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,

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

$A$ body is moving unidirectionally under the influence of a constant power source. Its displacement in time $t$ is proportional to :

The power of a water pump is $4 \,kW$. If $g = 10 \,m/s^2$,the amount of water it can raise in $1 \,minute$ to a height of $20 \,m$ is ........... $litres$.

Water falls from a waterfall onto the blades of a turbine at a rate of $100 \ kg/s$. If the height of the fall is $100 \ m$,what is the power generated by the turbine?

The force required to row a boat at constant velocity is proportional to the square of its speed. If a speed of $v \, km/h$ requires $4 \, kW$ of power,how much power does a speed of $2v \, km/h$ require?

$A$ pump is used to lift $7200 \ kg$ of water per hour from a depth of $100 \ m$. Calculate the power of the pump in $kW$. Assume its efficiency is $50\%$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module