Question

The relation between the time $t$ and distance '$x$' is given by $t=p{x}^2+qx$, where $p$ and $q$ are constants. The relation between velocity $v$ and acceleration $f$ is

• A. $f\propto v$
• B. $f\propto v^4$
• C. $f\propto v^2$
• D. $f\propto v^3$

Answer 1

The correct option is C $f\propto v^2$

On differentiating given equation, we get

$dt=(2px+q)dx$

$\therefore v=\frac{dx}{dt}=\frac{1}{2px+q}$ ...(i)

$\Rightarrow \frac{dv}{dt}=f=\frac{-2p}{{(2px+q)}^2}=-2p{v}^2$ [from equation (i)]

So, $f\propto v^2$