# Drag (physics)

1. Dec 10, 2016

### betty0202

1. The problem statement, all variables and given/known data
turning on the engine of a motorboat (v0=0),
K = constant force due to the engine
drag force of the water D = -cv
find v(t)=?
2. Relevant equations
integration
f=ma, a=dv/dt
3. The attempt at a solution

D+K = MA
K-cv = MA
(A=dv/dt)
K-cv=Mdv/dt
Mdv=dt(K-cv)
???
i want to do integration on both side of the
equation but I can't isolate V

thanks

2. Dec 10, 2016

### Staff: Mentor

You need only gather the v and t terms on opposite sides of the equation. Constants can be on either side. Divide both sides by (K - cv).

3. Dec 10, 2016

### betty0202

I hope I did the math right $\int_{}^{} (\frac{m}{k-cv})dv=\int_{}^{}dt$
$-m\ln(cv-k)=t$
$\ln(cv-k)=-\frac{t}{m}$
$v=\frac{e^{-\frac{t}{m}}+k}{c}$
??

4. Dec 10, 2016

### Staff: Mentor

You'll want to write the integrals as definite integrals (with specified limits of integration), otherwise you need to introduce and deal with the constants of integration. The starting limits for each integral are simple: they're both zero.

5. Dec 10, 2016

### betty0202

thank you

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