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So here's the question:

A train window is 1 meter high and 10 meters long. A raindrop rolls down the window at a vertical speed of 5 meters per second(when stationary). If the train is moving at 30 m s–1 how far will it move in the time it takes the raindrop to roll down the window?

My problem with this question that i don't believe it's as simple at realising it takes $0.2$ seconds for the rain to move down the window and work out the distance the train will travel at that speed, within that time. This is because i've read somewhere about as you faster, rain won't just go directly down unless your stationary. So i'm thinking that the angle of the rain will go up as the train increases it's velocity. Any help would be appreciated, maybe i'm missing something here?

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    $\begingroup$ The question states a 'vertical speed' - I believe you are overthinking this... $\endgroup$
    – Jon Custer
    Commented Jul 13, 2015 at 14:59
  • $\begingroup$ Yes however, i've been given a "background text" for this, and in it it states: "If the train is stationary during the storm you will see water drops running vertically down the window, but as soon as the train begins to pick up speed the drops run at an angle. If you knew the speed of the train you could take a few simple measurements to find the velocity of the rain." $\endgroup$
    – user85779
    Commented Jul 13, 2015 at 15:03
  • $\begingroup$ Just because the drops run at an angle does not mean their vertical velocity has changed. It could mean an irrelevant horizontal velocity was added $\endgroup$
    – Jim
    Commented Jul 13, 2015 at 15:12
  • $\begingroup$ I deleted my answer for the time being because I didn't realize that the rate of vertical descent is while the train is stationary. I think you need to compute the raindrop's angled path and the time it would take to traverse that path from top to bottom of the window. $\endgroup$
    – Ernie
    Commented Jul 13, 2015 at 15:26
  • $\begingroup$ @Ernie, i have no idea how i can do that, this is an AS level question, surely it isn't that hard? $\endgroup$
    – user85779
    Commented Jul 13, 2015 at 15:59

3 Answers 3

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If the raindrop's vertical velocity is constant as the train is both stationary and moving, the time taken for the raindrop to travel down the window would be: $$t = \frac{1\ \text{m}}{5\ \text{m}/\text{s}} = 0.2\ \text{s}$$ Remember, the time $t$ would not depend on the speed of the train. The exercise also specifically states that the raindrop's vertical velocity, not the total velocity, is $5\ \text{m}/\text{s}$, and this therefore allows calculating the time taken for the raindrop to travel down the window as easily as above.

Note, the raindrop will travel both vertically down the window and horizontally with the train.

We calculate $d$, the distance the train travelled in time $t$. In $0.2\ \text{s}$, the train will travel: $$d = 30\ \text{m}/\text{s} \times 0.2\ \text{s} = 6\ \text{m}$$

So the train travels 6 metres in 0.2 seconds.

Edit: Previously this answer calculated the distance the raindrop travels because I misread the question. Like WhatRoughBeast said, the "10 metre length" of the window may be a red herring so I've ignored it. WhatRoughBeast's view is also possible, that the velocity of the raindrop may be constant and therefore the time $t$ would be dependent on the length of the raindrop's 6.08 metre track as it moves both forward and down but I've kept this answer with the naive view.

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  • $\begingroup$ Thanks for the answer and explanation, makes a lot of sense now! Thanks a lot. $\endgroup$
    – user85779
    Commented Jul 13, 2015 at 16:36
  • $\begingroup$ Note that the question asks how far the train moves, not how far the raindrop moves. $\endgroup$
    – WhatRoughBeast
    Commented Jul 13, 2015 at 17:55
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This is one of those questions that can drive you crazy, since there is a great deal assumed and not stated. Let me try an alternate possibility.

Conceivably, the problem wants you to assume that, when moving, the overall speed of the raindrop remains fixed at 5 m/sec, but it travels in a straight line at an angle due to horizontal wind forces, and this may or may not be supported by the datum that the window is 10 meters long - this may be a red herring. Since only the stationary speed and the horizontal speeds are given (so there is no way to try to back out the aerodynamic forces on the raindrop), a very naïve view of the problem would say that the angle of inclination would be 6 horizontal units for every vertical unit of travel (ratio of 30 to 5). Then the length of the raindrop's track would be $$L = \sqrt{1^2 + 6^2} = \sqrt{37} = 6.083 meters$$

The fact that the horizontal track length (6 meters) is less than 10 meters suggests that this might be what the question is asking for.

Regardless of just what the question is intended to ask, the answer of 6.08 meters is not correct since the question asks how far the train will travel, not how far the raindrop travels.

Then the total travel time of the raindrop is 1.21 seconds, and the train travels 36.5 meters.

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  • $\begingroup$ Thanks for the alternate answer, both make total sense i just think at this point the question is quite ambiguous. $\endgroup$
    – user85779
    Commented Jul 13, 2015 at 23:26
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Another way of looking at this:

The vertical component of the raindrop's velocity vector is 5 m/sec downward, and the horizontal component is 30 m/sec across. By the Pythagorean method, the resultant velocity vector is 30.41 m/sec diagonal.

The vertical component of the raindrop's displacement vector is 1 meter, and as the raindrop is pushed horizontally 6 units for every unit it falls vertically, 6 meters is the horizontal component of the raindrop's displacement vector. By the Pythagorean method, the resultant displacement vector is 6.08 meters diagonal.

The raindrop travels 6.08 meters at a velocity of 30.41 m/sec, so its travel time diagonally down the window is 0.199 seconds. Therefore the train travels 30 m/sec * 0.199 seconds = 5.99 meters during the time it takes for the raindrop to move from top to bottom of the window.

That's close to 6 meters, which would be the distance traveled if the vertical travel time of the raindrop were the same whether the train were moving or standing still. I don't think this is right, and I question if you were given enough information to solve this problem.

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  • $\begingroup$ So i'm just overthinking this? Ahh.. $\endgroup$
    – user85779
    Commented Jul 13, 2015 at 15:11

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