Question

The relation between time $t$ and distance $x$ is $t = \alpha {x^2} + \beta x$ where $\alpha$ and $\beta$ are constants.Find out the relation between acceleration and velocity.
(A) $v^3$
(B) $v^2$
(C) $v^{3/2}$
(D) $v^{2/3}$

Answer 1

Hint
We will differentiate the equation twice. According to Newton's second law when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

Complete step by step answer
As we know the derivative form of acceleration= change of velocity with respect to time
$\Rightarrow a = \dfrac{{dv}}{{dt}}$
Derivative form of velocity = change of position with respect to tame $\Rightarrow v = \dfrac{{dx}}{{dt}}$ ......[where x is displacement]
$t = \alpha {x^2} + \beta x$
Now differentiate the equation with respect to t
$\Rightarrow \dfrac{d}{{dt}}(t) = \alpha 2x\dfrac{{dx}}{{dt}} + \beta \dfrac{{dx}}{{dt}}$
$\Rightarrow 1 = 2\alpha x + \beta (\dfrac{{dx}}{{dt}})$ ..........equation 1
We know $\dfrac{{dx}}{{dt}} = v$ ,putting the value of equation 1
$\Rightarrow 1 = (2\alpha x + \beta )v$
$\Rightarrow \dfrac{1}{v} = 2\alpha x + \beta$
Again differentiate the equation with respect to t
$\Rightarrow \dfrac{d}{{dt}}\dfrac{1}{v} = \dfrac{d}{{dt}}2\alpha x + \dfrac{d}{{dt}}\beta$
$\Rightarrow - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} = 2\alpha \dfrac{{dx}}{{dt}}$
We know that $\dfrac{{dv}}{{dt}} = a$ ,putting the value in the equation
$- \dfrac{1}{{{v^2}}}a = 2\alpha v$
$a = - 2\alpha {v^3}$
$a \propto {v^3}$ … [where -2 $\alpha$ is constant]
Option (A) is the correct answer.

Note
Velocity is the derivative of position with respect to time: $v=\dfrac{dx}{dt}$.
Acceleration is the derivative of velocity with respect to time: $a=\dfrac{dv}{dt}=\dfrac{d^2{x}}{dt^2}$.