I've been given this equation: $$\mathrm{velocity}, v = 2\,\mathrm{cm/s^3} \,t^2 + 5\,\rm cm/s.$$

If someone now asks me to tell the velocity at, say 4 sec, then before proceeding with the equation, do I need to differentiate it (which would give me $4t$) or should I proceed to put 4sec in that equation at place of $t$ which would then give me 37 cm/s.

So, which of the approach is correct? and since one of the approach is only right, so, also let me know that when do we use the other approach

  • $\begingroup$ You just put the 4s in. Differentiating would only make sense if the function you are looking at, would be an expression of distance travelled throug in dependence of time. $\endgroup$ – Nontriviality Sep 20 '15 at 19:59
  • $\begingroup$ Did you underestand why you usually take derivatives? $\endgroup$ – Nontriviality Sep 20 '15 at 19:59
  • 3
    $\begingroup$ I've been told that the flavors of sodas are printed on the cans. If someone now asks me what flavor is in a can marked Grape, do I need to first drop the first letter and rearrange the others (which would give me Pear) or can I just tell him it's Grape? $\endgroup$ – WillO Sep 20 '15 at 20:14
  • $\begingroup$ The $2 t^2$ term should have been written as 2 cm/s $(t/s)^2$ $\endgroup$ – Count Iblis Sep 20 '15 at 22:21

we first define velocity with the following equation $$v=\int{a}dt$$ Differentiating the equation would yield acceleration which is incorrect based on what the question is asking You stated your equation for velocity is velocity$$v=f(t)=2cm/s^3t^2+5cm/s$$ Therefor by simple substitution we just sub $v=f(4)$ $$f(t)=2\times(4)^2+5=37cm/s$$ Verifying the equation we also find it is dimensionally accurate



When you have a function $f(t)$ that expresses a quantity (say the velocity $v$), then you just evaluate that function at $t$ in order to get the function (velocity) at $t$.

However, if I give you the position at $t$, $x(t)$, and I ask you for the velocity, then you first have to recall that velocity is the derivative of position with time:

$$v = \frac{dx}{dt}$$

I hope this clears up your confusion. If you are being given a function that expresses one thing, and I ask you for another thing, then it may be appropriate to differentiate (or integrate, or...) that function in order to get the answer. But if the function expresses the thing you want - then no further operation is needed.


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.