The velocity v reached by a car of mass m at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Power $P$ is defined as the rate of change of work done,$P = \frac{dW}{dt} = Fv = (ma)v = m \left(\frac{dv}{dt}\right) v$.Since $P$ is constant,we

Vedclass · Accepted Answer

$v^3 \propto \frac{2 P}{m}$

Vedclass · Answer

(C) Power $P$ is defined as the rate of change of work done,$P = \frac{dW}{dt} = Fv = (ma)v = m \left(\frac{dv}{dt}\right) v$.Since $P$ is constant,we have $P dt = mv dv$.Integrating both sides from $t=0$ to $t$ and $v=0$ to $v$,we get $Pt = \frac{1}{2}mv^2$.Thus,$v^2 = \frac{2Pt}{m}$.Also,$v = \frac{dx}{dt}$,so $\frac{dx}{dt} = \sqrt{\frac{2P}{m}} t^{1/2}$.Integrating with respect to $t$,$x = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2}$,which implies $t^{3/2} \propto x \Rightarrow t \propto x^{2/3}$.Substituting $t$ back into the velocity equation: $v^2 \propto t \propto x^{2/3}$,so $v \propto x^{1/3}$.However,looking at the relationship between $v$ and $P$ at a fixed distance $x$,we have $v^3 \propto P$. Thus,$v^3 \propto \frac{P}{m}$.

The velocity $v$ reached by a car of mass $m$ at a certain distance from the starting point,driven with constant power $P$,is such that:

Similar Questions

The average work done by a human heart while it beats once is $0.5 \, J$. Calculate the power used by the heart if it beats $72$ times in a minute.

$A$ body is moving in a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to:

An electric motor exerts a force of $40\, N$ on a cable and pulls it by a distance of $30\, m$ in one minute. The power supplied by the motor is .............. $W$

$A$ cylindrical well of radius $2.5 \ m$ has water up to a height of $14 \ m$ from the bottom. If the water level is at a depth of $6 \ m$ from the top of the well,then the time taken (in minutes) to empty the well using a motor of $10 \ HP$ is approximately,(Take $g = 10 \ m/s^2$)

$A$ force applied by an engine on a train of mass $2.05 \times 10^6 \ kg$ changes its velocity from $5 \ m/s$ to $25 \ m/s$ in $5 \ minutes$. The power of the engine is (in $MW$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module