Asked by melissa
                a boat that travels at a speed of 6.75 m/s in still water is to go directly across a river and back . the current flows at 0.50 m/s (a) at what angle(s) must the boat be steered. (b)How long does it take to make a round trip? (assume that the boats speed is constant at all times and neglect turn around time)
            
            
        Answers
                    Answered by
            bobpursley
            
    If it is to go directly across,then it must be headed upstream.
Theta=arcsin(.5/6.75)
Then velocity across= 6.75*arccosTheta
If you need more assistance, ask a followup question.
    
Theta=arcsin(.5/6.75)
Then velocity across= 6.75*arccosTheta
If you need more assistance, ask a followup question.
                    Answered by
            Damon
            
    say the boat travels at angle T from straight across. Then the upstream component of the boat velocity relative to water must be .5 m/s to counteract current
so 6.75 sin T = .5 m/s
sin T = .5/6.75 = .0741
so T = sin^-1 (.0741) = 4.25 degrees from straight across toward upstream both ways.
The component of velocity across the river is then
6.75 cos 4.25 deg = 6.73 m/s
You did not say how wide the river is so I will call it D
time = 2 D/6.73
    
so 6.75 sin T = .5 m/s
sin T = .5/6.75 = .0741
so T = sin^-1 (.0741) = 4.25 degrees from straight across toward upstream both ways.
The component of velocity across the river is then
6.75 cos 4.25 deg = 6.73 m/s
You did not say how wide the river is so I will call it D
time = 2 D/6.73
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