0
$\begingroup$

This is probably very stupid, but today at a lecture, our professor solved a problem where we had to find the time taken to travel from 0-5m, where $v = 3/x$ (velocity is a function of position.)

Then he integrated, $$\int_0^5 x dx = \int_0^t 3 dt$$

The integral of x represents the area under the position time curve which is giving the time taken. How is this possible? Where am I going wrong here?

$\endgroup$
  • 2
    $\begingroup$ "The integral of x represents the area under the position time curve" This is wrong. $\endgroup$ – lucas Jun 11 '16 at 12:27
  • $\begingroup$ See also this answer for how to handle non uniform acceleration under various scenarios. You will see that $$t = \int \frac{1}{v(x)}\,{\rm d}x$$ $\endgroup$ – ja72 Jun 11 '16 at 13:39
2
$\begingroup$

The displacement is equal to the area under a velocity time graph so in your example the change in displacement $dx$ in a time $dt$ is equal to $v\; dt$ with $v=\frac 3 x$

So $dx = v\; dt = \frac 3 x \; dt \Rightarrow x\; dx= 3\; dt$ and then you do the integration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.