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

Answer:

a)$V_b=6.1m/s$

b) $\alpha=66.59 \textdegree$

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

 $V_b=\sqrt{(4.8-7.2sin33)^2+(7.2cos33j)^2}$

 $V_b=6.1m/s$

b)

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

 $\alpha=tan^{-1}\frac{y}{x}$

 $\alpha=tan^{-1}\frac{7.2cos33)}{4.8-7.2sin33}$

 $\alpha=66.59 \textdegree$