Calculating Time Length of Train and Wall Intersection

[dnt]

dont know how else to word the title but here's the problem:

you are in a train traveling on a straight and flat track (just assume its going right to left for my explanation) at 3.0 m/s. your window is 0.9 m high by 2.0 m wide and as you look out, the wall is going up at a 12 degree slope. the question is what is the time length between when the wall appears at the bottom of your window til it leaves at the top?

the way i did it was i created a triangle (with angle 12) and assumed that the wall had to "travel" 0.9 m up my window of sight.

from there i calculated the bottom part of this triangle using trig to get 4.23 m.

however, i also needed to add in the length of the window itself since you could see the wall from either end (i assume you start looking at the window from the far left end and as it travels look out the right end to maximize how long you can see it). so i added 2.0 m to that length to get 6.23 m.

then just divide by 3.0 m/s to get the time 2.078 s. answer in book states 2.1 sec.

i just don't think my logic is right however. i think i just got lucky and actually didnt do it the right way. the question is so difficult for me to visualize so that i can correctly solve it.

help. thanks.

 

[ZapperZ]

Even with a thread telling people not to post things like this in the General Physics section, we are still getting them there. What else is there to do when a simple and obvious message like that is either ignored or missed?

Zz.

 

[dnt]

sorry. won't do it again.