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Find the velocity of the stock car

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A professional race-car driver buys a car that can accelerate at 5.7 m/s^2. The racer decides to race against another driver in a souped up stock-car. Both start from rest, but the stock-car driver leaves 1.4s before the driver of the sports car. The stock-car moves with a constant acceleration of +3.6m/s^2.

    A) Find the time it takes the two drivers to travel before they are side by side.
    B) Find the distance the two drivers travel before they are side by side.
    C) Find the velocity of the race car when the two drivers are side by side.
    D) Find the velocity of the stock car when the two drivers are side by side.


    2. Relevant equations

    v=u+at

    X= ut + 1/2at

    quadratic formula

    3. The attempt at a solution

    My teacher assigns our homework online and we are only allowed a certain number of tries. I used all my tries up for part A so now I cant find B-D without the answer to A.

    First I found out what each car was doing in the first 1.4 seconds using x= ut + 1/2at then set them both to equal eachother and pluged the times and accelerations into the quadratic formula.

    Ive been working on this one problem all week its driving me crazy. PLEASE HELP !!
     
  2. jcsd
  3. Feb 25, 2009 #2

    LowlyPion

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    I trust you mean that

    X= ut + 1/2at²

    Remember there is more to that equation, namely

    X = Xo + Vo*t + 1/2*a*t²

    So your method is ok, but you need to have both the initial speed and position for the car that starts first. I'm not sure you did this by your description.
     
  4. Feb 25, 2009 #3
    Yes I left out the squared and Xo would be zero since they both start from the same pos. and Vo would be zero in one of the cars and the other Vo would be the velocity of the car 1.4s after they started but I still cant get the correct answer
     
  5. Feb 25, 2009 #4

    gabbagabbahey

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    So after 1.4 secs, the second car is still at the same positions?:wink:
     
  6. Feb 25, 2009 #5

    LowlyPion

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    Figure where the first car is at t = 0 of when the second car is going to start. If you don't do that, you will never get the right answer. (Oh and ² the time for the acceleration term.)
     
  7. Feb 25, 2009 #6
    is there a difference in accelertation and constant acceleration ??
     
  8. Feb 25, 2009 #7

    LowlyPion

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    Just use the acceleration in the equations for each car as given.

    These accelerations are constant.

    It doesn't have to be constant, but they want constant acceleration here.
     
  9. Feb 25, 2009 #8
    Ive taken this problem http://zebu.uoregon.edu/~probs/mech/1dkin/constacc5/node2.html [Broken] and put in the info from my problem and got a wrong answer did i miss something or would it not work if i did ?
     
    Last edited by a moderator: May 4, 2017
  10. Feb 25, 2009 #9

    LowlyPion

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    Write the equations for this problem. Don't worry about other problems.

    It's important you learn to build the equations from scratch. Otherwise you won't understand them. And come exam time I'm thinking that's something you will want to be doing.
     
  11. Feb 25, 2009 #10
    so ive found the stock car has moved 3.528m in those first 1.4s and its velocity is 5.04m/s. So thats Xo and Vo. Would the other cars Xo and Vo be 0 and 0 ?
     
  12. Feb 25, 2009 #11
    I got 9.47s for an answer to part A but I cant check if its right because ive run out of tries. could you please tell me if its right
     
  13. Feb 25, 2009 #12

    gabbagabbahey

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    Yes, this looks promising:smile:
     
  14. Feb 25, 2009 #13

    gabbagabbahey

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    No, this is incorrect. Show us your work and we'll be able to see what you are doing wrong now.
     
  15. Feb 26, 2009 #14
    0 + 1/2(5.7)t^2 = 3.528 + 5.04t + 1/2(3.6)t^2

    2.85t^2 = 3.528 + 5.04t + 1.8 t^2

    1.05t^2 - 5.04t - 3.528 = 0

    -5.04 +- sqaure root of ( 5.04^2 - 4(1.05)(-3.528) / 2(1.05)
     
  16. Feb 26, 2009 #15

    gabbagabbahey

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    Close, b=-5.04 not +5.04 so,

    t=[-(-5.04) +- sqaure root of ( (-5.04)^2 - 4(1.05)(-3.528)] / 2(1.05)
     
  17. Feb 26, 2009 #16
    I get 8.37 for a time. is this correct ?
     
  18. Feb 26, 2009 #17

    gabbagabbahey

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    No, you must be punching things into your calculator wrong....be careful with your brackets!:smile:
     
  19. Feb 26, 2009 #18
    and to find the distance i got 199.6m by taking the racer cars info and plugged it into the equation.


    x= 0 + 0 + 1/2(5.7)(8.37)^2
     
  20. Feb 26, 2009 #19

    gabbagabbahey

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    Right idea, but wrong time value...
     
  21. Feb 26, 2009 #20
    i keep getting 8.37

    im plugging this in excatly

    (-(-5.04)+square root((-5.04)^2-4(1.05)(-3.528))/2(1.05)
     
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