# Do I need to differentiate this equation or not?

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

• 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. – Nontriviality Sep 20 '15 at 19:59
• Did you underestand why you usually take derivatives? – Nontriviality Sep 20 '15 at 19:59
• 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? – WillO Sep 20 '15 at 20:14
• The $2 t^2$ term should have been written as 2 cm/s $(t/s)^2$ – 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:
$$v = \frac{dx}{dt}$$