Q2.11 Relative velocities 1 (S)

A train travels due south at 30 ms-1, relative to the ground. Rain is falling (it is Scotland), and is swept south by the wind. According to an observer standing on the ground, by the track, the rain makes an angle of 70 with the vertical. An occupant of the train, however, sees the rain fall vertically. Determine the speed of the rain relative to the ground.
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Hint

You will need Key Point 1.9. The vectors you need to relate are $\vec{v}_{TG}$, $\vec{v}_{RT}$ and $\vec{v}_{RG}$ where R, T and G stand for rain, train and ground.
 

Solution

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Solution

The task we have is to use the various bits of information provided to construct the vector triangle relating the vectors $\vec{v}_{TG}$, $\vec{v}_{RT}$ and $\vec{v}_{RG}$ where R, T and G stand for rain, train and ground, and

  • $\vec{v}_{TG}$ means the velocity of the train relative to the ground
  • $\vec{v}_{RT}$ means the velocity of the rain relative to the train
  • $\vec{v}_{RG}$ means the velocity of the rain relative to the ground;

We know from Key Point 1.9 that these three vectors must satisfy the vector triangle relationship

\[ \vec{v}_{RG}= \vec{v}_{RT} +\vec{v}_{TG} \]

Since we know the magnitude and the direction of $\vec{v}_{TG}$ we probably draw that vector first, to the right, say (choosing that as south), and horizontal.

Graphic - No title - trainrain

Now we recognise that although we don’t know the magnitude of $\vec{v}_{RG}$ we do know its direction (the angle it makes with the vertical). So we draw a line, to represent $\vec{v}_{RG}$, making the correct angle with the vertical, and with its tip lying on the tip of $\vec{v}_{TG}$. (You will see why we position it that way in a moment).

We don’t know the magnitude of the remaining vector $\vec{v}_{RT}$ either. But we do know that it is ‘vertical’ (this means that if there are x and y axes drawn on a window pane of the train a passenger sees the rain running down the y-axis). So we draw a line to represent $\vec{v}_{RT}$. It has to be vertical; and we position it so that its tip lies on the start of $\vec{v}_{TG}$. Now, if we chop the two vectors (whose length we don’t know) at the point where they cross, we have a vector triangle that satisfies all the data AND the triangle rule given above. (The choices we made in positioning the vectors were needed to bring us to this point).

The rest is easy. We can read off from the triangle that vRG (that is, the speed of the rain relative to the ground, and hence the magnitude of the vector $\vec{v}_{RG}$) is

\[ v_{RG} = \frac{v_{TG}} {\sin{70^{\circ}}} = 32 \, ms^{-1} \hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]