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I have a question. Falling object graph is curve shape right? And instantaneous velocity is tangent line but how does this velocity make displacement in distance? Because suppose instantaneous velocity is 10 m/s but it's just one point or instant of time it's not make any displacement if it make it should 2*t( time more than zero) but in graph it's just one point.

Qmechanic
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Heroz
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    Related post by OP: https://physics.stackexchange.com/q/746579/2451 – Qmechanic Jan 23 '23 at 11:27
  • Velocity does not make (cause) displacement. Instantaneous velocity is just a description of the current state of things - it tells the current speed of the object, and its current traveling direction. So, it's not the cause of motion. Two things will cause the displacement of the object in this example: (1) Newton's first law - if the object is already moving, and nothing prevents it (or aids it), it will continue to move in just the same way on its own, forever, and (2) the the force of gravity, which will pull the object down, and bend its path, thus causing a change in inst. velocity. – Filip Milovanović Feb 02 '23 at 16:55

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Your question echoes Zeno's paradox, if I understand you correctly. Basically you are asking how come a derivative can have a non zero value if it is defined at a point. Well from elementary calculus, we learn that it is not quite true that the derivative depends only on a single point, but also on its neighborhood, for the case of a velocity, assuming $y(t)$ is the function that relates time $t$ to the height of a falling object $y(t)$:

$$ v(t) = y'(t) = \lim_{\Delta{t}\to 0}\frac{y(t+\Delta{t})-y(t)}{\Delta{t}} $$

So you see, for $y'(t)$ to be defined at any point $t$, it must be true that $y$ has a defined limit when it approaches the value $y(t)$ which is saying something about the neighborhood of $y(t)$. There's a lot more that can be said about this, but this is only to give you a feel that the derivative doesn't really depend only on a single point.

Now the next thing to realize is that a displacement of the falling object which occurs between nearby times $t=t_1$ and $t=t_2$ is the velocity (speed in this case, say at $t_1$) times this small time interval: $y'(t_1)\cdot(t_2-t_1)$, which also explains how such a displacement is generated by this instanteneous speed.

Note: This is only a good approximation for when the speed doesn't vary significantly (or at all) between times $t_1$ and $t_2$. Otherwise you will need to integrate the speed (or velocity for the general case) function across the time interval to calculate the precise displacement!

Amit
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