# Free falling with air resistance question

1. Jul 10, 2016

### Clara Chung

1. The problem statement, all variables and given/known data
An object falls under gravity and air resistance force bv. It is known there is a characteristic time τ=m/b such that the object comes close to the terminal velocity for t>>τ. Derive an exact expression for the distance d travelled at time t.

2. Relevant equations
not sure, dv/dt=g-γv....(1) ,where γ=b/m, or v=v∞(1-e^(-γt))....(2) , v∞=g/γ...(3)
3. The attempt at a solution
First, I don't understand why cant I integrate equation (1) this way.....
dv/dt=g-γv
v=gt-γd
d=(gt-v)/γ =gτ^2-gτ^2 =0 which is wrong

Then, I tried to integrate equation (2),
d=v∞(t+γe^(-γt))+C
put t=0,v=0,d=0
C=-v∞γ=-g
so,
d=v∞(t+γe^(-γt))-g
putting the information inside
d=gτ(τ+(1/τe))-g
=gτ^2 + g/e -g
=(gτ^2e+g-ge) /e

2. Jul 10, 2016

### cnh1995

dv/dt=g-γv is a differential equation which describes velocity as a function of time i.e. v(t). Velocity is not constant w.r.t. time but is a function of time .So, you can't integrate the RHS as ∫(g-γv)dt. You'll have to separate the appropriate variables.

3. Jul 10, 2016

### cnh1995

The answer given for d is true for t=τ and not t>>τ.

4. Jul 10, 2016

### Delta²

You can integrate that way (taking definite integrals in both sides with time points t=0 and $t=\tau$) but $v(\tau)$ is not equal to $g\tau$ because the object doesn't do free fall exactly.

5. Jul 10, 2016

### Clara Chung

Thanks. But at time t>>τ, I set dv/dt = 0, then dv/dt = g-γv , so itsn't v=gτ (1/γ = τ)?

6. Jul 10, 2016

### Delta²

yes that's correct. But this doesn't mean that d=0, it is still $d=(gt-v)/\gamma$ where $t>>\tau$ not $t=\tau$

7. Jul 10, 2016

### cnh1995

t>>τ implies that the velocity at time t is g/γ. At time τ, the velocity is still changing.

8. Jul 10, 2016

### Clara Chung

Oh I got it.:) But how to solve the question with equation (2) now?

9. Jul 10, 2016

### Delta²

You can continue with the initial approach and plug in $v(\tau)$ from the equation $v(t)=v_{\infty}(1-e^{-\gamma t})$.

Or as you go by integrating (2) I see a mistake in the integration, it is $\int -e^{-\gamma t}=\frac{e^{-\gamma t}}{\gamma}+C$

But you got me confused, you want d at time $\tau$ or d at time $t>>\tau$?

10. Jul 10, 2016

### Clara Chung

Oh my god.... thanks for reminding me, now I've solved it

11. Jul 10, 2016

### Clara Chung

I think the question need me to find t tends to τ, as v at τ is the same as t>>τ ,so my dv/dt can't set to be 0 as the curve hasn't level off. Not sure about it

12. Jul 10, 2016

### Delta²

No sorry velocity at time $t=\tau$ is quite different from velocity at time $t>>\tau$.

The velocity at time $t=10\tau$ is probably close enough to the final velocity $v_{\infty}$

13. Jul 10, 2016

### cnh1995

No. Terminal velocity is attained at t>>τ. As Delta2 said, it could be 10τ. In fact, the output is approximately equal to the steady state value after 5τ (99%). So practically, for any value of t after t=5τ, the output can be said to have reached steady state.