Phys Problem, I tried to do this problem by taking the derivative and it did not get me the right answer. I also tried the Vdv=Ads integration and could not get the right answer. Please Help

  • $\begingroup$ What's with the down votes? This is a clearly asked question. $\endgroup$
    – Keith
    Commented Jan 29, 2014 at 5:36
  • $\begingroup$ @Keith: the aim of this site is to become a reservoir of knowledge of interest to future students of physics (or anyone else interested in physics). Doing someone's homework for them does not contribute to this goal. $\endgroup$ Commented Jan 29, 2014 at 8:04

1 Answer 1


The correct formula to use would be $a = dv/dt = dv/ds \cdot ds/dt = dv/ds \cdot v$. Integration is not the right way to go about using this. Since you are given a formula for $v(s)$, simply compute the derivative at $s=5m$, and multiply that derivative by the value of v itself at $s=5m$.

  • $\begingroup$ That did get me the right answer. but I do not know why you did this, can you please explain? $\endgroup$
    – user38498
    Commented Jan 29, 2014 at 4:37
  • $\begingroup$ The fact that $v$ is a function of $s$ means there are certain regions of space where go slowly and others where you go quickly. If you move into a region where your speed is higher, you will accelerate. Your acceleration will be equal to the product: (how much faster you go once you are in the new region) * (how quickly you are moving into the new region). If either of these things are zero, you would have zero acceleration. $\endgroup$ Commented Jan 29, 2014 at 5:18
  • $\begingroup$ Mathematically, the chain rule was used. $\endgroup$
    – BMS
    Commented Jan 29, 2014 at 5:37
  • $\begingroup$ How was the chain rule applied? $\endgroup$
    – user38498
    Commented Jan 29, 2014 at 5:44
  • $\begingroup$ $dv/dt = dv/ds \cdot ds/dt$ is the chain rule. You can find it in the first sentence of the answer. $\endgroup$ Commented Jan 29, 2014 at 5:45

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