Question

One ship leaves a port and sails at 17km/h on a bearing of 024 degrees. A second ship leaves the same port at the same time and sails at 21km/h on a bearing of 079 degrees. How far apart are 2 ships after 2 hours.

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Answer to a math question One ship leaves a port and sails at 17km/h on a bearing of 024 degrees. A second ship leaves the same port at the same time and sails at 21km/h on a bearing of 079 degrees. How far apart are 2 ships after 2 hours.

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Neal
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58 Answers
Step 1: Find the displacement vector for each ship after 2 hours.
Let d_1 be the displacement vector for the first ship and d_2 be the displacement vector for the second ship.
The displacement vector is calculated using the formula:
\textbf{d} = \textbf{v} \times \textbf{t}
where:
\textbf{v} = velocity vector,
\textbf{t} = time vector.

For the first ship:
\textbf{v}_1 = 17 \, \text{km/h}
\textbf{t} = 2 \, \text{hours}

Calculating the displacement vector for the first ship:
d_1 = 17 \times 2 = 34 \, \text{km} \, (\text{bearing} \, 024^\circ)

For the second ship:
\textbf{v}_2 = 21 \, \text{km/h}
\textbf{t} = 2 \, \text{hours}

Calculating the displacement vector for the second ship:
d_2 = 21 \times 2 = 42 \, \text{km} \, (\text{bearing} \, 079^\circ)

Step 2: Find the distance between the two ships.
To find the distance between the two ships, we need to find the vector sum of the two displacement vectors:
\textbf{D} = \sqrt{(d_{1x} - d_{2x})^2 + (d_{1y} - d_{2y})^2}
where:
The x-component of d_1 = 34 \cos(24^\circ) ,
The y-component of d_1 = 34 \sin(24^\circ) ,
The x-component of d_2 = 42 \cos(79^\circ) ,
The y-component of d_2 = 42 \sin(79^\circ) .

Calculating the x and y components:
d_{1x} = 34 \cos(24^\circ)
d_{1y} = 34 \sin(24^\circ)
d_{2x} = 42 \cos(79^\circ)
d_{2y} = 42 \sin(79^\circ)

Step 3: Find the distance between the two ships using the formula.
Substitute the x and y components into the formula:
\textbf{D} = \sqrt{(34\cos(24^\circ) - 42\cos(79^\circ))^2 + (34\sin(24^\circ) - 42\sin(79^\circ))^2}

Step 4: Calculate the distance and find the final answer.
\textbf{D} = \sqrt{(34\cos(24^\circ) - 42\cos(79^\circ))^2 + (34\sin(24^\circ) - 42\sin(79^\circ))^2}
\textbf{D}=\sqrt{(23.05)^2+(-27.4)^2}
\textbf{D}=\sqrt{531.3+750.76}
\textbf{D}=\sqrt{1282.06}
\textbf{D}\approx35.8\text{ km}

Therefore, the two ships are approximately 35.8 km apart after 2 hours.

\boxed{35.8\text{ km}}

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