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Bonus Physics Problem

  1. Oct 30, 2003 #1
    Yea well I've run into this bonus physics problem thats really hard.

    Ok first off the reason its bonus is cause it involves using integration (which we haven't covered yet)

    But our teacher says thats what makes it a bonus problem.

    Ok here is the problem.

    A 1000 kg boat is traveling at 90 km/h where its engine is shut off. The magnitude of the frictional force f(k) between boat and water is proportional to the speed v of the boat: f(k) = 70v, where v is in meters per second and f(k) is in newtons. Find the time required for the boat to slow down to 45 km/h.

    1000 kg - boat
    90 km/h - 25 m/s
    45 km/h - 12.5 m/s

    f(k) = 70v

    a = dv/dt

    f(k) = ma
    f = (1000 kg)(dv/dt)

    Integration Rule.

    f(V) = V^X -> f(V)dt = (V^X+1/X+1) + C

    dt = 1000kg (70v^2/2) + C
    dt = 1000 kg 35v^2 + C
    dt = ?

    (yea....i'm not sure where to from here...)

    Um, yea. Any help would be appreciated.


    Tony Zalles
  2. jcsd
  3. Oct 30, 2003 #2


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    Staff Emeritus
    Science Advisor

    Yes, f= ma= m dv/dt. In this situation the friction force is "proportional to the speed v of the boat: f(k) = 70v"

    (Actually, that should be f= -70v since the force is always opposite to the direction of motion. I wouldn't actually write "f(k)" since f depends on v, not k. This is telling you that k= 70.)

    f= ma becomes -70v= 1000 dv/dt or dv/dt= -0.07v

    You might try rewriting this as (1/v)dv= -0.07 dt or
    v-1dv= -0.07 dt and integrating. Unfortunately, the integration rule you give: the integral of xndx= 1/(n+1)xn+1+ C doesn't work here: n= -1 so n+1= 0 and you can't divide by 0!

    The formulas you have:
    aren't correct. You have the v in the denominator.
    (By the way, if the v were in the numerator, you get that result by integrating on BOTH sides. You would have "t= ...", not dt.)
  4. Nov 1, 2003 #3
    ehh....sorry I couldn't reply earlier.

    um when I typed in f(k) I meant f subscript k, which is the coefficient of friction. Not f(k) as in, f*k.

    So sorry about that secondly...um I well picked up where I left off and came down to.

    .035v^2 + C = dt

    But my issue here is what is v? and C?

    See cause the answer to this problem (from that back of our text) is 9.9s.

    I know also that dv is 12.5 m/s

    and that the forces evaluated by, f(k) = 70v, at 25 m/s and 12.5 m/s, are: 1750 N and 825 N

    I'm still a bit stumped....

    perhaps maybe you set 70v = m(dv/dt)....but then I don't know how to resolve my constant when integrating or what to do about v, since dv is already 12.5 m/s. I don't believe then in this case dv = v.

    Yea...again, sorry for the mistype earlier and any help would ofcourse be appreciated.

    By the way thans for the reply hallsofivy

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