0$

Now integrating both sides we get,

$\frac{dv}{dt}=c$ [ Where $c$ is constant]

or, acceleration $=c$.