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Green light

  1. Aug 26, 2005 #1
    At the instant the traffic light turns green, an automobile starts with a constant acceleration [itex] a_{x} [/itex] of 6.0 [itex] \frac{ft}{sec^{2}} [/itex]. At the same instant, a truck traveling with a constant speed of 30 [itex] \frac{ft}{sec} [/itex], overtakes and passes the automobile. (a) How far beyond the starting point will the automobile overtake the truck? (b) How fast will the car be traveling at that instant?

    (a) So as time increases the speed of the first car increases while the speed of the second car stays constant. After five seconds the first car is 186 feet from the starting line while the second car is 150 feet from the starting line. Is this correct?
    (b) Would the car be traveling 96 ft/sec?




    Any feedback is appreciated.

    Thanks
     
  2. jcsd
  3. Aug 26, 2005 #2

    Doc Al

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    True.
    How did you conclude this?
    Right.

    Approach this problem systematically. Can you write equations that describe:
    (1) The position of the car (from the starting point) as a function of time?
    (2) The position of the truck (from that same point) as a function of time?​

    You can set the final positions equal to each other to find the time when the car overtakes the truck. Then you can use that time to find the distance and speed of the car at that point.
     
  4. Aug 27, 2005 #3
    Ok so I can use [itex] x = x_{0} + v_{x}_{0}t + \frac{1}{2}a_{x}t^{2} [/itex] or [itex] x = 3t^{2} [/itex] (acceleration) and [itex] x = x_{0} + \frac{1}{2}(v_{x}_{0}+v_{x})t [/itex] or [itex] x = 30t [/itex]. So [itex] t = 10 [/itex]. So the car would be 363 feet from starting line traveling at 66 ft/sec.

    Is this correct?
     
    Last edited: Aug 27, 2005
  5. Aug 27, 2005 #4

    Fermat

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    I'm afraid not :(
    What did you get for the time ?

    Sorry - you got 10 s - that's correct.
    So what is the distance travelled ??
     
  6. Aug 27, 2005 #5
    363 feet for car and 330 feet for truck
     
  7. Aug 27, 2005 #6

    Fermat

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    you have x = 3t²

    if t = 10 then x = 3*10² = 3*100 = 300 ft.
     
  8. Aug 27, 2005 #7

    Doc Al

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    Exactly right for the car.
    No need to find the average speed of the truck, since its speed is constant. But correct nonetheless.
    Right.


    Realize that you found the time by setting the distances equal; so if you calculate different distances for car and truck--something must be wrong!

    But the car starts from rest, not at 66 ft/sec. (That's why [itex] x = 3t^{2} [/itex]; the initial speed is zero.)
     
    Last edited: Aug 27, 2005
  9. Aug 27, 2005 #8
    no but dont you have to go to x = 11? Because at x = 10, the two vehicles are at the same position. The question asked the distance from the starting line when the car overtakes the truck.

    is this corrrect?

    thanks
     
    Last edited: Aug 27, 2005
  10. Aug 27, 2005 #9

    Fermat

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    the two vehicles are at the same position at x = 0. This is where the traffic lights are, i.e the starting line.

    How did you get x = 10, or 11?
     
  11. Aug 27, 2005 #10
    I meant t = 11
     
  12. Aug 27, 2005 #11

    Doc Al

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    To solve the problem, you have to pretend that the point where they are at the same position is the point where the car overtakes the truck.

    But I think I know what you're thinking... that until the body of the car passes the front of the truck, it hasn't really overtaken it. If that's what you think, good for you! But since you have no idea how big the car is, you have to pretend they are points! :smile: (And get used to problems being poorly worded.)

    In any case, why x = 11??? (Do you mean t = 11?) That's just a guess--never a good idea.
     
  13. Aug 27, 2005 #12
    I used [itex] v_{x}^{2} = v_{x}_{0}^{2} + 2a_{x}(x-x_{0}) [/itex] and [itex] v_{x} = 66 ft/sec [/itex]
     
  14. Aug 27, 2005 #13

    Doc Al

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    Find the speed that the car is traveling at the moment it passes the truck. (Which is at t = 10 sec, not t =11.) And recalculate the distances as Fermat suggests.
     
    Last edited: Aug 27, 2005
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