Free falling with air resistance question

[Clara Chung]

Homework Statement

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 traveled at time t.

Homework Equations

not sure, dv/dt=g-γv...(1) ,where γ=b/m, or v=v∞(1-e^(-γt))...(2) , v∞=g/γ...(3)

The Attempt at a Solution

the answer is d=mgτ/eb.
First, I don't understand why can't 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

PLEASE HELP , THX

 

[cnh1995]

First, I don't understand why can't I integrate equation (1) this way...
dv/dt=g-γv
v=gt-γd

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.

 

[cnh1995]

Derive an exact expression for the distance d traveled at time t.

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

 

[Delta2]

1.
3. The Attempt at a Solution
the answer is d=mgτ/eb.
First, I don't understand why can't 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

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.

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

PLEASE HELP , THX

 

[Clara Chung]

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.

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

 

[Delta2]

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

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

 

[cnh1995]

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

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

 

[Clara Chung]

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

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

 

[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 $t>>\tau$?

 

[Clara Chung]

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 $t>>\tau$?

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

 

[Clara Chung]

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

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

 

[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 $v_{\infty}$

 

[cnh1995]

as v at τ is the same as t>>τ

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.