Question

You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that slopes upward at a 12° angle with the horizontal. As you face the window ( 0.830 m high, 2.34 m wide) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner A and eventually disappears at window corner B. How much time passes between appearance and disappearance of the upper edge of the wall?

Answer 1

Answer:

Total time passed = 2.085 seconds

Explanation:

In the question,

Speed of the train to the left, v = 3 m/s

The slope of the wall (with horizontal) = 12°

Height of the window = 0.83 m

Width of the window = 2.34 m

Now,

The time is needed to be find out for the top edge of the wall to reach the point B (top right) from A (bottom left).

So,

First calculating the time taken by the Wall's top point to reach B in he horizontal is,

$Time=\frac{Distance}{Speed}\\Time=\frac{2.34}{3}\\Time=0.78\,s$

Now,

Time taken to reach the top point of the wall up to the height of the window to reach point B is,

Time = Distance traveled by the train in the horizontal until wall reaches B/ Speed of the train

So,

$tan12=\frac{0.83}{D}\\D=3.915\,m$

So,

Time taken by the train to cover this distance is,

$Time=\frac{D}{v}\\Time=\frac{3.915}{3}\\Time=1.305\,s$

Therefore, the total time for the point to reach B from A is,

Time = 0.78 + 1.305 = 2.085 seconds

Therefore, Total Time = 2.085 s