Question

A boat travelled from its home port to an island at the distance of $l = 4$ km directly towards south. It trajectory consisted of three straight segments the directions of which were not recorded. During each of the segments, the boat maintained a constant velocity; however, a different speed was maintained for different segments. During the travel time, wind speed and direction was measured from the boat. The travel time on the first segment was $t_1 = 3$ min, the measured wind speed was $v_1 = 15$ m/s and the wind blew directly from east. The travel time on the first segment was $t_2 = 1.5$ min, the measured wind speed was $v_2 = 10$ m/s and the wind blew directly from southeast. The travel time on the first segment was $t_3 = 1.5$ min, the measured wind speed was $v_3 = 5$ m/s and the wind blew directly from southwest. What was the wind speed? It is known that the wind speed and direction were constant during all the travel time.

• A. 12 m/s
• B. 24 m/s
• C. 36 m/s
• D. 6 m/s

Answer 1

Answer: (A)

12 m/s

Answer 2

Solution

Let the true constant wind vector be $\vec{W}=(W_x, W_y)$ (in m/s). In each segment, the boat’s measured (apparent) wind is

$$\vec{u}_i=\vec{W}-\vec{v}_i,$$

so that

$$\vec{v}_i=\vec{W}-\vec{u}_i.$$

Step 1. Set up the relative wind vectors.

We use the convention: East $+$x, North $+$y.

1. Segment 1:
   Measured wind: 15 m/s “from east” means the apparent wind comes from east, i.e.

   $$\vec{u}_1=(-15,\,0).$$

   Thus,

   $$\vec{v}_1=\vec{W}-\vec{u}_1 = \vec{W}+(15,\,0) = (W_x+15,\,W_y).$$

2. Segment 2:
   Measured wind: 10 m/s “from southeast.”
   “From southeast” means the apparent wind is arriving from southeast, so it points toward northwest. Since the southeast direction makes a 45° angle, we have

   $$\vec{u}_2=(-10/\sqrt{2},\,10/\sqrt{2}).$$

   Hence,

   $$\vec{v}_2=\vec{W}-\vec{u}_2=\Big(W_x+\frac{10}{\sqrt{2}},\,W_y-\frac{10}{\sqrt{2}}\Big).$$

3. Segment 3:
   Measured wind: 5 m/s “from southwest” means the apparent wind comes from southwest (pointing toward northeast). Thus,

   $$\vec{u}_3=(5/\sqrt{2},\,5/\sqrt{2}).$$

   And,

   $$\vec{v}_3=\vec{W}-\vec{u}_3=\Big(W_x-\frac{5}{\sqrt{2}},\,W_y-\frac{5}{\sqrt{2}}\Big).$$

Step 2. Use the displacement constraint.

The boat’s overall displacement is 4 km due South, i.e.,

$$\Delta \vec{r}=(0,\,-4000) \text{ m}.$$

Convert times to seconds:

$$t_1=3\text{ min}=180\,s,\quad t_2=t_3=1.5\text{ min}=90\,s.$$

The total displacement is:

$$\vec{r} = \vec{v}_1\,t_1 + \vec{v}_2\,t_2 + \vec{v}_3\,t_3.$$

x-component:

$$\begin{aligned} 0 &= 180(W_x+15) + 90\Big(W_x+\frac{10}{\sqrt{2}}\Big) + 90\Big(W_x-\frac{5}{\sqrt{2}}\Big)\\[1mm] &= (180+90+90)W_x + 180\cdot15 + 90\Big(\frac{10}{\sqrt{2}}-\frac{5}{\sqrt{2}}\Big)\\[1mm] &= 360W_x + 2700 + 90\Big(\frac{5}{\sqrt{2}}\Big)\\[1mm] &= 360W_x + 2700 + \frac{450}{\sqrt{2}}. \end{aligned}$$

Thus,

$$W_x = -\frac{2700+\frac{450}{\sqrt{2}}}{360}.$$

y-component:

$$\begin{aligned} -4000 &= 180W_y + 90\Big(W_y-\frac{10}{\sqrt{2}}\Big) + 90\Big(W_y-\frac{5}{\sqrt{2}}\Big)\\[1mm] &= (180+90+90)W_y -90\Big(\frac{10+5}{\sqrt{2}}\Big)\\[1mm] &= 360W_y - \frac{1350}{\sqrt{2}}. \end{aligned}$$

So,

$$360W_y = -4000 + \frac{1350}{\sqrt{2}},\quad\text{and}\quad W_y = \frac{-4000 + \frac{1350}{\sqrt{2}}}{360}.$$

Step 3. Find the wind speed $|\vec{W}|$.

Calculate numerically (using $\sqrt{2}\approx1.414$):

$$\begin{aligned} W_x &=-\frac{2700+ \frac{450}{1.414}}{360} \approx -\frac{2700+318.2}{360} = -\frac{3018.2}{360} \approx -8.3839 \text{ m/s},\\[1mm] W_y &\approx \frac{-4000+\frac{1350}{1.414}}{360} = \frac{-4000+955}{360} = \frac{-3045}{360} \approx -8.4583 \text{ m/s}. \end{aligned}$$

Thus,

$$|\vec{W}|=\sqrt{W_x^2+W_y^2}\approx \sqrt{(8.3839)^2+(8.4583)^2} \approx \sqrt{70.3+71.5}\approx \sqrt{141.8}\approx 11.91\, \text{m/s}.$$

This is very close to 12 m/s.

Answer: 12 m/s (Option 1)

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Core Explanation:
Set up vector equations for each segment using $\vec{W}-\vec{v}_i=\vec{u}_i$. Write the boat’s displacement (summing $\vec{v}_i t_i$) and equate its x-component to 0 (since total displacement x = 0) and its y-component to $-4000$ m. Solve for $W_x$ and $W_y$ and then compute $|\vec{W}|$.