A physics student who is late for school is trying to catch her bus. The student is 10 m behind the bus as the bus starts from rest and begins to accelerate at 0.5 m/s2 away from her. If the student continues running at a steady 3 m/s, will she catch her bus?

Answer 1

distancestudent=3*time + 10
distancebus=1/2 .5 *time^2

set them equal, if she catches the bus..

3t+10=t^2/4

t^2-12t+40=0

t=(12+-sqrt(144-160))/2 notice the sqrt of a negative term...means the solution cannot be real, she does not catch the bus, except in imaginary time.

Answer 2

d bus = .5 a t^2 = .25 t^2

d student = d bus + 10 = 3 t

so
.25 t^2 = 3 t - 10

.25 t^2 -3 t + 10 = 0

t^2 - 12 t + 40 = 0

t = [ 12 +/- sqrt(144 -160) ] /2

oh my, complex t
student only catches bus in imagination

Answer 3

There is a small error. It should be -10, not +10, in the original equation of d=3t-10