How Do You Calculate Terminal Velocity Involving Drag?

[spacetimedude]

Hello PF,
I have once simple (well, not so simple for me) question.

I'm trying to derive an equation for the velocity of a falling body with accordance to terminal velocity.

The equation incorporates drag proportional to the speed.

m*dv/dt=mg-bv

and

mg/b=terminal velocity v_t

So the steps I took were:

m*dv/dt+bv=mg

(m/b)*(dv/dt)+v=v_t

dv/dt=(b/m)(v_t-v)

dv/(v_t-v)=(b/m)dt

Integrating both sides would give
ln[(v_t-v)/(v_t)]=(b/m)t

But the textbook says that I'm supposed to get negative (b/m)t on the left side.

Have I made a mistake on the integration part?

Any help will be deeply appreciated.

 

[spacetimedude]

hmm. I seem to have gotten the answer if I just divided the entire equation by -b in the beginning without bringing the bv to the left side. Have I made a mistake on the integration part?

 

[king vitamin]

You just missed the minus sign while integrating (chain rule)

 

[spacetimedude]

pshh. I can't believe I missed that. Thanks so much king vitamin!

 

[PJC01]

Hello,

Thank you for your question. It appears that you are on the right track with your derivation. The equation you have derived is known as the differential equation for terminal velocity, and the solution you have found is correct. However, the textbook may be referring to the final form of the equation, which involves taking the natural logarithm of both sides and then rearranging the terms to isolate the velocity. This would result in a negative term for the coefficient of time, as you mentioned.

To clarify, here is the full derivation:

m*dv/dt + bv = mg

Dividing both sides by m:

dv/dt + (b/m)*v = g

This is a first-order linear differential equation, which can be solved using the integrating factor method. The integrating factor is e^(b/m*t), so multiplying both sides by this factor:

e^(b/m*t)*(dv/dt) + (b/m)*e^(b/m*t)*v = e^(b/m*t)*g

Applying the product rule on the left side, we get:

d/dt(e^(b/m*t)*v) = e^(b/m*t)*g

Integrating both sides with respect to time:

∫ d/dt(e^(b/m*t)*v) dt = ∫ e^(b/m*t)*g dt

Using the fundamental theorem of calculus, the left side becomes e^(b/m*t)*v, and the right side becomes e^(b/m*t)*(g*t + C). So we have:

e^(b/m*t)*v = e^(b/m*t)*(g*t + C)

Dividing both sides by e^(b/m*t), we get:

v = g*t + C

To find the value of C, we can use the initial condition that at t=0, v=0. So plugging in these values, we get C=0. Therefore, the final equation for velocity as a function of time is:

v = g*t

To find the terminal velocity, we set the drag force equal to the weight of the object, as you did in your derivation:

mg = bv

Solving for v, we get:

v = mg/b = vt

Substituting this back into our original equation:

dv/dt + (b/m)*vt = g

Integrating both sides with respect to time:

∫ dv/dt dt + (b/m)*vt^2 =