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2 years ago

9

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?

1 answer:

Yuliya22 [10]2 years ago

7 0

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

$v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}$

Put the value into the formula

$v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}$

$v=\sqrt{18^2+26^2}$

$v=\sqrt{324+676}$

$v=10\sqrt{10}$

We need to calculate the distance between the ships

$d =v\times t$

Put the value into the formula

$d=10\sqrt{10}\times1.5\times1.852$

$d=87.84\ km$

Hence, The distance between the ships is 87.84 km.

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Answer:

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Explanation:

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2) The instantaneous velocity is $64t-260t^3$

Explanation:

1)

The average velocity of an object is given by

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d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

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While the position at time t = 12 sec is

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The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

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By differentiating with respect to t, we find the velocity vector:

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Learn more about velocity:

brainly.com/question/5248528

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