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Acceleration/speed question

  1. Jan 31, 2005 #1
    Could someone help me to solve this question?

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    Question:
    At the instant when the traffic light turns green, an automobile starts with a constant acceleration of [tex]2.00 m/s^2[/tex]. At the same instant, a truck travelling with a constant speed of 9.00 m/s overtakes and passes the automobile. How far beyond the starting point will the automobile overtake the truck?
     
    Last edited by a moderator: Jan 31, 2005
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  3. Jan 31, 2005 #2

    Andrew Mason

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    The condition for the car passing the truck is:

    [tex]d_{car} \ge d_{truck}[/tex]

    (1)The distance moved by the car as a function of time is: ________
    (2)The distance moved by the truck as a function of time is: _______

    Set (1) = (2) and you have the expression for the time at which the truck and car have travelled equal distances. Solve for the time.

    AM
     
  4. Jan 31, 2005 #3
    Sorry to bother you again, but what equations should I use for the car and the truck? I'm having 'one of those days' again, and at the same time I'm having a complete brain fart :)
     
  5. Jan 31, 2005 #4

    Andrew Mason

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    Plot the velocity as a function of time for the truck and for the car. What gives you the distance covered by the truck and car? (what is velocity in terms of distance and time?).

    AM
     
  6. Jan 31, 2005 #5
    okay thanks; [tex]40.5 m[/tex] is what I got.
     
  7. Jan 31, 2005 #6

    Andrew Mason

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    How did you get that? (hint: The area under the graph for the accelerating car is a triangle).

    AM
     
  8. Feb 1, 2005 #7
    I did:

    car

    [tex]2 = \frac {d} {t^2}[/tex]

    [tex]t = \sqrt{d/2}[/tex]

    truck

    [tex]d/9 = \sqrt{d/2}[/tex]

    [tex]2d^2 - 81d + 0 = 0[/tex]

    Then I just used

    [tex]x = \frac {-b +- \sqrt{b^2 - 4ac}} {2a}[/tex]

    To solve for x. I got [tex]x = 0[/tex] and [tex]x=40.5[/tex]. The second one makes sense.
     
    Last edited by a moderator: Feb 1, 2005
  9. Feb 1, 2005 #8

    Andrew Mason

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    This is not correct. The area under the v-t graph represents the distance travelled by the accelerating car. It is a triangle with base t and height v, so the distance (area) is [itex]\frac{1}{2}vt = \frac{1}{2}(at)t[/itex]

    So [itex]d_{car} = \frac{1}{2}at^2[/itex]

    AM
     
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