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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?

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    $\begingroup$ If you wanted to know the object's weight (or price) at time 2, would you still "plug into" $2t^3$ ? $\endgroup$ – WillO Nov 13 '19 at 12:37
  • $\begingroup$ what are dimensions of number 2 - is it $[m/s^3]$ or is it just a dimensionless number ? $\endgroup$ – Agnius Vasiliauskas Nov 13 '19 at 13:05
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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

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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

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  • $\begingroup$ Any reason of downvote ? Because I have explained main physics rule about checking any physics equation correctness ? $\endgroup$ – Agnius Vasiliauskas Nov 13 '19 at 12:50
  • $\begingroup$ 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. $\endgroup$ – Bob D Nov 13 '19 at 12:59
  • $\begingroup$ Yes, it should be stated as so, because it looks like just dimensionless number. $\endgroup$ – Agnius Vasiliauskas Nov 13 '19 at 13:03
  • $\begingroup$ Yes I agree with you. But it seemed to me the OP’s confusion was with concepts not units $\endgroup$ – Bob D Nov 13 '19 at 13:06
  • $\begingroup$ Now it looks like someone down voted me. Oh well c’est la vie $\endgroup$ – Bob D Nov 13 '19 at 13:08

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