Question

A self propelled vehicle of mass$m$whose engine delivers constant power $P$ has acceleration $a=\dfrac{P}{mv}$ (Assume that there is no friction). In order to increase its velocity from ${{v}_{1}}\,to\,{{v}_{2}}$ , the distance it has to travel will be-
(A). $\dfrac{3P}{m}({{v}_{2}}^{3}-{{v}_{1}}^{3})$
(B). $\dfrac{m}{3P}({{v}_{2}}^{3}-{{v}_{1}}^{3})$
(C). $\dfrac{2P}{m}({{v}_{2}}^{3}-{{v}_{1}}^{3})$
(D). $\dfrac{m}{2P}({{v}_{2}}^{3}-{{v}_{1}}^{3})$

Answer 1

The acceleration of the self-propelled vehicle is inversely proportional to its power and velocity, but power is constant. Substituting acceleration in terms of velocity and then integrating the equation will give us the relationship between distance travelled by the vehicle and change in its velocity.

Formulas used:
$a=\dfrac{dv}{dt}$

Complete step by step solution:
The rate at which work is done is called power ($P$). It is also the energy transmitted per unit time. Its SI unit is watt ($W$).
$P=\dfrac{E}{t}$
Here,$E$is the energy transmitted
$t$is time taken
We know that acceleration is the rate of change of velocity, therefore,

& a=\dfrac{dv}{dt} \\\ & a=\dfrac{dv}{dt}\cdot \dfrac{dx}{dx} \\\ \end{aligned}$$ $$\Rightarrow a=v\dfrac{dv}{dx}$$ - (1) Given, $$a=\dfrac{P}{mv}$$ - (2) From eq (1) and eq (2), we get, $$v\dfrac{dv}{dx}=\dfrac{P}{mv}$$ $${{v}^{2}}dv=\dfrac{P}{m}dx$$ Integrating on both sides, we get, $$\int\limits_{{{v}_{1}}}^{{{v}_{2}}}{{{v}^{2}}}dv=\dfrac{P}{m}\int\limits_{0}^{x}{dx}$$ $$\begin{aligned} & \left[ \dfrac{{{v}^{3}}}{3} \right]_{{{v}_{1}}}^{{{v}_{2}}}=\dfrac{P}{m}\left[ x \right]_{0}^{x} \\\ & \Rightarrow \dfrac{1}{3}({{v}_{2}}^{3}-{{v}_{1}}^{3})=\dfrac{P}{m}\cdot x \\\ & \therefore x=\dfrac{m}{3P}({{v}_{2}}^{3}-{{v}_{1}}^{3}) \\\ \end{aligned}$$ The vehicle will travel a distance of $$x=\dfrac{m}{3P}({{v}_{2}}^{3}-{{v}_{1}}^{3})$$ before changing its velocity from$${{v}_{1}}\,to\,{{v}_{2}}$$. **So the correct option is (B).** **Additional Information:** Differentiation deals with calculating the change in a function with respect to change in the arguments on which it depends. Integration deals with adding the whole part. It is used to calculate areas, volumes etc from respective functions. **Note:** When acceleration is,$$a=\dfrac{dv}{dt}$$ , here $$\dfrac{dv}{dt}$$ is instantaneous acceleration for a very small change in velocity in small very interval of time. Similarly velocity for very small change in distance in a very small time interval is $$v=\dfrac{dv}{dt}$$ . The work done or energy of an isolated system is always conserved; this means that the power of an isolated system is constant.