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?
Relevant image for the answer;
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.