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A particle moving in a straight line is acted on by a force which works at a constant rate and changes its velocity from u to v in passing over a distance x. Prove that the time taken is$ \frac{3(u+v)x}{2(u^2+uv+v^2)} $

What I got from the question is that the work done is constant, i.e., $ \frac{dW}{dt} = 0 $ where $ \ W$ denotes work done by force $ \ F$ in covering a distance $ \ x $.or, $ \frac{d(Fx)}{dt} = 0 $ or, $\ x\frac{dF}{dt} + F\frac{dx}{dt} $.

Please let me know if I have misunderstood the question in the first place. Also, I am unable to solve the differential equation resulting from above relation (if it is correct).

Edit: I am unable to solve the following differential equation( if my interpretation is correct): $\ x\frac{d^3x}{dt^3} + \frac{d^2x}{dt^2}\frac{dx}{dt} = 0 $.

Michael Lee
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Sharmi C
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  • I see I overlooked the misstep in the attempted solution. The work done per unit time is constant, so $\frac{dW}{dt}$ is constant, so $\frac{d^2W}{dt^2} = 0.$ The answer below takes a better approach, I think. – David K Feb 28 '19 at 19:17

1 Answers1

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Constant work rate means power is constant, i.e. $force \times speed = constant$ which means that $acceleration \times speed = constant$. Let $s$ be the speed. We have $s\frac{ds}{dt}=constant$ or $\frac{dt}{ds}=As$ where $A$ is a constant. Solving for $t$ you get $t=As^2+B$ where B is also a constant. This means that $s=\sqrt{\frac{t-B}{A}}$. Let the time taken be $T$. You can now solve for $T$ in terms of $u,v$ and $x$ by using the fact that at $t=0$, $s=u$, at $t=T$, $s=v$ and $\int_{0}^{T}sdt=x$.

Zaeem Hussain
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  • Thank you for your help. I have been able to solve. I committed a fundamental error in assuming that rate of work done was zero where clearly rate of work done is a constant. – Sharmi C Feb 28 '19 at 18:39
  • @SharmiC Don't forget to accept the answer if it helped you solve the problem. Usually we advise to wait a few hours to see if anyone else says something even more helpful, but I think this answer settles the question quite well. – David K Feb 28 '19 at 19:20