If a point moves in straight line in such a manner that its acceleration is proportional to its speed (a ∝ v) , what is the relation between distance covered and speed?

a) $x \propto v$ b) $x \propto {v}^{2}$ c) $x \propto {v}^{3}$ d) $x \propto \sqrt{v}$

Dec 29, 2017

I get (a)

Explanation:

Given that acceleration $a \propto v$, speed

From definition of acceleration and velocity
$\implies \frac{\mathrm{dv}}{\mathrm{dt}} \propto \frac{\mathrm{dx}}{\mathrm{dt}}$
Integrating both sides with time $\mathrm{dt}$, for LHS we get

$\int \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{dt} = v + c$
where $c$ is a constant of integration and can be found from initial conditions.

For RHS we get

$\int \frac{\mathrm{dx}}{\mathrm{dt}} \mathrm{dt} = x + {c}_{1}$
where ${c}_{1}$ is a constant of integration and can be found from initial conditions.

Assuming that initially at $t = 0$, acceleration, velocity and displacement all are $= 0$. Both $c \mathmr{and} {c}_{1}$ are$= 0$. We get

$v \propto x$
$\implies x \propto v$