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Car acceleration physics help

  1. Jul 18, 2005 #1
    Hello everyone! I just registered to this board, and need some help solving a physics problem. It reads:

    Car A leaves a city and travels along a straight road for 1.5 min at 60km/h. It then accelerates uniformly for 0.25 min until it reaches a speed of 80 km/h. It proceedes at that speed for 2.0 min, then decelerates uniformly for 0.50min until it comes to rest. Car B leaves the same city along the same road and accelerates uniformly for 1.6 min until it reaches a speed of 160km/h. It then decelerates uniformly until it comes to rest again after 1.6 min. How far will the two cars have traveled during the different stages?

    This is as far as I completely understand it:

    Car A:
    1 step: conversion 60km/h = 1 km/min
    therefore = 1.5 km

    2 step: conversion 80km/h = 1.33 km/min
    therefore v = vo + at
    1.33 - 1 = a = 1.32km/min
    --------
    .25

    here is confusion:
    x = x0 + v0t + 1/2at^2

    x = 1.5 + 1.33(.25 min) + (1/2)(1.32)(.25)^2
    or
    x = (1/2)(1.32)(.25)^2
    ????
     
  2. jcsd
  3. Jul 18, 2005 #2

    quasar987

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    x = x0 + v0t + 1/2at^2

    is the correct formula. But you gotta know what the symbols mean. x_0 is the position of the object at the moment it starts accelerating. You chose x_0 well. Similarily, v_0 is the speed of the object, also at the moment it starts accelerating. For some reason you set v_0 = 1.33 km/min, which is the FINAL speed. Of course, you need v_0 = 1 km/min.
     
  4. Jul 18, 2005 #3
    Thanks. What do I use for x_0 for car B?
     
  5. Jul 18, 2005 #4

    quasar987

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  6. Jul 18, 2005 #5
    I like this place, answers aren't just givin!
     
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