1. ## Finding the function

Hello,

I need help with this one.

"During a boat trip, the engine suddenly stops. If the breaking force is proportional to the velocity it is possible to set up a differential equation $\displaystyle m*dv/dt=c*v$ where m is the mass of the boat, v is the velocity, t is the time and c is the braking force"

a) Determine v as a function of time if t=0 yields v=$\displaystyle v_0$

How am i supposed to do this?

2. Hello, Jones!

During a boat trip, the engine suddenly stops.

If the braking force is proportional to the velocity,
it is possible to set up a differential equation: $\displaystyle m\cdot\frac{dv}{dt}\:=\:kv$
where $\displaystyle m$ is the mass of the boat, $\displaystyle v$ is the velocity,
$\displaystyle t$ is the time and $\displaystyle k$ is the braking force.

a) Determine $\displaystyle v$ as a function of time if $\displaystyle t=0$ yields $\displaystyle v = v_0$

We are given: .$\displaystyle m\cdot\frac{dv}{dt} \:=\:kv$

Separate variables: .$\displaystyle \frac{dv}{v}\:=\:\frac{k}{m}dt$

Integrate: .$\displaystyle \ln v \:=\:\frac{k}{m}t + c$

. . and we have: .$\displaystyle v \:=\:e^{\frac{k}{m}t + c} \:=\:e^{\frac{k}{m}t}\cdot e^c \:=\:e^{\frac{k}{m}t}\cdot C$

Hence: .$\displaystyle v(t)\:=\:Ce^{\frac{k}{m}t}$

We are told that: .$\displaystyle v(0) = v_o$
. . So we have: .$\displaystyle v_o \:=\:Ce^{\frac{k}{m}(0)}\quad\Rightarrow\quad C = v_o$

Therefore: .$\displaystyle \boxed{v(t)\;=\;v_oe^{\frac{k}{m}t}}$

3. Im i supposed to know this stuff?

Thanks for the help.

4. Originally Posted by Soroban
Hello, Jones!

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We are given: .$\displaystyle m\cdot\frac{dv}{dt} \:=\:kv$

Separate variables: .$\displaystyle \frac{dv}{v}\:=\:\frac{k}{m}dt$
Isn't $\displaystyle dt$ just with respect to the variable you are differentiating?

Why are we substituting for $\displaystyle dt$?

5. Originally Posted by Jones
Im i supposed to know this stuff?

Thanks for the help.
Originally Posted by Jones
Isn't $\displaystyle dt$ just with respect to the variable you are differentiating?

Why are we substituting for $\displaystyle dt$?
All Soroban is doing is the "separation of variables" method of solving a differential equation.

-Dan