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Finding Distance with Constant Acceleration

  1. Feb 5, 2014 #1

    B18

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    1. The problem statement, all variables and given/known data
    At the instant the traffic light turns green, a car starts with a constant acceleration of 5.00 ft/s^2. At the same instant a truck, traveling with a constant speed of 70.0 ft/s, overtakes and passes the car. How far from the starting point (in feet) will the car overtake the truck?



    2. Relevant equations
    Well I know the 4 kinematic equations for motion with constant acceleration.
    I know final velocity of the truck=final velocity of the car.


    3. The attempt at a solution
    To find the time it took for the car to catch the truck i divided 70ft/s by 5.0ft/s^2=14.0s
    t=14
    Xi=0
    Vi=0
    a=5.0ft/s^2

    Ive got the answers 490ft, 1015ft, 980 ft which are all incorrect. Am I using the wrong value for something here?

    I appreciate any help.
     
  2. jcsd
  3. Feb 5, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Hi B18! The question DOES NOT ask how many seconds until the car's speed matches that of the truck!

    The question asks: how many feet from their starting point are they seen to be side-by-side? Big difference
     
  4. Feb 5, 2014 #3

    B18

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    Hi NascentOxygen. Thanks for the reply. Im sorry I wasn't more specific. I found t=14 s. After plugging in the values I know I wasn't getting the correct answer.

    What I mean is the t value I found correct? Did it take the car 14 s to reach the truck??

    Wouldn't you need to know the duration of time in this problem??
     
  5. Feb 5, 2014 #4

    NascentOxygen

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    Staff: Mentor

    I haven't worked it out, but I can see that dividing that v by that a will give you the time when the car reaches a speed of 70' per sec. The problem does not require you to determine anything relating to the event of the car's speed matching that of the truck. The problem involves equal distances, not equal speeds.
     
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