Question

What is the magnitude and direction of the boat's velocity relative to the ground? (See problem)

A `220m` wide river flows due east at a uniform speed of `4.8m/s`. A boat with a speed of `8.1m/s` relative to the water leaves the south bank pointed in a direction `35^@` west of north.

Answer 1

Speed of the boat : `\qquad \quad v_{bg} = 6.63(2)` `ms^{-1}`
Orientation of its path : `\phi = 88.6^o` to the north of east.

Explanation:

Relative Velocities: `vec v_{bg} = vec v_{br} + vec v_{rg}` ...... (1)
`vecv_{bg}` : Velocity of the boat relative to ground;
`vecv_{br}` : Velocity of the boat relative to river;
`vecv_{rg}` : Velocity of the river relative to ground;

Coordinate System: Consider a cartesian coordinate system with its origin located at the boat's starting point and with its X-axis pointing east and Y-axis pointing north.

Write down the velocities in this coordinate system:
(i) River is directed east: `vec v_{rg} = (+4.8, 0)` `ms^{-1}`

(ii) Velocity vector of the boat is pointing `35^o` to the west of north, which is `125^o` counter-clockwise from east. Speed of the boat relative to the river is `v_{br} = 8.1` `ms^{-1}`.

`vec v_{br} = (v_{br}\cos125^o, v_{br}sin125^o ) = (-4.64, +6.63)` `ms^{-1}`

(iii) Use (1) to find the velocity vector of the boat relative to the ground:
`vec v_{bg} = vec v_{br} + vec v_{rg} = (-4.64+4.8, 6.63+0)` `ms^{-1}`.
`\qquad \qquad= (+0.16, +6.63)` `ms^{-1}`

`v_{bg} = \sqrt{(0.16)^2+(6.63)^2} = 6.63(2)` `ms^{-1}`
`\phi = arctan((6.63ms^{-1})/(0.16ms^{-1})) = 88.6^o`