Question

A 240-m-wide river flows due east at a uniform speed of 4.8m/s. A boat with a speed of 7.2m/s relative to the water leaves the south bank pointed in a direction 33o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground

Answer 1

a)
`V_b=6.1m/s`

b)
`\alpha=66.59 \textdegree`

Step-by-step explanation:

From the question we are told that:

Width of river
`W=240`

River speed
`V_r=4.8m/s`

Boat speed
`V_b=7.2m/s`

Boat Direction
`\theta= 33 \textdegree`

Generally the equation for Boat Velocity Vector is mathematically given by

By Resolving Co-Planar forces

`\=V_(br)=\=V_b-\=V_r`

`\=V_(br)=-7.2sin33i+7.2cos33j`

Therefore

`\={V_(b)}=\={V_(br)}-\={V_r}`
`\=V_(b)=\=V_(br)-\=V_r`

`\=V_(b)=4.8-7.2sin33i+7.2cos33j`

`\=V_b=4.8-7.2 sin33 i+7.2 cos33j`

Therefore Magnitude of Boat velocity is

`V_b=√((4.8-7.2sin33)^2+(7.2cos33j)^2)`

`V_b=6.1m/s`

b)

Generally the equation for Direction of the boat's velocity is mathematically given by

`\alpha=tan^(-1)(y)/(x)`

`\alpha=tan^(-1)(7.2cos33))/(4.8-7.2sin33)`

`\alpha=66.59 \textdegree`