# A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is 1.50 m/s due north relative to the ferry, andd 4.50 m/s at an angle of 30 degrees...

A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is 1.50 m/s due north relative to the ferry, andd 4.50 m/s at an angle of 30 degrees west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?

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Let us use the following notations.

M = Man

F = Ferry

W = Water

The given detail are found in the attached image.

Let us define the horizontal and vertical components of each velocity.

`uarrV_(M,F) = 1.5`

`larrV_(M,F) = 0`

`uarrV_(M,W) = 4.5cos30`

`larrV_(M,W) = 4.5sin30`

Using vector addition we know that;

`V_(F,W) = V_(F,M)+V_(M,W)`

We have to get the `V_(F,W) ` at horizontal and vertical components.

`uarrV_(F,W) = -V_(M,F)+V_(M,W)`

`uarrV_(F,W) = -1.5+4.5cos30 = 2.397`

`larrV_(F,W) = V_(F,M)+V_(M,W)`

`larrV_(F,W) = -V_(M,F)+V_(M,W)`

`larrV_(F,W) = 0+4.5sin30 = 2.25`

`V_(F,W) = sqrt[uarr(V_(F,W))^2+larr(V_(F,W))^2]`

`V_(F,W) = sqrt(2.397^2+2.25^2)`

`V_(F,W) = 3.288m/s`

`alpha = tan^(-1)(uarrV_(F,W))/(rarrV_(F,W)) = tan^(-1)(2.397/-2.25)`

`alpha = -46.81deg`

*So the velocity of ferry relative to the water is 3.288m/s and it is at the direction of 46.81deg west to due north.*

Relevant image for the answer;