# Find velocity at the given point

I'm fighting with a problem of a sad sort. It's when you face a task with allusion to physics in the pure mathematical book. Here it goes:

We have a boat which moves in the still water. We have one pointer about nature of this motion: "Resistance of environment causes slowdown proportional to velocity".
At the moment, when engine was turned off, velocity was 200 m/min. We know, that 30 sec since that event, velocity was 100 m/min. Task is to find what velocity will boat have in 2 minutes?

We are given an correct answer to check with - it's should be 12.5 m/min
(Sorry it's likely to be 12.5 but maybe 125 - the print is fuzzy)

Let's review results of my frustration:
As I get it - they were saying that acceleration changes linearly. Hence I assumed, that jerk is constant: $j = c$. Then $a(t) = \int j dt = jt$. Accordingly, $v(t)= \int a(t)dt=\int jtdt = \frac {jt^2} {2}$
Convert given velocities to standart units $v_0=3.333 \frac m s$ ,$v_1=1.666 \frac m s$.

$1.666 = 3.333 + \frac {jt^2} {2}$. But $j$ we receive makes no sense.

So, I would appreciate any help here. But guys, keep in mind that, after all, it is more of mathematical problem about derivatives, so physical model probably is simple.

"They were saying that acceleration changes linearly"

What they are actually saying is this:

The rate of change of velocity (=acceleration) is proportional to the velocity of the particle.
And the constant of proportionality is negative because the drag force is opposite the existing velocity.

$\vec{a} = \frac{d \vec{v} }{dt}= -\gamma \vec{v}$

This kind of an equation is classic. If the rate of change of something is proportional to the thing itself, then it's a sure sign of an exponential variation.
And it will turn out so here too.

Since the problem is one dimensional, we can switch over to scalar notation and solve the differential equation by separating the variables to get:

$ln(\frac{v_t}{v_o})=-\gamma(t-t_o)$

You now know the relation between velocities at any two instants of time.

You should be able to verify that the answer is indeed $\approx 12.5 ms^{-1}$