# Why do we differentiate displacement to get velocity if displacement is given as a function of time [closed]

Like if the displacement is given as S=(2t³)and if we are asked to find the velocity in 2 seconds then if we put t=2 in the expression we get 16 which isn't correct. The correct must be dS/dt=6t² and then if we plug 2 in t we get the velocity as 24. Someone plz explain. Why don't we get the correct answer if we don't differentiate and directly plug the value in the expression which was given?

• If you wanted to know the object's weight (or price) at time 2, would you still "plug into" $2t^3$ ? Nov 13, 2019 at 12:37
• what are dimensions of number 2 - is it $[m/s^3]$ or is it just a dimensionless number ? Nov 13, 2019 at 13:05

It’s because setting t=2 in the first equation tells you the distance the object goes in 2 seconds, not its velocity after 2 seconds.

I would add, per @Agnius Vasiliauskas, that there should be a statement given with the equation that says the coefficient 2 has units of m/s$$^3$$ in order to make sense.

Hope this helps

Your displacement equation is faulty, because left side has dimensions of $$[m]$$ and right side has dimensions of $$[s^3]$$ (dimensions of left and right side of equation must be the same). Thus it should be $$s = 2At^3$$ where constant $$A$$ has dimensions of $$[m/s^3]$$. In this case speed : $$v= \frac{ds}{dt} = 6 A t^2$$ or variable A is function of time $$A=A(t)$$ also, then speed is : $$v=\frac{ds}{dt} = 2 t^3 \frac{dA(t)}{dt} + 6 t^2 A(t)$$

In this last case calculus trick is used : $$\frac{d}{dk}\big(x(k) \cdot y(k)\big) = y(k) \frac{dx(k)}{dk} + x(k) \frac{dy(k)}{dk}$$

And in general calculus is used in Physics from a long time ago- the point Newton and Leibniz invented it. It's still a main tool in Physics nowadays, so you need to get used to it. Maybe grab some calculus book before jumping to Physical concepts ? Without this knowledge you will loose a lot in Physics. My gut feeling is that you strugle to understand speed and other concepts, because of this exact problem. Newton needed it, so you too

• Any reason of downvote ? Because I have explained main physics rule about checking any physics equation correctness ? Nov 13, 2019 at 12:50
• I didn’t down vote, but The equation can be used if you assume (as I did) that the coefficient 2 has units of m/s$^3$. Admittedly it should be stated so. Nov 13, 2019 at 12:59
• Yes, it should be stated as so, because it looks like just dimensionless number. Nov 13, 2019 at 13:03
• Yes I agree with you. But it seemed to me the OP’s confusion was with concepts not units Nov 13, 2019 at 13:06
• Now it looks like someone down voted me. Oh well c’est la vie Nov 13, 2019 at 13:08