Question

A passenger walks from one side of a ferry to the other as it approaches a dock .Passenger's velocity is 1.45 m/sm/s due north relative to the ferry, and 4.40 m/s at an angle of 30.0 ∘ west of north relative to the water.

part a) What is the magnitude of the ferry's velocity relative to the water?

part b) What is the direction of the ferry's velocity relative to the water?

Answer #1

a)

Assuming y-axis along the north-south direction with north being towards positive Y-axis.

Assuming x-axis along the east-west direction with east being towards positive x-axis.

v_{pf} = velocity of passenger relative to ferry = 0 i +
1.45 j

v_{pw} = velocity of passenger relative to water = -
(4.40 Sin30) i + (4.40 Cos30) j = - 2.20 i + 3.81 j

v_{fw} = velocity of ferry relative to water

v_{pf} = v_{pw} - v_{fw}

0 i + 1.45 j = - 2.20 i + 3.81 j - v_{fw}

v_{fw} = - 2.20 i + 3.81 j - 1.45 j

v_{fw} = - 2.20 i + 2.36 j

magnitude is given as

|v_{fw}| = sqrt((- 2.20)^{2} +
(2.36)^{2})

|v_{fw}| = 3.23 m/s

b)

Direction :

= tan^{-1}(2.36/2.20)

= 47.01 deg north of west.

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