A motorboat heads due west at 10 m/s. The river has a current that travels south at 6 m/s.

a) what is the resultant velocity?

b) If the river is 200 m wide, how long doe sit take the boat to cross the river?

c) How far downstream is the boat when it reaches the other side?

Please, please, please help me and explain.
I have a test on all kinds of problems like this tomorrow morning.

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  1. b)
    I am going to do b first because it is easy
    I assume the river is North to South so west is straight across.
    Our velocity component straight across is 10m/s
    so time to cross = 200m / 10m/s = 20 seconds

    Now c) because we know how long we went at 6 m/s downstream
    6*20 = 120 meters downstream

    Now finally a, the long one
    magnitude of resultant velocity = sqrt (10^2 + 6^2) = sqrt (136) = 11.662 m/s
    tan angle below straight across = 6/10
    so angle south of west = 30.964 degrees

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  2. Okay so when we were learning this, we had to add the components and do sine and cosine. Do i have to do that in this problem? If so, where?

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  3. Well, I used tangent instead of sine and cosine but if you wanted to use sine or cos do this
    We have the hypotenuse = 11.662
    so sine angle = 6/11.662
    so angle = sin^-1 .5145
    which of course is 30.964 again
    or we could use
    cosine angle = 10/11.662 = .8575
    so angle = cos^-1 .8575
    which lo and behold is 30.963 close enough :)

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  4. You probably gave me a problem that was too easy because you were already given the west and south components instead of a direction and a speed.

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  5. Ah got it. Thank you. So another kind of problem that is confusing me, probably because i'm making it harder than it is, is when forces are acting on a point:

    Three forves act on a point concurrently. Forve a is 80N at 220 degrees, force b is 70N at 90 degrees, and forve c is 70N at 0 degrees. Find:

    a) resultant force

    b) equilibrant force

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