# Horizontal Motion with Quadratic Drag

## Homework Statement

Consider an object that is coasting horizontally (positive x direction) subject to a drag force f = -bv - cv^2 . Write down Newton's second law for this object and solve for v by separating variables. Sketc the behaviour of v as a function of t. Explain the time dependence for t large. (Which force term is dominant when t is large?)

## The Attempt at a Solution

m(dv/dt) = -bv -cv^2

m(dv/dt) = -v(cv + b)

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

dv/v + (cv/m)dt = (-b/m)dt

Then... do I integrate with respect to v for the dv terms and t for the dt terms? I can't figure out what this will give me with the second term on the right...

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You need all the 'v' terms together with 'dv'. But you subtracted 'cv' in line 3 which is a mistake. Instead divide the whole equation in line 2 by 'v(cv+b)'. Then solve for it.

I get to the point where all my v terms are on one side so that I have:

(m dv)/[v(cv+b)] = -dt

From here when I try and integrate with respect to v on the left and t on the right, I'm doing a definite integral setting the limits between v and v0 and t and t0 so I get:

m[ln(cv^2 + bv) - ln(cv0^2 - bv0^2)] = -t

ln(cv^2 + bv) = -t/m + ln(cv0^2 - bv0^2)

Then if I take the exponential of both sides I still end up in a situation where I can't solve for v because there will be v on both sides... i.e

cv^2 + bv = e^(-t/m) + cv0^2 - bv0^2

How can I solve this for v?

You need to check your integration, it is incorrect.

How exactly do I integrate m(dv)/[v(cv+b)] ? The way I do it I consistently get natural log expressions which I am guessing is wrong because the function should tend towards 0 when I plot it...

gabbagabbahey
Homework Helper
Gold Member
How exactly do I integrate m(dv)/[v(cv+b)] ? The way I do it I consistently get natural log expressions which I am guessing is wrong because the function should tend towards 0 when I plot it...
$$\frac{d}{dv}\ln|cv^2+bv|=\frac{2cv+b}{cv^2+bv}\neq\frac{1}{cv^2+bv}$$

To integrate your expression properly, try completing the square on the denominator and then making an appropriate substitution.

Another trick is to break the fraction up into to fractions that look like this:

$$\frac{1}{v(cv+b)} = \frac{A}{v} + \frac{B}{cv+b}$$

Then solve for A and B. Integrating the two fractions separately will be much easier.

Ah that clears things up. I see where I went wrong there, thanks for your help!