# Still Friction

So I posted this earlier and it got to a point where differentiation is neccessary. I am a little familiar with differentiation, but not to the point where I know how to apply the concepts I have recently learned in Calculus. i would greatly appreciate somebody walking me through the process...it should be fairly simple in this particular problem. Again, this problem is for my own personal practice, not for a class, and I am very interested in finding out how to apply these concepts.
Thanks~Casey

Original Post:

## Homework Statement

A 1000kg Boat is traveling 90km/h when its engine is cut. The magnitude of the frictional force fk is proportional to the boat's speed v: fk=70v, where v is in m/s and fk is in newtons. Find the time required for the boat to slow to 45 km/h.

## Homework Equations

Newton's Second
V^2=Vo^2+2a(X-Xo)
X-Xo=VoT+1/2at^2
V=Vo+aT

## The Attempt at a Solution

Vo=25m/s
V=12.5m/s
fk=70v=1750N

I drew a FBD and it seems that since the engine was cut, there is only fk in the x direction. Thus, fk=ma--->1750=-1000a-->a=-1.75
Then I used V=Vo+at---> t=(V-Vo)/a
-->t=(12.5-25)/-1.75=7.1
But this is not correct....9.9seconds is the correct solution.
~Casey

...It was pointed out that "a" is not constant. But I am not sure where to go from here as I have only dealt with problems dealing with constant acceleration.....what am I differentiating? I am not sure of the equation...or how to derive one. hollah.

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Yeah. The retardation is not constant since the magnitude of the frictional force depends on the speed of the boat.

So, at a particular instant of time, assume the speed to be v. At that moment, the force acting on it is kv in the -x direction (k=70). Which means that the instantaneous acceleration is v(k/m).

This means, $$\frac{dv}{dt}=\frac{kv}{m}$$
Integrating this equation, you get the expression, $$ln(v)=\frac{kt}{m}+c$$ where c is the constant of integration.

Now, at t=0, the engine was cut and the boat had a speed of 90km/h.
Putting that into the equation, you get c=ln(90).
$$ln(v)=\frac{kt}{m}+ln(90)$$, where ln is natural logarithm.