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Finding Velocity

  1. Sep 14, 2014 #1
    1. The problem statement, all variables and given/known data
    A car is traveling along a straight road at a velocity of +34.2 m/s when its engine cuts out.
    For the next 2.09 seconds, the car slows down, and its average acceleration is a1. For the
    next 6.49 seconds, the car slows down further, and its average acceleration is a2. The velocity of the car at the end of the 8.58-second period is +17.5 m/s. The ratio of the
    average acceleration values is a1/a2 = 1.67. Find the velocity of the car at the end of the initial 2.09-second interval.

    2. Relevant equations
    Using the kinematic equations

    part 1:
    vo = 34.3m/s

    part 2:
    3. The attempt at a solution

    1. using v=vo + at for part 1, v1 =34.2m/s + (1.67a1)(2.09s)
    2. plug in what v1 equal to part 2
  2. jcsd
  3. Sep 14, 2014 #2
    what is the link between a1 and a2? Can you think of a way to replace a2 once you plug in what v1 equal to part 2?
  4. Sep 14, 2014 #3
    well i initially made a1=1.67(a2) but i'm having a hard time finding out how to plug in V1 into part 2.

    so far my equation for part 2 is: 17.5m/s = V1+6.49 s (a2)
  5. Sep 14, 2014 #4
    If you make a1 = 1.67a2, then when you replace the a1, you'd be solving for a2 and that acceleration is in the 6.49s part not the 2.09s part the question is asking for.

    V=vo +at


    V2=vo + a2t2
    What equation does the vo represent?

    To help you see the connection better:

    v1 =34.2m/s + (a1)(2.09s)
    17.5m/s = V1+(a2)(6.49 s )

    In other words, Can you connect these two equations into one?
    Last edited: Sep 14, 2014
  6. Sep 14, 2014 #5
    Since V0=V1, does V1 = 34.2m/s +(1.67a2)(2.09s) = 34.2m/s + 3.4903a2
  7. Sep 14, 2014 #6
    Yes vo=v1

    So substitute the equation for v1 into vo

    (use the equations under my edit: because your part 1 equation is wrong.)
    (You had acceleration in your part 1 as 1.67a1 when it just needs a1)
  8. Sep 14, 2014 #7
    I'm sorry if i'm making some algebra mistake but would putting part 1 equation into part 2 be:

    17.5m/s=(34.2m/s +2.09s a1) +6.49s a2
  9. Sep 14, 2014 #8
    Thats correct.

    You can replace a2 with a1/1.67. Solve for a1 and plug it into the first equation for your speed after the 2.09s mark.

    (34.2m/s + 2.09s a1)
  10. Sep 14, 2014 #9
    so far i have:

    17.5m/s = (34.2m/s +2.09s (1.67a2)) + 6.49s a2

    -16.7m/s = 2.09s(1.67a2) + 6.49s a2

    -16.7m/s = 3.4903sa2 + 6.49sa2

    -16.7m/s = 9.9803sa2
    -1.67 = a2
  11. Sep 14, 2014 #10
    You need to replace a2 with (a1/1.67) not the other way around. (i explained this in a earlier post)

    17.5 = (34.2 + 2.09(a1)) + 6.49((a1)/1.67)

    Now solve for a1 and once you find that use this equation (from part 1) to get your velocity after the 2.09s mark.

    (34.2m/s + 2.09s a1)

    If you need help with this lmk. :)
  12. Sep 14, 2014 #11
    so would a1 = -3.2504662

    17.5 = 34.2 + 2.09a1 + 6.49 (a1/1.67)
    subtracted 17.5 by 34.2
    -16.7 = 2.09a1 + 6.49 (a1/1.67)
    multiply both side by 1.67
    -27.889 = 2.09a1 + 6.49a1
    add both a1 and divide the number by-27.889
    a1= -3.2504662
  13. Sep 14, 2014 #12
    -16.7 = 2.09a1 + 6.49 (a1/1.67)
    multiply both side by 1.67
    -27.889 = 2.09a1 + 6.49a1

    Here you made a mistake. If you choose to multiply 1.67, you have to do that to all terms. You missed the 2.09 term. I would just take 6.49/1.67 a1. Pull the a1 out and add the 2.09 and (6.49/1.67) together and divide.

    a1 should be something like -2.79


    -16.7 = a1(2.09+(6.49/1.67))

    should look something like that if I didn't explain that clear enough.
  14. Sep 14, 2014 #13
    Correct a1= -2.79264...

    so from here it would be:

    v1= 34.2+2.09(-2.79) = 28.3689
  15. Sep 14, 2014 #14
    yep, that is what I got, dont forget units and significant digits :approve:
  16. Sep 14, 2014 #15
    Big Thanks. really appreciated
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