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Relative velocity at a junction

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Two straight roads intersect at point O and cross at an angle Θ to one another, such that tan Θ = 3/4.Two cars, A and B are travelling towards O on these roads,A at 5 m/s and B at 8 m/s

    At a certain moment.A is 100 m from the junction and B, on the other road, is 200 m from the junction. Find
    (i)the time at which A reaches O
    (ii) The distance between A and B at this time.
    (iii) The magnitude and direction of the velocity of A with respect to B
    (iv) The shortest distance between them.
    (v)the time at which they are nearest to one another
    (vi)the time when they are equidistant from O


    2. Relevant equations

    Vab = Va - Vb

    3. The attempt at a solution

    I got every part except (vi)

    (i) 100/5 = 20 seconds

    (ii) 200 - 8(20) = 40m

    (iii) Vab = ( -5(4/5)i -5(3/5)j ) -(-8i) = 4i -3j m/s

    |Vab| = 5m/s
    inverse tan 3/4 = 36.87 degrees south of east

    (iv)
    40(sinΘ) = 40(3/5) = 24 m = shortest distance

    (v)
    Pythagoras theorem (40^2 -24^2)^(1/2) = 32m

    32/5 =6.4 seconds

    20 + 6.4 = 26.4 seconds
    (vi)
    This is the part im stuck on.I know I need the distance from O when they are equidistant but I just can't seem to think of any way that I could get none of the examples in my book
    covered this any help would be appreciated.
     
  2. jcsd
  3. Mar 2, 2013 #2

    TSny

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    See if you can write an equation for how far A is from O for any time t. Similarly for how far B is from O at any time t.
     
  4. Mar 2, 2013 #3
    Alright thanks, I just jotted down this down randomly
    5t - 100 = 8t - 200............ 3t = 100

    t = 33.33 seconds

    This is the right answer according to the back of my book but for some reason I feel as though the method I just used isn't the right one since if I picked any time before A passed O I would get a negative distance.
    Any thoughts?
     
  5. Mar 2, 2013 #4
    Since displacement is a vector quantity would the negative sign just be telling you that its displacement is at the opposite side of O?
     
  6. Mar 2, 2013 #5

    TSny

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    OK, you set up equations such that positive positions are interpreted as positions beyond O while negative positions are before reaching O. That's fine.

    The question is a little ambiguous. Usually "distance" is interpreted as positive independent of which side of O you're on. You have found a solution for when both cars are past O. Can you find another solution where one car is beyond O while the other has not yet reached O (but they are the same distance from O)?
     
    Last edited: Mar 2, 2013
  7. Mar 2, 2013 #6
    Would it be 5t -100 = 200 - 8t

    t=23.077 seconds

    bearing in mind the previous answer I got is the only one that is given as a solution at the back of my book.
     
  8. Mar 2, 2013 #7

    TSny

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    Yes. That looks good. I guess the book is ignoring that answer.
     
  9. Mar 2, 2013 #8
    Alright thank you.
     
  10. Mar 2, 2013 #9
    Alright, thank you.
     
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