1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple 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

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple kinematics problem
Loading...