Solving Boat's Motion Under Retarding Force

[cscott]

Homework Statement

boat turns off engine when it's attained v_0 at t = 0. Starting from that moment the boat is slowed by retarding force F = Ce^{-kv}

The Attempt at a Solution

m\dot{v} = -Ce^{-kv}

Throwing this in Maple gives me

v(t) = \frac{1}{k}\ln\left(\frac{-kC(t + C^{'})}{m}\right)

This doesn't make sense to me--you can't take the log of a negative number and even if I remove that negative from the initial DE v(t)->inf as t->inf

 

[HallsofIvy]

What makes you think this involves the logarithm of a negative number? Whether -kC(t+ C') is negative or not depends on C'. In particular, you know that when t= 0, v= v₀. Putting this into your formula
v(0)= v_0= \frac{1}{k}ln\left(\frac{-kC(C')}{m}\right)
which gives
\frac{-kC(C')}{m}= e^{kv_0}
or
C'= -\frac{m}{C}e^{kv_0}
which is negative. You will be taking the logarithm of a positive number as long as t< -C'. But when t= -C', what happens?


That's the problem with using Maple, or a computer or any such rather than doing it yourself! If you have done the rather simple integral yourself you wouldn't have had that question.

 

[cscott]

EDIT: thinking

 

[cscott]

You missed the 'k' when solving for C'

C&#039; = -\frac{m}{kC}e^{kv_0}

 

[cscott]

Think I got it

 

[cscott]

Thanks