Question

A train is travelling at $v$ m/s along a level straight track. Very near and parallel to the track is a wall. On the wall a naughty boy has drawn a straight line that slopes upward at 37° angle with the horizontal. A passenger in the train is observing the line out of window (0.90 m high, 1.8 m wide as shown in figure). The line first appears at window corner A and finally disappears at window corner B. If it takes 0.4 sec between appearance at A and disappearance of the line at B, what is the value of $v$ (in cm/s)?

• A. 150 cm/s
• B. 75 cm/s
• C. 300 cm/s
• D. 225 cm/s

Answer 1

Answer: (A)

150 cm/s

Answer 2

Let the train's velocity be $v$ m/s. The window has height $H = 0.90$ m and width $W = 1.8$ m. The line on the wall slopes upwards at an angle $\theta = 37°$. The slope of the line is $m_w = \tan(37°) \approx 3/4$.

We consider the frame of reference of the train. In this frame, the window is stationary, and the line on the wall appears to move horizontally with speed $v$. Let A be the origin (0,0) and B be the point (1.8, 0.90) in the window's coordinate system.

The equation of the line in the window's frame, as it moves horizontally with speed $v$, can be represented as $y = m_w (x - vt)$.

The line first appears at A(0,0) at $t=0$. Substituting these values: $0 = m_w (0 - v \cdot 0)$ $0 = 0$ This confirms our coordinate system and initial condition.

The line finally disappears at B(1.8, 0.90) at $t=0.4$ s. Substituting these values: $0.90 = m_w (1.8 - v \cdot 0.4)$

Using $m_w \approx 3/4$: $0.90 = \frac{3}{4} (1.8 - 0.4v)$

Now, we solve for $v$: Multiply both sides by $4/3$: $0.90 \times \frac{4}{3} = 1.8 - 0.4v$ $1.2 = 1.8 - 0.4v$

Rearrange the equation: $0.4v = 1.8 - 1.2$ $0.4v = 0.6$

Solve for $v$: $v = \frac{0.6}{0.4} = 1.5$ m/s.

The question asks for the speed in cm/s: $v = 1.5 \text{ m/s} \times 100 \text{ cm/m} = 150 \text{ cm/s}$.