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Motion Question

  1. Sep 24, 2004 #1
    Hello,

    I'm having some difficulty in a physics question.

    The question is:
    To save fuel, some truck drivers try to maintain a constant speed when possible. A truck traveling at 81.0 km/h approaches a car stopped at the red light. When the truck is 94.6 meters from the car the light turns green and the car immediately begins to accelerate at 2.80 m/s^2 to a final speed of 117.0 km/hr. How close does the truck come to the car assuming the truck does not slow down? How far from the stop light has the car travelled when the truck reaches its closest distance?

    My answer so far:
    First thing I did, I drew a position time graph with the car beginning its acceleration at the origin and the origin being when the light turns green. The truck's constant speed intercepts the position axis at -94.6 m.

    Truck arrives at the lights: when t = 4.20 s
    (The Eqn of the truck is y = 22.5x - 94.6)
    Therefore the car is 25 m away from the lights when the truck arrives at the lights. (d = 1/2at^2)

    The car reaches its final velocity (17.0 km/h or 32.5 m/s) 1450 m away from the lights. Or about 32.5 seconds after the light turns green.


    I'm not sure on how to figure out how close the car gets. I started a table and calculated when each vehicle was at different time points, but this is getting time consuming and I am sure there is a simpler way.

    Thanks for the assistance, any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 24, 2004 #2
    **Correction: The car reaches its final velocity when it is 1480 m away from the lights. (not 1450 m)
     
  4. Sep 24, 2004 #3
    I think you need to calculate the distance of the accelerating car when it reaches a velocity that is equal to that of the truck. v =at so we get for this the following : for the car : 22.5(in meter per seconds)=2.8*t so t = 8. Now suppose the car starts in x = 0 then the distance after 8 seconds is via
    x = at²/2 : x = 89.6 meters.

    In eight seconds the truck does x = vt : x = 180 meters starting from where the truck was when the car was at rest. So 94.6 meters behind the car. from the zero-point (the point where the car started to move) the truck has done a distance of 180 - 94.5 = 85.5 meters and the car has done 89.6 meters so the closest distance must be 89.6 - 85.5 = 4.1 meters.

    this is the closest distance because after the eight seconds the car will have a velocity that exceeds the one of the truck, thus the car will drive away from the truck...

    marlon
     
  5. Sep 24, 2004 #4
    Thank you so much, Marlon.
    It was greatly appreciated.
     
  6. Sep 24, 2004 #5

    My pleasure...keep on asking if you wish...

    marlonissimo
     
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