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Velocity and acceleration problem

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data
    The car goes with a initial velocity of 234km/h, and accelerates 9m/s2 for time of 20s. Find the displacement of the car in accelerating uniformly motion and the final velocity. Also the find the change in velocity !##Δv##

    2. Relevant equations
    ##v_i=234\frac{km}{h}=234\frac{1m}{3.6s}##
    ##a=9\frac{m}{s^2}##
    ##t=20s##
    ##d=v_0t+\frac{1}{2}at^2##(Find)
    ##v_f=v_0+at##(Find)
    ##Δv=v_f-v_i## (Find)

    3. The attempt at a solution
    Just waiting an answer :) thanks! (i know you may not answer because you don't do our homework but you help us, but i really need help for this)
     
    Last edited: Feb 14, 2012
  2. jcsd
  3. Feb 14, 2012 #2
    You have the equations to use. The speed is given in km/hr but the acceleration rate is in m/sec^2. You must convert to like units before adding the terms in the displacement equation.
     
  4. Feb 14, 2012 #3
    I see, but
    , what does that mean, does anything change?
     
  5. Feb 14, 2012 #4
    It should be stated "accelerating uniform motion". Motion is a noun so it needs an adjective not an adverb as a modifier.

    That said, the term means that the acceleration is constant. The equation is not valid unless the acceleration is constant over the specified time, t.
     
  6. Feb 14, 2012 #5
    Oh thanks so:
    ##v=234 \frac{km}{h}=234 \frac{1m}{3.6s}=65 \frac{m}{s}##
    ##v_f=v_i+at=65 \frac{m}{s} + (9 \frac{m}{s^2})(20s)=65 \frac{m}{s} + 180 \frac{m}{s}=245 \frac{m}{s}##
    ##d=v_0 t + \frac{1}{2} at^2=(65 \frac{m}{s})(20s)+ \frac{1}{2}(9 \frac{m}{s^2})(20s)^2=1300m+1800m=3100m=3.1km##
    Is that correct?
     
    Last edited: Feb 14, 2012
  7. Feb 14, 2012 #6
    Yes, it is correct.
     
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