Question

The speed \[v\] reached by a car of mass m in travelling a distance \[x\], driven with constant power P, is given by
(A) \[v = \dfrac{{3xP}}{m}\]
(B) \[v = {(\dfrac{{3xP}}{m})^{1/2}}\]
(C) \[v = {(\dfrac{{3xP}}{m})^{1/3}}\]
(D) \[v = {(\dfrac{{3xP}}{m})^2}\]

Answer 1

Hint Directly use the formula of the mechanical power to find the equation for velocity. Mechanical power is defined as the rate of doing work given as the ratio of work done to time. Using this formula, find the suitable expression for velocity.

Complete Step By Step Solution
Mechanical power is defined as the amount of work done per unit time. We know that work done on an object is given as the product of force applied on the object and the displacement experienced by the object when the force is applied. Mathematically given as,
\[W = F \times s\]
Now, mechanical power is given as ratio of work done by time, which can also be written as,
\[P = \dfrac{W}{t} = \dfrac{{F \times s}}{t} = Fv\] , Since displacement per time is equal to velocity of the object. Using this formula in our question, it is given that the car moves \[x\] distance in speed \[v\]. Now , power can be calculated as ,
\[ \Rightarrow P = Fv\]
We know that force experienced by an object can be written as the product of its mass and acceleration. Using this in above equation we get,
\[ \Rightarrow P = mav\]
We know that acceleration of a body can be defined as the rate of change of velocity per unit time. Using this in the above equation we get,
\[ \Rightarrow P = m\dfrac{{dv}}{{dt}} \times v\]
\[ \Rightarrow \dfrac{P}{m} = v\dfrac{{dv}}{{dt}}\]
Now, splitting the \[\dfrac{{dv}}{{dt}}\]term as a product of \[dx\], we obtain
\[ \Rightarrow \dfrac{P}{m} = v\dfrac{{dv}}{{dx}} \times \dfrac{{dx}}{{dt}}\]
Now we know that the rate of change of displacement with respect to time is velocity. Using this, we get
\[ \Rightarrow \dfrac{P}{m} = {v^2}\dfrac{{dv}}{{dx}}\]
Taking the \[dx\] to the other side and integrating on both sides we get,
\[ \Rightarrow \int {\dfrac{P}{m}dx} = \int {{v^2}dv} \]
Post integrating, we obtain
\[ \Rightarrow \dfrac{P}{m}x = \dfrac{{{v^3}}}{3}\]
\[ \Rightarrow {v^3} = \dfrac{{3Px}}{m}\]
Applying cube root on both sides, we get
\[ \Rightarrow v = {(\dfrac{{3xP}}{m})^{1/3}}\]


Thus, option (c) is the right answer for the given question.

Note In general kinematics, a work is said to be done on the object when there is a force acting upon a body causing it to get displaced from its original point. According to the work energy theorem, the work done by all forces acting on a body is equal to the change in kinetic energy of the body.