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Help with Acceleration question

  1. Apr 23, 2012 #1
    A car starts from rest and moves in a straight line with constant acceleration. After it has started moving it covers 60m in 5s and 170m in the next 6 seconds. Find the acceleration of the car?

    I thought to use v1=d/t then once I have the velocity for the first 5s to then use a=v minus u/t

    Please help
     
  2. jcsd
  3. Apr 23, 2012 #2
    Hint:

    You can compute the average speed in each leg of the displacement. Because acceleration is constant, the average speed is also the sum of the initial and final speeds divided by 2 for each leg. The final speed in each leg can also be represented by Vf = Vi + at.
    The final speed at leg 1 is intial speed at leg 2.
     
  4. Apr 23, 2012 #3
    Thank you so much for your reply..I am not 100percent sure I understand, but nevertheless here's what I understood

    Speed1= 60/5=12 12/2= 6m/s

    V=u minus at
    a= 6 minus 6/5 = 4.8m/s/s

    Speed2=170/6 = 28.3 28.3/2 = 14.16m/s
    v=u minus at
    a=14.16 minus 4.8/6 = 1.593m/s/s

    This that correct by any chance?
     
  5. Apr 24, 2012 #4
    Here is some more help.

    Average speed leg 1 is 12 m/sec. Since this is constant acceleration, the average speed may be represented as:

    Vbar = (Vi + Vf)/2 = 12

    Therefore Vi= 24 - Vf

    But the final velocity can also be determined by:

    Vf = Vi + a * t

    Substitute in for Vi from first equation and using time:

    Vf = 12 + 2.5 * a

    So now you have the acceleration in terms of the final speed in leg 1. Now do the same for leg 2 realizing that Vf for leg1 is Vi for leg 2. Eliminate the velocity terms leaving only the acceleration a to solve.
     
  6. Apr 25, 2012 #5
    Thank you :approve:
     
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