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A 200-m-wide river flows due east at a uniform speed of 2.0 m/s. A boat with a speed of 8.0 m/s relative to the water leaves the south bank pointed in a direction 30° west of north. What are the (a) magnitude and (b) direction of the boat’s velocity relative to the ground? (c) How long does the boat take to cross the river?

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Answer:

Step-by-step explanation:

Given

Width of river=200 m

Speed of river
(v_r)=2 m/s

Speed of Boat relative to river
(v_(br))=8 m/s

Boat leaves the bank
30^(\circ) west of north


v_(br)=-8cos30\hat{i}+8sin30\hat{j}


v_(r)=2\hat{i}

Therefore velocity of boat w.r.t ground


v_b=v_(br)+v_r=-8cos30\hat{i}+8sin30\hat{j}+2\hat{i}


v_b=-4.928\hat{i}+4\hat{j}

Magnitude of boat speed
=√(4.928^2+4^2)=6.34 m/s

For direction


tan\theta =(4)/(4.928)


\theta =39.06^(\circ) w.r.t to west

to cross the river its north component will help so

time taken
t=(200)/(4)=50 s

User MrShemek
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