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Linear Kinematics problem

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Two cars are at the start lines of two roads at right angles to each other and pointing to the intersection 500 metres away. Car A starts from rest with uniform accelereation of 0.5m/s2 and accelerates for 20 seconds. It then continues at the speed reached. Car B starts 5 seconds later and 20 metres further away with an acceleration of 0.7m/s2.
    Calculate wrt the start time of Car A:

    a) Time taken for B and A to be the same distance from the cross-road.

    b) The distance from the cross roads at the above time

    c) The average velocity of Car A at this distance.

    2. Relevant equations

    v=u+at
    v2=u2+2as

    3. The attempt at a solution

    First i used v=u+at and plugged the results into a table showing velocity against time for car A and B, for upto 60 seconds at ten second intervals.
    Then i transposed the second equation for s to work out the distance s=v2-u2[tex]/2a[/tex] for car a and s=(v2-u2[tex]/2a[/tex])-20 for car B.
    This was then plugged into a second table showing distance against time.
    From this table it was deduced that the area of interest was between 50 and 60 seconds.
    I carried on with this process until i narrowed it down.
    The answer i got for a) was 57.67 seconds.
    b) was just working the time back through the previous equations which gave 23.3m (500-476.7).
    c) i calculated change in distance over change in time: 476.7/57.67 = 8.267m/s2

    However, this was time consuming and i am wondering if there was a simpler way to tackle this(if indeed the above solution is correct!), possibly as a simultaneous equation?

    Any help is greatly appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 14, 2009 #2

    ideasrule

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    Homework Helper

    Rest assured that you don't need to use tables for any physics problem. When the two cars are at the same distance from the cross-road, two things are the same for both cars: distance travelled and time taken. Which equation relates distance, acceleration, and time? Start there.
     
  4. Oct 14, 2009 #3

    Delphi51

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    Homework Helper

    Your method looks pretty good. I did it a different way and got a different answer! Kind of complicated and easy to make a little mistake, which I have a habit of doing.
    For car A I did the first 20 seconds as v = at = .5*20 = 10
    d = .5*a*t^2 = 100 m
    Then for times larger than 20 we have d = 100 + vt = 100+10(t-20)

    For car B, d = do + vi*t + .5*a*t^2
    d = -20 +.5*.7*(t-5)^2

    for (a) I set the two distance equations equal, collected like terms and got this quadratic: .35*t^2 - 13.5*t + 88.75 = 0
    It has two solutions, one less than 20 so no good, and t = 30.17 s
    I checked the distances at that time and they are equal (if my equations are correct!).
     
  5. Oct 14, 2009 #4
    Car B starts 20 m further away. That's why i went the table route.
     
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