How does instantaneous velocity cause displacement in just one point? 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.
 A: 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!
