Find the function of velocity of a person that accelerates to a constant v

[diazdaiz]

Homework Statement

how to find velocity function of accelerating object toward constant velocity?

Relevant Equations

S= a(1-r^n)/(1-r)

for example, I want to know velocity of a person when time is equal to t, that person start running from 0m/s (t=0s) to max velocity of 1m/s (t=1s). I am thinking that this is like rain droplet that affected by gravity and drag force, where force is directly proportional to its velocity, to make it simple, I am ignoring mass in this equation because this equation might also have different proportionality with drag force. Here I will just state that this is function of velocity, so when the person have velocity of 1m/s, he will no longer accelerating (a=0m/s^2), but when the person is not yet reach 1m/s, the person will have less and less acceleration as the person velocity approaching 1m/s. the problem is, what is the person acceleration function proportional to?. If we use discrete function of the person velocity with dt=1s, it would be easy since we know its based and max velocity of that person, but when we take smaller dt, it would be hard since we know the acceleration is not constant

(to be honest, I am not sure with my own solution) the first idea that's come up in my mind is to use geometric series summation "S=a(1-r^n)/(1-r)" for finding dv (acceleration*dt) on each step, since we know that S (in this case velocity) is max velocity (1m/s), and n is how many discrete step to reach 1s (in this case dt=1s/n), and a is the first acceleration when v=0m/s, so the dv we add from step to step is acceleration*dt. But the problem is I don't know what is acceleration proportional to, therefore I don't know what the first value in the geometric series. I also having problem to separate r from the summation equation.

is the solution above heading in the right direction?, if it is, how can I find value of acceleration in each step?, and can someone might help me in separating r from equation. If my solution not right, can someone give me a clue to right equation

 

[haruspex]

The standard way to approach such a question is to write a differential equation to represent the process, then try to solve it.

 

[diazdaiz]

The standard way to approach such a question is to write a differential equation to represent the process, then try to solve it.

I am actually new to differential equation, but I will try to write what I know, please help to correct it:
since acceleration get smaller as it get closer to max velocity and become 0 when it get equal or more than max velocity, I think the equation will become look like this(?):
$dv = (a_0 - kv)*dt$
I add $a_0$ because when v=0m/s, we will still have acceleration.
and we want to know the velocity when time is equal to t, t=0s to t=t is a boundary, so if we integral both side..., wait I realize that we need to have v in left side of the equation right?. Also I realize that $a_0$ and $k$ will be unknown(?)

[edit]
k is a function of v? when v>=max velocity, k should make $kv= a_0$?

 

[haruspex]

I am actually new to differential equation, but I will try to write what I know, please help to correct it:
since acceleration get smaller as it get closer to max velocity and become 0 when it get equal or more than max velocity, I think the equation will become look like this(?):
$dv = (a_0 - kv)*dt$
I add $a_0$ because when v=0m/s, we will still have acceleration.
and we want to know the velocity when time is equal to t, t=0s to t=t is a boundary, so if we integral both side..., wait I realize that we need to have v in left side of the equation right?. Also I realize that $a_0$ and $k$ will be unknown(?)

[edit]
k is a function of v? when v>=max velocity, k should make $kv= a_0$?

That's a good start. More generally, it could be $\frac{hdv}{dt}=a_0-kv^{\alpha}$.
In the case of drag in air, $\alpha$ is 1 at low speeds but increases to 2 at higher speeds.
Can you solve such an equation (for constant $\alpha$?).