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Acceleration, distance, time. help.

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data

    vehicle 1 is traveling at a constant speed of 90mph (132 f/s) 5 seconds before vehicle 2 begins to accelerate from 0-90 mph taking 17.09 seconds to do so. how far a distance will vehicle 2 travel while accelerating to 90mph. and how long will it take for vehicle 2 to catch up to vehicle 1 while accelerating at approximately 5.2 m/s/s or 15.6 f/s

    2. Relevant equations

    The relationship between the 2 vehicles is dependant.

    what is the total distance traveled by vehicle 2 before it catches up to vehicle 1?


    3. The attempt at a solution

    ok, it's too hard to explain and argue at the same time, so i'll leave it at this:
    find a function of position for car #1 and car #2, call them c_1(t) and c_2(t), set them equal and solve for t. what does this mean exactly?
    This should help you a bit, at constant acceleration, call it a, the position of car 2 at time t is at^2/2
    the first car, meanwhile is moving at a constant velocity v, so its position at time t is vt
    does car 2 start from rest?
    so solve vt=at^2/2 -> at^2-vt=0 -> use quadratic formula
    this will give you the time t, when car 2 catches up to car 1, ?? to answer part 2 of the Q
     
  2. jcsd
  3. Jan 20, 2009 #2

    jgens

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    Gold Member

    For your first question you could use the equation Vf^2 = Vi^2 + 2(a)(d) where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the displacement.

    For number 2, since the first car is traveling at constant speed we can label the distance it travels as d = V(t + 5) while we may use the equation df = (a)(t^2)/2 + (Vi)t + di. Equate the two and solve for the time.
     
  4. Jan 20, 2009 #3
    did you get a distance of 4044.8 ft for the first part of the question?

    or 558.46 ft?
     
    Last edited: Jan 21, 2009
  5. Jan 21, 2009 #4
    anyone get an answer for this?
     
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