a bit more basic than the usual thread, but..

Problem (unit adjusted):

A 1000 kg boat is travelling at 25 m/s when its engine is shut off. The magnitude of the frictional force fk between the boat and water is proportional to the speed v of the boat: fk=70v, where v is in meters per second and fk is in newtons. Find the time required for the boat to slow to 12.5 m/s.

Solution:

I assume the forces in the y-direction have nothing to do with the problem (Fn=Fg).

forces in the x-direction once engine is shut off:

fk=-70v

fk=fnet

so,

fnet=1000a

therefore,

-70v=1000a

or

-0.07v=a

if

a=(v-vo)/(t-to)

then

-0.07v=(v-vo)/(t-to)

if

vo=25 m/s

v=12.5 m/s

t-to=dt

then

-0.07(12.5)=(12.5-25)/dt

so

dt=(12.5-25)/-0.07(12.5)

dt=14.29 sec.

--------------

So my problem is that the book gets 9.9 seconds as the answer.

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If anyone can let me know where I went wrong it would be greatly appreciated!

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Thanks,
James

 

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\\(a = \frac{v_{1}-v_{0}}{t_{1}-t_{0}\) gives you the average acceleration but that isn't what you're working with, you have a varying acceleration.

If acceleration is proportional to v then you have \(ma = F = -kv\) where k = 70 (minus sign since its decelerating). Thus \(a = \frac{-70}{m}v\). This is a differential equation and you use the identity \(a = \frac{dv}{dt}\) and so you have \(\frac{dv}{dt} = \frac{-70}{m}v\). Solve this with the initial conditions and you'll get your answer.

 

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thanks for the quick reply!

Dear AlphaNumeric,

Thanks a lot for the quick reply! It's much appreciated.

I did a bit of work from your advice, unfortunately I got the same answer as before! Perhaps, I have done something wrong once again. Below is my work:

-0.07v=a

a=dv/dt

-0.07v=dv/dt

-0.07 integral(v)dt = integral(dv)

-0.07(vt+so)=v

so=0

-0.07vt=v

t=-0.07^-1

t=14.29 sec.

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Thanks ahead a time for any help!!

 

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You've solved the equation incorrectly. Suppose you have the following equation : \(kv = \frac{dv}{dt}\)

Bring all the v's on one side and all the t's on the other, \(kdt = \frac{dv}{v}\). Integrate this and you get \(kt + C = \ln v\) where C is the integration constant. Use that in your methods instead.

 

... thanks a lot!

So, that's quite embarrassing.

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Thanks though!

 

I've done far worse myself in my time.