Friction and nonconstant acceleration?

  • Thread starter Zorodius
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A problem in my book reads as follows:

A 1000 kg boat is traveling at 90 km/h when its engine is shut off. The magnitude of the frictional force [itex]\vec{f_k}[/itex] between boat and water is proportional to the speed v of the boat: [itex]f_k = 70v[/itex], where v is in meters per second and [itex]f_k[/itex] is in newtons. Find the time required for the boat to slow to 45 km/h.

My question with this is: g'nhuh? If the magnitude of the frictional force is a function of velocity, that seems to imply that the acceleration is not constant. I was under the impression that the equations for motion and friction that I had been given so far applied only to constant acceleration. I tried to solve this by converting the measurements into meters per second (25 m/s when the engine is shut off, slows to 12.5 m/s) and then guessing that, since a=f/m, then a=70v/1000, and perhaps I could say v = 25 - 70 v / 1000 * t. I solved this for v, and graphically found that v = 12.5 when t is about 14.6. Unfortunately, that wasn't the right answer, which isn't particularly surprising since I'm unsure where to go with this from the very start.

A little help?
 

Answers and Replies

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Doc Al
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integrate!

Zorodius said:
My question with this is: g'nhuh? If the magnitude of the frictional force is a function of velocity, that seems to imply that the acceleration is not constant. I was under the impression that the equations for motion and friction that I had been given so far applied only to constant acceleration.
The force and the acceleration are not constant. You can't apply the simple kinematic equations for uniform acceleration to this problem. You have to apply F=ma to set up a simple differential equation. You'll need to integrate! I hope you've covered a little calculus.

[tex] F = ma = m \frac{dv}{dt}[/tex]
[tex] 70v = m \frac{dv}{dt}[/tex]

Etc...
 

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