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 (0.86 m high, 2.1 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?
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To determine the time it takes for the top edge of the wall to appear and disappear at the window corners, we need to consider the relative motion between the train and the wall.
Let's break down the problem step by step:
1. Find the vertical height of the wall that is visible from the train compartment.
- Given the height of the window (0.86 m) and the angle of the wall (12°), we can use trigonometry to calculate the visible height.
- Using the formula: visible height = window height / cos(angle), we have:
visible height = 0.86 m / cos(12°) ≈ 0.883 m
2. Calculate the horizontal distance between window corners A and B.
- Given the width of the window (2.1 m) and the angle of the wall (12°), we can use trigonometry to find this distance.
- Using the formula: horizontal distance = window width * tan(angle), we have:
horizontal distance = 2.1 m * tan(12°) ≈ 0.448 m
3. Determine the relative speed between the train and the wall.
- Since both the train and the wall are moving, we need to consider their relative speed.
- The train is moving at 3.0 m/s to the left, and the wall is stationary in its reference frame.
- Therefore, the relative speed is 3.0 m/s.
4. Calculate the time it takes for the top edge of the wall to move from A to B.
- To find the time, we need to divide the horizontal distance (0.448 m) by the relative speed (3.0 m/s).
- Therefore, the time it takes is 0.448 m / 3.0 m/s ≈ 0.149 seconds.
So, the time it takes for the upper edge of the wall to appear and disappear at the window corners is approximately 0.149 seconds.