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A boat's acceleration is proportional to its velocity

  1. Sep 25, 2005 #1
    A 1000kg boat is traveling at 90km/h when its engine is shut off. The magnitude of the frictional force f between the boat and water is proportional to the speed v of the boat: f=70v where v is in meters per second and f is in newtons. Find the time required for the boat to slow to 45 km/h.

    This problem wasn't assigned so I might be trying something I'm not supposed to know how to do. I have a FBD and have defined the x-axis in the direction of the boat's motion.

    therefore v(t)=[tex]-\int .070vdt [/tex]

    I wish I could show more work but I'm don't know where to go next. I think this employs some calculus I'm not familiar with so if someone could just point out a concept I need to look at I would appreciate it.
  2. jcsd
  3. Sep 25, 2005 #2


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    Write a=(dv/dt).
    So, you have (dv/dt)=-kv, where k is a constant.

    This is a "differential equation for v".
    To solve for v, you have to find a function v that satisfies this equation. You don't need a class in differential equations to solve this, however.

    Can you think of a function of t whose derivative is proportional [with a negative constant] to itself? If you can't you can try a technique called "separation of variables" to obtain such a function.
  4. Sep 25, 2005 #3
    Thanks I got it.
  5. Jan 28, 2010 #4
    To keep the problem real use drag force proportional to the velocity squared as:

    Water Drag = 1/2 xCd x A x V^2 or proportional to V^2 with Cd and A constant.

    Same functional relation for air drag except different Cd and A
    Last edited: Jan 28, 2010
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