MHB Kinematics, time taken to a wall disappear

[Fantini]

I'm posting this because I disagree with the answer. Here's the question.

You are in a train that is traveling at $3.0$ m/s along a straight horizontal railroad. Very close and parallel to the railroad there exists a wall with upwards inclination of $12^{\circ}$ with the horizontal. Looking through the window ($0.9$ m tall and $2.0$ m wide) from its compartment, the train is moving to the left. The superior face of the wall appears first at edge $A$ of the window and finally disappears at edge $B$ of the window. How much time passes between the appearance and disappearance of the superior face of the wall?

The book gives the answer as $2.1$ s, but I find $3.5$ s. There is a picture but I'll have to upload it later. I'll add it to this post together with my thoughts on the problem after I sleep. :)

 

[I like Serena]

I'm posting this because I disagree with the answer. Here's the question.

You are in a train that is traveling at $3.0$ m/s along a straight horizontal railroad. Very close and parallel to the railroad there exists a wall with upwards inclination of $12^{\circ}$ with the horizontal. Looking through the window ($0.9$ m tall and $2.0$ m wide) from its compartment, the train is moving to the left. The superior face of the wall appears first at edge $A$ of the window and finally disappears at edge $B$ of the window. How much time passes between the appearance and disappearance of the superior face of the wall?

The book gives the answer as $2.1$ s, but I find $3.5$ s. There is a picture but I'll have to upload it later. I'll add it to this post together with my thoughts on the problem after I sleep. :)

After making a guess as to the shape and positioning of the wall, I'm getting $2.1\text{ s}$.
Hmm. I can also see how $3.5\text{ s}$ can come out.
It all depends on the shape of the wall and the wording superior face.

I'll explain after you add the picture and post your thoughts. (Wink)

 

[Fantini]

I know the image is too big. Sorry about that.

The distance point $A$ travels is $d$, which is the lower side of the right triangle given by point $B$, the line drawn and the direction traveled by $A$, plus the width of the train, $2$ m. Therefore the time taken will be $$t = \frac{d+2}{v} = \frac{\frac{0.9}{\tan(12^{\circ})}+2}{3} \approx 3.5 \text{ s}.$$ I've checked and the answer given coincides with a situation where we discount the time taken to travel the width of the train, that is, $$t = \frac{d-2}{v} \approx 2.1 \text{ s}.$$ Is my interpretation correct?

 

[I like Serena]

Looks fine, except that I find that:

$$t = \frac{\frac{0.9}{\tan(12^{\circ})}+2}{3} \approx 2.1 \text{ s}$$

See W|A.

 

[Fantini]

Could I have already become insane? ;) I think I typed the wrong commands in Mathematica. I'm not that bad at kinematics afterall. Thanks ILS!

 

[I like Serena]

Could I have already become insane? ;) I think I typed the wrong commands in Mathematica. I'm not that bad at kinematics afterall. Thanks ILS!

Yes. I'm a wizard.
But mind you, I'm trying to keep a low profile. ;)