Question

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?

Answer 1

The distance between the ships is 87.84 km.

Step-by-step 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=√(18^2+26^2-2*18*26*\cos90)`

`v=√(18^2+26^2)`

`v=√(324+676)`

`v=10√(10)`

We need to calculate the distance between the ships

`d =v* t`

Put the value into the formula

`d=10√(10)*1.5*1.852`

`d=87.84\ km`

Hence, The distance between the ships is 87.84 km.