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two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing of 130 degrees at 26 knots. How far apart are they after 1.5 hours?

Respuesta :

Answer:

The distance between the ships is 87.84 km.

Explanation:

Given that,

Angle of first ship= 40°

Speed of first ship = 18 knots

Angle of second ship= 130°

Speed of second ship = 26 knots

We need to calculate the resultant velocity

Using cosine rule

[tex]v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}[/tex]

Put the value into the formula

[tex]v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}[/tex]

[tex]v=\sqrt{18^2+26^2}[/tex]

[tex]v=\sqrt{324+676}[/tex]

[tex]v=10\sqrt{10}[/tex]

We need to calculate the distance between the ships

[tex]d =v\times t[/tex]

Put the value into the formula

[tex]d=10\sqrt{10}\times1.5\times1.852[/tex]

[tex]d=87.84\ km[/tex]

Hence, The distance between the ships is 87.84 km.

Ver imagen CarliReifsteck