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Homework Help: Simple motorboat kinematics problem

  1. Sep 13, 2010 #1
    I'm stuck doing this problem in my first year mechanics class. I feel sort of dumb for not being able to get it. The teacher wants us to just memorize the formulas but I really really really don't want to do that - I'm trying to just do it using the basics and calculus. I can obviously do this problem by plugging in numbers, but I'm looking for a way to do it from scratch.

    1. The problem statement, all variables and given/known data
    A motorboat traveling on a straight course slows dow uniformly from 75 km/h to 40 km/h in a distance of 50 m. What is the magnitude of its acceleration?

    2. Relevant equations
    The one that you're "supposed" to used to get the right answer is v^2=vo^2+2a(x-xo).
    I'm just trying to figure out why one would derive it this way - why the square speed? Obviously it's dimensionally needed to get m/s^2 for the answer... but I'm just looking for any explanation I can get.

    3. The attempt at a solution
    several approaches I've tried...
    v(t)=(75-at)km/h - but don't know how to find what t would be, since only the distance (50m) is given

    average acceleration = (75-40)/t... again, the problem is that I can't find t.

    I would appreciate any help you guys could give me. And if I can provide any more information about my thoughts so far let me know. Thanks
  2. jcsd
  3. Sep 13, 2010 #2


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    Gold Member

    From scratch eh? Okay:

    You start with the acceleration as a function of time:

    a(t) = a = const. (1)​

    Integrate this once with respect to time to get the velocity as a function of time :

    v(t) = v0 + at (2)​

    Here, v0 is the constant of integration, and it is obviously equal to the initial velocity (which you'll see if you set t = 0). Integrate this function with respect to time to get the position function:

    x(t) = x0 + v0t + (1/2)at2 (3)​

    Now you've derived all but one of the kinematics formulae. The last one makes no explicit mention of time. To eliminate t from the equations, solve for t in (2):

    t = (v - v0) / a​

    Substitute this into (3), rearrange, and you will end up with the result that:

    v2 - v02 = 2a(x - x0) (4)​

    An easier way to derive (4) is to use the work-energy theorem. As soon as your prof teaches you what that is, try it!
  4. Sep 13, 2010 #3
    Thank you so much! Eliminating t by setting it equal to (v-vo)/a was just what I was missing. Thanks!
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