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March 25, 2017

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Jack is 12m from the bus and is running towards the bus at a constant 4.0m/s. If the bus accelerates away from the bus stop at .8m/s^2, can he catch the bus? If so where?

Vi= initial velocity
vf=final velocity
a= acceleration
d=distance
t=time



I found out that it takes Jack 3 seconds to reach where the bus was originally at the 12m. But after that i have no clue how to finish it.

  • Physics? clueless on what to do - ,

    bus goes distance d
    Jack goes distance (d+12)

    d = (1/2) a t^2 = .4 t^2
    d+12 = 4 t so d = 4 t - 12

    4 t - 12 = .4 t^2

    .4 t^2 - 4 t + 12 = 0

    t = [ 4 +/- sqrt (16 -.16*12) / .8

    t = [ 4 +/- 3.75 ] /.8
    t = 9.69 or .313
    the .313 solution is impossible because he has not run 12 meters by then
    so he catches the bus after 9.69 seconds.

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