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How can you tell if the interaction is isotropic?

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A teenager has a car that accelerates at 3.00 m/s2 and decelerates at -4.50 m/s2. On a trip to the store, he accelerates from rest to 16.5 m/s, drives at a constant speed for 5.00 s, and then comes to a momentary stop at the corner. He then accelerates to 18.0 m/s, drives at a constant speed for 15.0 s, and decelerates for 2.67 s. Then he continues for 4.00 s at this speed and comes to a stop.


    2. Relevant equations
    (a) How long does the trip take?
    (b) How far has he traveled?
    (c) What is his average speed for the trip?
    (d) How long would it take to walk to the store and back if he walks at 1.50 m/s?


    3. The attempt at a solution
    I don't know which equations and what numbers to use. Any help on how I can find out which numbers are the ones I need and which equation I should use?
     
  2. jcsd
  3. Feb 13, 2008 #2

    mgb_phys

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    Get some graph paper and draw a diagram.
    Put time along the bottom and speed up the side.
    The area under the graph is the distance.

    Then when you understand it you can use s=ut + 1/2 at^2 to solve each section.
     
  4. Feb 13, 2008 #3
    I've actually done that but I'm still confused.
     
  5. Feb 13, 2008 #4

    mgb_phys

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    You should be able to easily plot the speed at each point in time - it's just a series of straight lines.

    So we assume all accel is 3m/s^2 and all braking is -4.5m/s^2 (actually if you are going to say deceleration and have a negative vallue that cancels out and implies aceleration ...)

    Use v = u + a t with 3.0 m/s^2 to get 't'


    Easy

    Use v=u + a t again with -4.5m/s^2

    Use v = u + a t with 3.0 m/s^2 to get 't'

    Easy

    Use v=u + a t again with -4.5m/s^2

    Easy

    Use v=u + a t again with -4.5m/s^2

    Then the time you can just read of the axis.
    The distance is the integral (ie the area under the curve)
    And the average speed is just distance/time
     
  6. Feb 13, 2008 #5
    Is it the same thing as finding the X and Y components? If not I'm confused. I was never really good at physics anyways.
     
  7. Feb 14, 2008 #6
    I got the time the trip takes. All I really need now is how far he traveled, but I am having trouble figuring out how to get this. Any help?

    Edit: I solved it :D
     
    Last edited: Feb 14, 2008
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