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As everyone know that the integral of acceleration respect to time will give the function of velocity respect to time. However, my question concern the different case when the function of a(t) is not given but a(v) instead. That means, the acceleration varies with the value of velocity. The time t is not the parameter here. Of course, the acceleration change the velocity and the velocity change acceleration. You can think of it like a feedback system which the velocity is the input that give the the output of acceleration and output is fed back again to change the input

So how can people can derive the function a(t) while they have only the function a(v). Supposed that the initial velocity is already known. We can't integrate a(v). If we had the function of v(t) then it would be easier, just derivate v(t) then we will know a(v) but now we have nothing but initial v and a(v).

For example, something with mass m is falling from sky and its speed is v. The gravitational force have maximized its speed but now the drag force slow it down. The height is not enough for it to reach the terminal speed so we must calculate its speed when it hit the ground here. We know the drag force so we can calculate the acceleration but as the acceleration decrease the velocity, it also decrease the acceleration as well. So how can a(t) can be derived from a(v)

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  • $\begingroup$ Why don't you try plotting the relationships on a graph? Can you plot velocity vs. time or velocity vs. distance? That might yield a recognizable shape whose formula can be discerned. $\endgroup$ – Inquisitive Feb 2 '15 at 23:27
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In the case you described the function $a(v)$ gives you the differential equation $$ \frac{d v}{d t}=a(v). $$ You just have to solve this differential equation for $v(t)$ with some initial condition $v(t=0)=v_0$. After solving the equation you get the acceleration by differentiating the solution $a(t)=\dot{v}(t)$.

For the example you mentioned: The drag force for an object of cross section area $A$ in a fluid with density $\rho$ is $F=\frac{1}{2}\rho v^2 C A$ where is $C$ is the drag coefficient coefficient, therefore the acceleration of a falling object is $$ a(v)=g-\frac{1}{2m}\rho v^2 C A, $$ where $g$ is the gravitational acceleration. Using the abbreviation $\xi=\frac{1}{2m}\rho C A$, we get $$ \frac{d v}{d t}=g-\xi v^2\\ \Rightarrow \int_{v_0}^{v(t)}\frac{dv'}{g-\xi v'^2}=\int_{0}^{t}dt'\\ \Rightarrow \left.\frac{1}{\sqrt{g \xi}}\operatorname{arctanh}\left(\sqrt{\frac{\xi}{g}}v'\right)\right|_{v_0}^{v(t)}=t\\ \operatorname{arctanh}\left(\sqrt{\frac{\xi}{g}}v(t)\right)-\operatorname{arctanh}\left(\sqrt{\frac{\xi}{g}}v_0\right)=t. $$ Now, you can invert that relation to get $v(t)$: $$ v(t)=\sqrt{\frac{g}{\xi}}\operatorname{tanh}\left[t+\operatorname{arctanh}\left(\sqrt{\frac{\xi}{g}}v_0\right)\right]. $$ Then $$ a(t)=\frac{dv}{dt}=\frac{g(g-v_0^2 \xi)}{\left(\sqrt{g}\cosh\left(\sqrt{g \xi}t\right)+v_0\sqrt{\xi}\sinh\left(\sqrt{g \xi}t\right)\right)^2}. $$ Well, the last few lines were some unpleasant (hopefully correct) algebra, but I hope the idea is clear. The situation you described leads you to solving a differential equation. The function $a(v)$ determines how difficult that task is.

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