Question

A self propelled vehicle of mass $m$ whose engine delivers a constant power $P$ has an acceleration $\alpha = \dfrac{P}{{mv}}$ .What is the distance traveled by it (assuming no friction) to increase the velocity of the vehicle from ${v_1}$ to ${v_2}$?
A.$x = \dfrac{{3m}}{P}\left( {v_2^3 - v_1^3} \right)$
B. $x = \dfrac{m}{{3P}}\left( {v_2^3 - v_1^3} \right)$
C. $x = \dfrac{{2m}}{P}\left( {v_2^3 - v_1^3} \right)$
D. $x = \dfrac{m}{{2P}}\left( {v_2^3 - v_1^3} \right)$

Answer 1

Hint: Acceleration is the rate of change of velocity. It can be written as $\alpha = \dfrac{{dv}}{{dt}}$ In another form acceleration can also be written as $\alpha = v\dfrac{{dv}}{{dx}}$. By substituting the given value of $a$ in the equation and integrating we can arrive at the equation for distance $x$.

Complete step by step answer:
Acceleration is the rate of change of velocity. It can be written as $\alpha = \dfrac{{dv}}{{dt}}$. This equation can be written in another form as
$\alpha = \dfrac{{dv}}{{dt}}$
$ \Rightarrow \alpha = \dfrac{{dv}}{{dt}} \times \dfrac{v}{v}$
$ \Rightarrow \alpha = \dfrac{{dv}}{{dt}} \times \dfrac{v}{{\left( {\dfrac{{dx}}{{dt}}} \right)}}$
$ \Rightarrow \alpha = v \times \dfrac{{dv}}{{dx}}$
Since, $v = \dfrac{{dx}}{{dt}}$.
It is given that $\alpha = \dfrac{P}{{mv}}$.Now substitute this value in the above equation. Then we get,
$\dfrac{P}{{mv}} = v\dfrac{{dv}}{{dx}}$
On rearranging this equation we get
$\Rightarrow \dfrac{P}{m}dx = {v^2}dv$
Now suppose the particle starts from origin with a velocity ${v_1}$.At a distance $x$ the velocity becomes ${v_2}$.thus we should integrate the above equation for the limit $x = 0\,$to $x = x$and $v = {v_1}$to $v = {v_2}$
$\int_0^x {\dfrac{P}{m}dx = \int\limits_{{v_1}}^{{v_2}} {{v^2}dv} } $
$\Rightarrow \dfrac{P}{m}\left[ x \right]_0^x = \left[ {\dfrac{{{v^3}}}{3}} \right]_{{v_1}}^{{v_2}}$
$ \Rightarrow \dfrac{P}{m}x = \dfrac{{v_2^3 - v_1^3}}{3}$
$ \Rightarrow x = \dfrac{{m\left( {v_2^3 - v_1^3} \right)}}{{3P}}$

This is the distance traveled by the vehicle to increase the velocity of the vehicle from ${v_1}$ to ${v_2}$ Therefore the correct answer is option (B).

Note:
Acceleration is the rate of change of velocity. It can be written as $\alpha = \dfrac{{dv}}{{dt}}$.this is the equation that we use commonly. But in this question for doing the integration we should have only two variables in the equation, distance $x$ and velocity $v$. That is why we changed the form of the equation for acceleration into $\alpha = v\dfrac{{dv}}{{dx}}$, which contains the required variables distance $x$ and velocity $v$.