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.
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.
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