Area under the displacement-time graph giving time?

Question

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?

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

score 2 · Accepted Answer · answered Jun 11 '16 at 12:49

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.

Area under the displacement-time graph giving time?

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