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Homework Help: Distance Problem

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data
    This problem has already been solved but i got a quite few clarifications:

    "a boy on his bicycle intends to arrive at a certain time to a town that is 30 km away from his home .after riding 10 km, he rested for half an hour and as a result he was obliged to ride the rest of the trip 2km/hr faster ."
    2. Relevant equations
    v = d/t
    t = d/v

    3. The attempt at a solution
    Why did he disregard the 10km?
    i think the formula should've of look like this:
    (total time riding on original speed for the whole trip) = (time riding on original speed (for 10km))+(time rested)+(faster velocity on the rest (for 20km))

    the formula he used:
    Last edited: Feb 22, 2013
  2. jcsd
  3. Feb 22, 2013 #2

    Simon Bridge

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    He has to go 20km in the time remaining after travelling 10km at the slow speed, and waiting half an hour.

    The time remaining is ##20/s = 0.5 + 20/(s+2)## ... agreeing with your formulation.

    ##40(s+2) = (s+2)s + 40s##


    by quadratic equation:
    ##s \in \{ 10,-8 \}##

    possible answers are 10hr and -8hr ... pick the positive one.
    This is the same answer.

    So your question is, "how did he know that his version would be correct?"
    Try plotting the velocity-time graph.
    Last edited: Feb 22, 2013
  4. Feb 22, 2013 #3
    The solutions to the quadratic equation are +8 and -10.
    The solution for the problem is then 8km/h.

    In the link given he is solving a different problem. I suppose he did not read the problem carefully.
  5. Feb 22, 2013 #4

    Simon Bridge

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    Did I set up the quadratic incorrectly ...

    the discriminant is 324
    so ##s =\frac{1}{2}(-2\pm\sqrt{324}=1\pm 9## ... Oh I see: I misplaced a minus sign!
    <mumble mumble grzzl>
    ... time for bed!
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