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Time and Acceleration

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data
    A race car driver buys a car that can accelerate at 5.9m/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 an acceleration of 4.1m/s^2. What is the time it takes the sports car driver to overtake the stock car driver.

    i have tried solving it using the formula x= 1/2at+Vi(t) but i get 4.68s which is wrong what is it that iam not doing correctly?
     
  2. jcsd
  3. Oct 23, 2008 #2

    tiny-tim

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    Hi bigzee20! :smile:

    x= 1/2at2 + Vi(t) :redface:
     
  4. Oct 23, 2008 #3
    Hey tiny-tim thanks for the reply

    I used that formula on both cars then i divided their positions and thats how i got 4.68sec which is wrong. I dont know what iam doing wrong??
     
  5. Oct 23, 2008 #4

    tiny-tim

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    ok … show us what you did, and then we can see where the mistake is. :smile:
     
  6. Oct 23, 2008 #5
    Car a 1/2(5.9)(2.015)+0(2.015)= 5.94425
    Car b 1/2(4.1(0.615)+0(0.615)=1.26075

    Then 5.94425/1.26075 = 4.71sec
     
  7. Oct 23, 2008 #6

    tiny-tim

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    Where did 2.015 and 0.615 come from?

    What happened to t2?

    And what happened to 1.4s? :confused:
     
  8. Oct 23, 2008 #7
    For Car a what do i put in as time? i know car b is 1.4s?

    Car a = 1/2 (5.9)(?)^2+0(?)
    Car b = 1/2(4.1)1.4^2+0(1.4) = 4.018
     
    Last edited: Oct 23, 2008
  9. Oct 23, 2008 #8

    tiny-tim

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    You put t as time.

    t, in the formula, is the unknown!

    t0 for car b is 1.4s
     
  10. Oct 23, 2008 #9
    Iam confused man :confused:
     
  11. Oct 23, 2008 #10
    can someone help me solve this?
     
  12. Oct 23, 2008 #11
    Anybody?
     
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