Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Man Vs Bus

  1. Mar 1, 2005 #1
    I was running for the bus when I thought this one up... (Maybe it is not so clever...)

    Ok. A man is running at his top speed for the bus that is stopped at a stop light. The light turns green and the bus accelerates away from him at an acceleration of a metres per second per second. At what time T does the man catch the bus (assuming that he does catch the bus). When you solve the problem you have two Times. Are both valid? If so what do they represent....

    Have fun.
    Last edited: Mar 1, 2005
  2. jcsd
  3. Mar 1, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    There's a condition which has to be met,in order for the problem to have sense...Let's denote by x_{0} the distance between the man and the bus,when the bus begins 2 accelerate.Then,assuming the acceleration to be +1ms^{-2},the condition reads:
    v_{man}^{2}>=2x_{0}.For the equality case,the problem admits uniques solution,the man reaches the bus after exactly 2 x_{0}.
    For the ">" case,the problem admits 2 solutions.The man reaches the bus in the interval (x_{0},2x_{0}) and overtakes and later,because of the acceleration,the bus overtakes the man.
    For the "<" case,the problem has no solution,the man never reaches the bus...

  4. Mar 1, 2005 #3
    call the initial distance between man and bus x
    then the distance between them as a function of time is
    [tex]d = x + 0.5at^2 - vt [/tex]

    the question is when is this distance equal to zero, so we must solve:
    [tex]0 = x + 0.5at^2 - vt [/tex]

    which gives:

    [tex] t = \frac{v+\sqrt{v^2 - 2ax}}{a} \vee t = \frac{v-\sqrt{v^2 - 2ax}}{a}[/tex]

    since the acceleration is 1 m/s2 this gives:
    [tex] t = v+\sqrt{v^2 - 2x} \vee t = v-\sqrt{v^2 - 2x}[/tex]

    v2 must be equal or larger than 2x if it is equal there is just one solution: t = v
    if v2 is larger than 2x there are two solutions: t1 = v - something and t2 = v + something

    Both are valid, you will reach the bus after t1 seconds. If you keep running you will get ahead of the bus, then some time later at t2 the bus will catch up with you again.
  5. Mar 1, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    1.You're 2 hours too late...:tongue:
    2.Your solution does not obey Brain Teaser Forum standards,namely hiding your sollution,correct or not,it doesn't matter.

    To the OP.This is not a Brain Teaser,this is a 9-th grade problem...

  6. Mar 1, 2005 #5
    1. I did not find your explanation very clear
    2. You cannot make the tex white, and since it is not a real "puzzle" I thought this would be ok...
  7. Mar 1, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    1.I think it was more than clear.Any more clear than that,and the kindergarten kids would have gotten it.
    2.I knew that.:tongue: :tongue2:

  8. Mar 2, 2005 #7
    :tongue: Hey I never said it was that clever! :tongue:
  9. Mar 2, 2005 #8


    User Avatar
    Science Advisor
    Homework Helper

    Nobody said anything about "clever"...Just to tease the brain...:wink:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Man Vs Bus
  1. Man on mars. (Replies: 5)

  2. Man vs Wild (Replies: 43)

  3. The This Man phenomenon (Replies: 20)

  4. Bullied bus monitor (Replies: 48)