Why Does Rain Look Slanted from a Moving Train?

You look out of a moving train and notice that the rain seems to fall diagonally towards the back of the carriage. Yet someone standing on the platform might see those same raindrops falling straight down. Has the train somehow changed the direction of the rain?

No. The two observers are measuring the rain’s motion from different reference frames. Rain falling vertically relative to the ground appears to move diagonally down and backwards relative to a train travelling forwards. The faster the train moves, the greater that apparent backward motion becomes.

Here’s a prediction to make before going further: if the train doubles its speed while the rain keeps falling at the same downward speed, will the rain appear more vertical or more diagonal?

More diagonal. The train moves a greater horizontal distance while each drop falls the same vertical distance.

01Why does rain appear to fall at an angle from a moving train?

Imagine standing beneath a dripping tap. The drops fall straight down, one after another.

Now imagine walking forwards beneath the same stream of drops. As you move, you encounter drops that were previously in front of you. Relative to your movement, those drops appear to travel backwards as they fall.

The tap hasn’t moved. You’ve changed your own motion.

The same idea explains rain viewed from a train.

The analogy describes relative movement, but not the complete behaviour of falling water. Real raindrops are affected by gravity, air resistance, and wind. We’ll initially ignore changes in their velocity over the short interval being measured.

Two observers, two reference frames

We’ll use two reference frames throughout:

Assume the train travels horizontally east on straight, level tracks.

Choose east as the positive horizontal direction and upwards as the positive vertical direction. Therefore, a downward velocity has a negative vertical component.

For now, suppose the raindrops fall vertically at a constant speed of \(10\,\mathrm{m\,s^{-1}}\) relative to the ground.

The platform observer measures:

\[ \vec v_{R/G}=(0,-10)\,\mathrm{m\,s^{-1}} \]

The first component represents horizontal velocity, and the second represents vertical velocity.

But a train moving east at \(12\,\mathrm{m\,s^{-1}}\) has velocity:

\[ \vec v_{T/G}=(12,0)\,\mathrm{m\,s^{-1}} \]

To find the rain’s velocity relative to the train, subtract the train’s velocity from the rain’s velocity:

\[ \vec v_{R/T}=\vec v_{R/G}-\vec v_{T/G} \]

Here, \(R\) means rain, \(T\) means train, and \(G\) means ground. The first letter identifies the moving object, and the second identifies the observer’s reference frame.

Substituting:

\[ \begin{aligned} \vec v_{R/T} &=(0,-10)-(12,0)\\ &=(-12,-10)\,\mathrm{m\,s^{-1}} \end{aligned} \]

So the passenger sees each raindrop moving at \(12\,\mathrm{m\,s^{-1}}\) towards the rear and \(10\,\mathrm{m\,s^{-1}}\) downwards.

The rain hasn’t acquired an extra force towards the back. Its backward motion is a consequence of observing it from a moving train.

Why the direction of the velocity vector matters

Speed tells us how quickly something moves. Velocity includes both speed and direction.

This means we cannot simply subtract the rain’s scalar speed from the train’s speed.

For example, subtracting \(10\,\mathrm{m\,s^{-1}}\) from \(12\,\mathrm{m\,s^{-1}}\) to get \(2\,\mathrm{m\,s^{-1}}\) would be wrong. Those speeds describe motion in perpendicular directions.

We must subtract the velocity vectors, then calculate the magnitude of the result.

For a wider explanation of why direction changes relative velocity, see Can Two Objects Have the Same Speed but Different Velocities?.

02How could two cameras test the explanation?

A slanted photograph of rain isn’t enough to prove that the apparent angle comes from the train’s movement. Wind could also move the droplets sideways.

We need a way to distinguish those explanations.

The first design: one camera on the train

Imagine mounting a level camera inside the train, pointing sideways through a window.

The camera records successive positions of raindrops as the train moves east.

If the rain is falling vertically relative to the ground, we predict that freely falling droplets will move diagonally backwards relative to the camera.

But this design has a problem.

A diagonal streak could be caused by the train’s motion, horizontal wind, or both. Water running along the window could also create streaks that don’t represent the movement of airborne droplets.

The first recording tells us how the rain appears from the train, but not how the rain moves relative to the ground.

The revised design: compare stationary and moving cameras

A better controlled demonstration would use a straight horizontal track, a falling water source, and two cameras.

One camera stays fixed beside the track, facing a narrow vertical region where droplets fall. A second camera is mounted on a small motorised carriage moving horizontally along the track.

Both cameras are level and oriented in the same direction. A calibrated background provides a distance scale. Water falls from a source above the viewing region towards a collection tray below it.

The cameras record successive positions of airborne droplets at known time intervals. We use the measured horizontal and vertical displacements to estimate velocity.

This is a proposed experimental comparison, not a report of a particular experiment.

What should the revised experiment reveal?

If the water has no horizontal ground-frame velocity, the stationary camera should measure zero horizontal displacement. The moving camera should measure a backward horizontal displacement equal to the distance travelled by the carriage over the same time interval.

The calculation below shows the predicted result using illustrative train-scale values.

03How do you calculate the apparent angle of rain?

Now suppose the rain has some horizontal motion because of wind.

We’ll describe the rain using two components:

• \(w\): the rain’s horizontal velocity relative to the ground, in \(\mathrm{m\,s^{-1}}\), positive towards the east.
• \(d\): the magnitude of the rain’s downward velocity relative to the ground, in \(\mathrm{m\,s^{-1}}\).
• \(u\): the train’s horizontal velocity relative to the ground, in \(\mathrm{m\,s^{-1}}\), positive towards the east.

The rain’s ground-frame velocity is:

\[ \vec v_{R/G}=(w,-d) \]

The train’s ground-frame velocity is:

\[ \vec v_{T/G}=(u,0) \]

Subtracting gives:

\[ \vec v_{R/T}=(w-u,-d) \]

This is the general relationship for our two-dimensional model.

To find the angle \(\theta\) measured from the downward vertical:

\[ \tan\theta=\frac{|w-u|}{d} \]

The absolute value makes the ratio positive when calculating the size of the angle. The sign of \(w-u\) tells us which way the rain appears to move.

The apparent speed is:

\[ v_{\mathrm{rel}}=\sqrt{(w-u)^2+d^2} \]

These equations work for any constant horizontal rain velocity and horizontal train velocity within this model.

Three possible appearances of rain

Consider the following illustrative cases. In every row, the rain falls downwards at \(10\,\mathrm{m\,s^{-1}}\) relative to the ground.

The second row is particularly useful.

Rain can appear to fall vertically to a moving passenger even while it is travelling diagonally relative to the ground.

The condition is simply:

\[ w=u \]

The rain and train must have equal horizontal velocities relative to the ground.

It’s also worth separating where the rain moves from where it comes from. Rain moving backwards relative to a train approaches the train from the front. That’s why a forward-facing windscreen can become wet even when the ground observer sees rain falling vertically.

04What the simplified model does and doesn’t explain

There are three limits worth keeping in mind.

Real falling drops need not move at constant speed

A raindrop experiences gravity and air resistance. It may accelerate initially and approach a terminal downward speed as drag balances its weight.

Our straight-line rain trajectories assume that each drop’s ground-frame velocity stays approximately constant over the interval of interest.

The instantaneous relative-velocity equation still works when velocities change. But if the train accelerates, the drop’s path in the accelerating train frame need not remain straight.

Window streaks are not necessarily airborne rain trajectories

A drop hitting a window can stick, spread, merge with another drop, or move along the glass.

Its motion is then controlled partly by the window surface and airflow. It is no longer behaving like a freely moving raindrop.

A long-exposure photograph also records displacement over a finite time, rather than necessarily showing the instantaneous velocity at one moment.

The observer must measure motion in the correct plane

Our calculations concern a side-on view along the train’s direction of travel, using horizontal east-west and vertical components.

A camera pointing at a different angle may show a different two-dimensional projection. Perspective and camera tilt can change the apparent angle in a photograph.

For an HSC-style question, follow the stated observer frame and coordinate directions rather than estimating angles from an uncalibrated image.

05Questions and solutions

The questions begin with a standard relative-velocity calculation and progress towards interpreting measurements, predicting collisions, and understanding motion from an accelerating train.

Unless otherwise stated, assume straight, level tracks, constant ground-frame rain velocity, negligible changes in airflow near the train, and raindrops that can be treated as particles until they strike a surface. Use east as positive horizontally and up as positive vertically.

Question 1

Rain falls vertically at \(8.0\,\mathrm{m\,s^{-1}}\) relative to the ground. A train travels east at \(6.0\,\mathrm{m\,s^{-1}}\).

Determine:

a) The rain’s velocity components relative to the train.

b) Its apparent speed.

c) Its angle of motion from the downward vertical, including direction.

Solution 1

The rain moves at \(10.0\,\mathrm{m\,s^{-1}}\) relative to the train, directed \(36.9^\circ\) towards the rear from the downward vertical. Its horizontal and vertical components are \(-6.0\) and \(-8.0\,\mathrm{m\,s^{-1}}\), respectively.

Part a) Subtract the train’s velocity vector from the rain’s ground-frame velocity:

\[ \begin{aligned} \vec v_{R/T} &=(0,-8)-(6,0)\\ &=(-6,-8)\,\mathrm{m\,s^{-1}} \end{aligned} \]

The negative horizontal component represents westward motion, towards the train’s rear.

Part b) Find the magnitude:

\[ \begin{aligned} v_{\mathrm{rel}} &=\sqrt{6^2+8^2}\\ &=10.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part c) Find the angle:

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{6}{8}\right)\\ &=36.9^\circ \end{aligned} \]

A common mistake is to subtract the scalar speeds and obtain \(2.0\,\mathrm{m\,s^{-1}}\). That ignores the perpendicular directions of the rain and train velocities.

The correct relative speed is greater than either component because both contribute to the magnitude of the resultant vector.

Question 2

A stationary platform camera records an airborne raindrop moving \(1.20\,\mathrm{m}\) east and \(2.40\,\mathrm{m}\) down during \(0.20\,\mathrm{s}\).

During the same interval, a level camera attached to an eastbound train records the same drop moving \(0.80\,\mathrm{m}\) towards the rear and \(2.40\,\mathrm{m}\) down.

Assume both cameras measure displacements along the same calibrated axes, and all relevant velocities are constant.

a) Determine the rain’s ground-frame velocity components.

b) Calculate the train’s eastward speed.

c) If the train slows to \(4.0\,\mathrm{m\,s^{-1}}\) east while the rain’s ground-frame velocity remains unchanged, calculate the rain’s new apparent angle and direction.

d) Explain why the original train recording does not prove that the wind is blowing towards the west.

Solution 2

The rain moves at \(6.0\,\mathrm{m\,s^{-1}}\) east and \(12.0\,\mathrm{m\,s^{-1}}\) down relative to the ground. The train initially travels east at \(10.0\,\mathrm{m\,s^{-1}}\). After slowing to \(4.0\,\mathrm{m\,s^{-1}}\), the rain appears to move \(9.46^\circ\) towards the front from the downward vertical.

Part a) Divide each ground-frame displacement by the \(0.20\,\mathrm{s}\) interval:

\[ \begin{aligned} v_{Rx}&=\frac{1.20}{0.20}=6.0\\ v_{Ry}&=\frac{-2.40}{0.20}=-12.0 \end{aligned} \]

Both components are in \(\mathrm{m\,s^{-1}}\).

Therefore:

\[ \vec v_{R/G}=(6.0,-12.0)\,\mathrm{m\,s^{-1}} \]

Part b) The rain’s horizontal velocity relative to the train is:

\[ v_{Rx/T}=\frac{-0.80}{0.20} =-4.0\,\mathrm{m\,s^{-1}} \]

Using \(v_{Rx/T}=w-u\):

\[ \begin{aligned} -4.0&=6.0-u\\ u&=10.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part c) At the slower train speed:

\[ \begin{aligned} v_{Rx/T}&=6.0-4.0\\ &=+2.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

The vertical component remains \(-12.0\,\mathrm{m\,s^{-1}}\).

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{2.0}{12.0}\right)\\ &=9.46^\circ \end{aligned} \]

The positive horizontal component means the rain now appears to move towards the front.

Part d) The original recording only measures motion relative to the train. Rain moving east more slowly than the train will appear to move west in the train frame.

The platform data settle the question: the rain itself has a positive eastward horizontal velocity.

The tempting conclusion is to treat the direction seen by the moving camera as the ground-frame direction. But those are different measurements. The reversal at \(u=6.0\,\mathrm{m\,s^{-1}}\) confirms why the observer’s frame must be included.

Question 3

A train travels east at a constant \(12\,\mathrm{m\,s^{-1}}\). Its front windscreen lies in a vertical plane perpendicular to the direction of travel, extending from \(1.0\,\mathrm{m}\) to \(3.0\,\mathrm{m}\) above track level.

Rain has a constant ground-frame velocity of \(2.0\,\mathrm{m\,s^{-1}}\) east and \(10.0\,\mathrm{m\,s^{-1}}\) down.

At the same instant, two raindrops are each \(5.0\,\mathrm{m}\) ahead of the train’s front:

• Drop P is \(7.0\,\mathrm{m}\) above track level.
• Drop Q is \(3.5\,\mathrm{m}\) above track level.

Assume the drops are aligned laterally with the windscreen, the ground is level, and each drop stops moving as an airborne particle when it hits the ground or train.

Determine whether each drop strikes the windscreen. Calculate any impact height or ground-landing position relative to the train’s front at that instant.

Solution 3

Drop P strikes the windscreen at a height of \(2.0\,\mathrm{m}\) above the track. Drop Q does not strike the windscreen: it reaches the ground \(1.5\,\mathrm{m}\) ahead of the train’s front before the train catches up horizontally.

The key is that matching horizontal positions is not enough. The raindrop must also be at windscreen height when that happens.

First calculate the rain’s horizontal velocity relative to the train:

\[ \begin{aligned} v_{Rx/T}&=2.0-12\\ &=-10.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

Each drop initially lies \(5.0\,\mathrm{m}\) ahead, so the horizontal meeting time would be:

\[ \begin{aligned} t&=\frac{5.0\,\mathrm{m}} {10.0\,\mathrm{m\,s^{-1}}}\\ &=0.50\,\mathrm{s} \end{aligned} \]

For drop P, the height after \(0.50\,\mathrm{s}\) is:

\[ \begin{aligned} y_P&=7.0-10.0(0.50)\\ &=2.0\,\mathrm{m} \end{aligned} \]

Since \(2.0\,\mathrm{m}\) lies between the windscreen’s lower and upper edges, P hits the windscreen.

Now consider drop Q.

Blindly using the same \(0.50\,\mathrm{s}\) gives:

\[ y_Q=3.5-10.0(0.50)=-1.5\,\mathrm{m} \]

But a raindrop cannot continue falling through the ground. This negative height signals that we have extended the airborne-motion model beyond its valid interval.

Drop Q actually reaches the ground at:

\[ \begin{aligned} t_Q&=\frac{3.5\,\mathrm{m}} {10.0\,\mathrm{m\,s^{-1}}}\\ &=0.35\,\mathrm{s} \end{aligned} \]

During that time, the horizontal separation ahead of the moving train decreases by:

\[ 10.0(0.35)=3.5\,\mathrm{m} \]

Therefore, its landing position relative to the front is:

\[ 5.0-3.5=1.5\,\mathrm{m} \]

Drop Q lands \(1.5\,\mathrm{m}\) ahead of the train.

This is why relative velocity predicts the direction of motion but doesn’t, by itself, guarantee a collision. The geometry and any earlier physical contact also matter.

Question 4

An eastbound train initially travels at \(6.0\,\mathrm{m\,s^{-1}}\) and then accelerates uniformly east at \(1.5\,\mathrm{m\,s^{-2}}\).

Rain falls vertically relative to the ground at a constant \(10.0\,\mathrm{m\,s^{-1}}\).

At the instant acceleration begins, a passenger sees a raindrop \(15.0\,\mathrm{m}\) ahead and \(30.0\,\mathrm{m}\) above their position.

Take \(t=0\) at this instant.

a) Determine the drop’s horizontal and vertical positions relative to the passenger after \(2.0\,\mathrm{s}\).

b) Calculate the rain’s instantaneous velocity components relative to the train at that time.

c) Would the passenger observe a straight or curved path as the drop moves across their view during those two seconds? Explain.

d) A student argues that because the drop’s horizontal position changes by \(15.0\,\mathrm{m}\) in two seconds, its backward velocity must have remained \(7.5\,\mathrm{m\,s^{-1}}\) throughout. Identify the mistake.

Solution 4

After \(2.0\,\mathrm{s}\), the drop is directly above the passenger’s horizontal position and \(10.0\,\mathrm{m}\) higher. Its instantaneous relative velocity is \(9.0\,\mathrm{m\,s^{-1}}\) backwards and \(10.0\,\mathrm{m\,s^{-1}}\) downwards. The path appears curved because the train is accelerating.

Part a) The drop has no horizontal ground-frame movement. The train’s horizontal displacement is found using:

\[ s=ut+\frac12at^2 \]

Here, \(s\) is displacement, \(u\) is initial speed, \(a\) is acceleration, and \(t\) is time.

\[ \begin{aligned} s_T&=6.0(2.0) +\frac12(1.5)(2.0)^2\\ &=12.0+3.0\\ &=15.0\,\mathrm{m} \end{aligned} \]

The train travels exactly the drop’s initial horizontal separation, so the drop is now at horizontal position \(x=0\,\mathrm{m}\) relative to the passenger.

Its vertical position is:

\[ \begin{aligned} y&=30.0-10.0(2.0)\\ &=10.0\,\mathrm{m} \end{aligned} \]

Part b) The train’s speed after two seconds is:

\[ \begin{aligned} u_T&=6.0+1.5(2.0)\\ &=9.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

The instantaneous rain velocity relative to the train is therefore:

\[ \vec v_{R/T}=(-9.0,-10.0)\, \mathrm{m\,s^{-1}} \]

The apparent horizontal motion is towards the rear.

Part c) The passenger sees a curved path because the train’s horizontal velocity changes while the drop’s ground-frame horizontal velocity stays zero.

For any time \(t\) measured in seconds, the relative coordinates are:

\[ \begin{aligned} x(t)&=15.0-6.0t-0.75t^2\\ y(t)&=30.0-10.0t \end{aligned} \]

Both coordinates are in metres.

The squared-time term in the horizontal position produces a curved trajectory in the train frame.

The relative horizontal acceleration is \(-1.5\,\mathrm{m\,s^{-2}}\). This doesn’t mean the drop has acquired a westward ground-frame acceleration. It arises because the observer’s frame is accelerating east.

Part d) The student has calculated an average horizontal velocity, not an instantaneous velocity.

The drop’s horizontal displacement relative to the train is \(-15.0\,\mathrm{m}\) over \(2.0\,\mathrm{s}\), so its average horizontal velocity is:

\[ \bar v_x=\frac{-15.0}{2.0} =-7.5\,\mathrm{m\,s^{-1}} \]

But its instantaneous horizontal velocity changes from \(-6.0\,\mathrm{m\,s^{-1}}\) to \(-9.0\,\mathrm{m\,s^{-1}}\).

The average correctly describes the total displacement per unit time. It does not describe the velocity at every instant.

Question 5

Three trains travel east along straight, level tracks through the same steady rainstorm.

Train A travels at a constant \(9.0\,\mathrm{m\,s^{-1}}\). Its passenger sees airborne rain move down and towards the front, shifting \(1.0\,\mathrm{m}\) horizontally for every \(5.0\,\mathrm{m}\) vertically downward.

Train B travels at a constant \(13.0\,\mathrm{m\,s^{-1}}\). Its passenger sees the rain move down and towards the rear, also shifting \(1.0\,\mathrm{m}\) horizontally for every \(5.0\,\mathrm{m}\) vertically downward.

Assume the rain has one uniform, constant ground-frame velocity.

Train C begins travelling at \(9.0\,\mathrm{m\,s^{-1}}\) east and accelerates uniformly to \(13.0\,\mathrm{m\,s^{-1}}\) east over \(2.0\,\mathrm{s}\).

Attached to its roof is a narrow vertical collection tube. The tube is \(1.0\,\mathrm{m}\) tall and has an internal east-west width of \(0.30\,\mathrm{m}\). Its opening is centred directly above a fixed point on the train’s roof.

At \(t=0\), a raindrop is directly above the centre of this tube, at a height of \(20.0\,\mathrm{m}\) above the roof.

Assume the tube walls are rigid and negligibly thin, the drop has negligible size, and airflow near the train does not alter the rain’s velocity.

a) Determine the rain’s horizontal and vertical velocity components relative to the ground.

b) At what time does the rain appear to fall vertically to a passenger on train C? Where is the particular drop relative to the centre of the tube at that instant?

c) A student argues that the drop must be collected because it starts directly above the tube and will again be directly above its centre when it reaches roof height. Determine whether the drop actually enters the tube. If it doesn’t, find where its first collision occurs.

d) Would shortening the tube to \(0.50\,\mathrm{m}\), without changing its width, allow the same drop to enter? Justify your conclusion.

Solution 5

The rain moves at \(11.0\,\mathrm{m\,s^{-1}}\) east and \(10.0\,\mathrm{m\,s^{-1}}\) down relative to the ground. It appears vertical to train C after \(1.0\,\mathrm{s}\), but the selected drop is then \(1.0\,\mathrm{m}\) ahead of the tube.

The drop does not enter the original \(1.0\,\mathrm{m}\)-high tube. It strikes the tube’s front exterior wall approximately \(0.78\,\mathrm{m}\) above the roof. Shortening the tube to \(0.50\,\mathrm{m}\) would allow it to enter.

The important complication is that arriving at the correct horizontal position at roof height is not sufficient. The drop must pass through the opening without first crossing a solid wall.

Part a): Recover the rain’s ground-frame velocity.

Let \(w\) be its eastward ground-frame velocity and \(d\) its downward speed, both in \(\mathrm{m\,s^{-1}}\).

Train A sees a positive horizontal component:

\[ \frac{w-9}{d}=\frac15 \]

Train B sees a negative horizontal component of the same magnitude:

\[ \frac{w-13}{d}=-\frac15 \]

Therefore:

\[ \begin{aligned} w-9&=\frac d5\\ w-13&=-\frac d5 \end{aligned} \]

Subtract the second equation from the first:

\[ \begin{aligned} 4&=\frac{2d}{5}\\ d&=10.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

Substitute back:

\[ \begin{aligned} w-9&=2.0\\ w&=11.0\,\mathrm{m\,s^{-1}} \end{aligned} \]

So:

\[ \vec v_{R/G}=(11.0,-10.0)\, \mathrm{m\,s^{-1}} \]

Both train observations are consistent with this same rain velocity. Train A moves east more slowly than the rain’s horizontal component, whereas train B moves east faster.

Part b): Find when the rain appears vertical.

Train C accelerates from \(9.0\) to \(13.0\,\mathrm{m\,s^{-1}}\) over two seconds.

Its acceleration is:

\[ \begin{aligned} a&=\frac{13.0-9.0}{2.0}\\ &=2.0\,\mathrm{m\,s^{-2}} \end{aligned} \]

At time \(t\), its eastward speed is:

\[ u_T=9.0+2.0t \]

The rain appears vertical when the train’s horizontal speed equals the rain’s:

\[ \begin{aligned} 9.0+2.0t&=11.0\\ t&=1.0\,\mathrm{s} \end{aligned} \]

But appearing vertical at that instant doesn’t mean the drop has remained above the opening.

To find its position, calculate horizontal displacements from \(t=0\).

The rain moves east by:

\[ x_R=11.0t \]

The train moves east by:

\[ x_T=9.0t+1.0t^2 \]

Here, positions are measured in metres relative to their respective starting positions, and time \(t\) is in seconds.

Because the drop begins directly above the tube, its relative horizontal position is:

\[ \begin{aligned} x_{\mathrm{rel}}&=x_R-x_T\\ &=2.0t-1.0t^2 \end{aligned} \]

At \(t=1.0\,\mathrm{s}\):

\[ \begin{aligned} x_{\mathrm{rel}}&=2.0(1.0)-(1.0)^2\\ &=1.0\,\mathrm{m} \end{aligned} \]

Its height above the roof is:

\[ \begin{aligned} y&=20.0-10.0(1.0)\\ &=10.0\,\mathrm{m} \end{aligned} \]

So the drop is \(1.0\,\mathrm{m}\) ahead of the tube and \(10.0\,\mathrm{m}\) above the roof at the precise moment its apparent motion is vertical.

This is the first conceptual trap: instantaneous relative velocity and relative position are different quantities.

Part c): Check whether the drop enters the tube.

If there were no obstruction, the drop would reach roof height at:

\[ \begin{aligned} 0&=20.0-10.0t\\ t&=2.0\,\mathrm{s} \end{aligned} \]

At this time:

\[ \begin{aligned} x_{\mathrm{rel}}&=2.0(2.0)-(2.0)^2\\ &=0\,\mathrm{m} \end{aligned} \]

That appears to confirm the student’s argument. The drop starts and ends directly above the centre.

But the collection tube is \(1.0\,\mathrm{m}\) tall. The drop encounters the height of its upper opening before reaching roof level.

At the height of the tube’s mouth:

\[ \begin{aligned} 1.0&=20.0-10.0t\\ t&=1.90\,\mathrm{s} \end{aligned} \]

Calculate its horizontal position relative to the tube:

\[ \begin{aligned} x_{\mathrm{rel}} &=2.0(1.90)-(1.90)^2\\ &=3.80-3.61\\ &=0.19\,\mathrm{m} \end{aligned} \]

The opening extends from \(x=-0.15\,\mathrm{m}\) to \(x=+0.15\,\mathrm{m}\).

At the height of the opening, the drop is \(0.19\,\mathrm{m}\) ahead of the centre, so it misses the opening by \(0.04\,\mathrm{m}\).

The train is now moving faster than the rain’s horizontal component, so the tube continues to catch up with the drop.

The drop first touches the tube’s front exterior wall when:

\[ x_{\mathrm{rel}}=0.15\,\mathrm{m} \]

Thus:

\[ 2.0t-t^2=0.15 \]

Rearranging and solving the quadratic:

\[ t^2-2.0t+0.15=0 \]

The two solutions are approximately \(0.078\,\mathrm{s}\) and \(1.922\,\mathrm{s}\).

At the earlier time the drop is still far above the tube, so there is no collision. The later time is the relevant one.

Its height then is:

\[ \begin{aligned} y&=20.0-10.0(1.922)\\ &\approx0.78\,\mathrm{m} \end{aligned} \]

This lies between roof height and the top of the \(1.0\,\mathrm{m}\)-high tube.

Therefore, the drop strikes the front exterior wall approximately \(0.78\,\mathrm{m}\) above the roof.

It never reaches the tube’s interior.

Part d): Test the shorter tube.

For a \(0.50\,\mathrm{m}\)-high tube, the drop reaches the opening height when:

\[ \begin{aligned} 0.50&=20.0-10.0t\\ t&=1.95\,\mathrm{s} \end{aligned} \]

Its relative horizontal position is:

\[ \begin{aligned} x_{\mathrm{rel}} &=2.0(1.95)-(1.95)^2\\ &=3.90-3.8025\\ &=0.0975\,\mathrm{m} \end{aligned} \]

This lies within the opening:

\[ -0.15<0.0975<0.15 \]

From \(t=1.95\,\mathrm{s}\) to \(t=2.0\,\mathrm{s}\), the drop continues moving towards the tube’s centre in the train frame. Its horizontal position decreases from \(0.0975\,\mathrm{m}\) to zero.

It therefore remains between the walls and reaches the bottom without striking either side.

The shorter tube collects the drop.

This is a counterintuitive result: reducing the tube’s height makes collection possible, even though the rain’s velocity and the train’s motion have not changed.

The governing constraint is the drop’s position when it crosses the height of the opening, not just its position at roof level.

The problem combines three ideas that must be checked separately: relative velocity determines the apparent direction, relative displacement determines where the drop moves, and physical boundaries determine whether that predicted motion can continue.

06What should you do with this idea next?

Whenever rain looks slanted from a moving train, start by identifying the observer.

If the rain’s ground-frame velocity is known, subtract the train’s velocity as a vector. The result tells you the apparent direction and speed. If you’re predicting whether a drop hits a window, roof, or opening, use that relative motion to calculate positions and check the geometry as well.

The same reference-frame reasoning also explains why an aircraft’s direction through the air may differ from its direction over the ground. In How to Calculate an Aircraft’s Required Heading from Wind Data, the task changes from interpreting apparent motion to deliberately choosing a velocity that compensates for wind.