# Numerical: Work Power Energy [closed]

A particle moves along a straight line. A force acts on the particle which produces a constant power. It starts with initial velocity 3 m/s and after moving a distance 252 m its velocity is 6 m/s. Find the time taken.

Hint : You assumed the force to be constant by using $W=Fx$ which is wrong. It is $W=\int Fdx$
Use $P=\frac{dW}{dt}$
Use work energy theorem. Use calculus. Find distance covered when speed is $6ms^{-1}$. That should eliminate your variables if the question has sufficient information.
Hint:Another possible way to solve it would be to observe that $$a=\frac{dv}{dt}=\frac{dv}{dx}*\frac{dx}{dt}$$ Hence $$a=v\frac{dv}{dx}$$ Now according to question power is constant Hence P=k(say) $$Fv=k$$ $$\Rightarrow mav=k$$ $$\Rightarrow mv^2\frac{dv}{dx}=k$$ Solve the differential equations with the given limits to get the an equation of v in terms of x.
For full solution: $$mv^2\frac{dv}{dx}=k$$ $$\Rightarrow mv^2dv=kdx$$ Integrating both sides $$\frac{v^3}{3}=x\frac km+c$$ Given v=3 when x=0 which gives c=9 Also v=6 when x=252 which gives $\frac km$=4 Therefore we get the relationship between v and x as $$v=4x+9$$ Rewriting v as $\frac {dx}{dt}$ we get $$\frac {dx}{dt}=4x+9$$ Solving the differential equation again and putting initial condition as x=0,t=0 we get $$\frac 14 ln(4x+9)=t$$ Put x=252 to get the final answer.