# A Woman Running Towards a Bus Which Pulls away

1. Oct 9, 2012

### Phonetic

1. The problem statement, all variables and given/known data
A woman is running at a constant velocity of 5.0m/s in an attempt to catch a bus. When she is 11m from the bus, it pulls away with a constant acceleration of +1.0m/s2. From this point, how much time does it take her to reach the bus if she keeps running with the same velocity?

2. The attempt at a solution
Finding the time
d = di + vit+1/2at2
0 = 11 + 0t + 1/2(1)t2
0 = 11 + .5t2
t = sqrt of 11.5
t = 3.39

Distance traveled by woman
d = v*t
d = 5*3.39
d = 16.95m

Distance traveled by bus
vf = vi + at
= 0+1*3.39
vf = 3.39

d = 1/2 (vi + vf)t
d = 1/2 (3.39) 3.39
d = 1/2 * 11.5
d = 5.75
The bus traveled 5.75m, add in the initial distance of 11m
16.75m

That's the closest I can get to the answer. Any guidance on where I want wrong? Thanks!

2. Oct 9, 2012

### stauber28

Set your initial position to Zero for woman, and initial position to 11 for the bus. Set these two equations equal to solve.

Woman d1=0+5t
Bus d2=11+.5t2

When both positions are equal she will reach the bus
d1=d2

5t = 11+.5t2
10t-t2 = 22
t=5±√3

______________________________
If everyone is thinking alike, then somebody isn't thinking

Garrett Stauber​

Last edited: Oct 10, 2012