Posted by Soph on Sunday, September 23, 2012 at 7:06pm.
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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