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Daemidra

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Posts posted by Daemidra

  1. This question can be solved using one of the Equations of Motion, namely v^2 = u^2 + 2ax where 
    v is the final velocity, u is the initial velocity, a is the acceleration and x is the displacement. 

    Since the motorcycle starts from zero u =0 and can be discarded from the equation. 
    The equation is now v^2 = 2ax 
    If the displacement changes to 3x our equation becomes (new v)^2 = 2a3x 
    (new v)^2 = 6ax 
    but the velocity in the first case was given by v^2 =2ax: (new v)^2 = 3 v^2 
    Square root both sides new v = square root of 3 times v 
    Square root of 3 is 1.73

    From - https://answers.yahoo.com/question/index?qid=20110917100043AAuhgCE

    • Like 1
    1. A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a final velocity v. Then t is increased so that the displacement is

      3x. In this same increased time interval, what final velocity does the motorcycle attain? 

    The book says the answer is 1.73 v, but my teacher gave a bunch of variables when he solved it.

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