A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 15.0 m/s due north relative to the water.
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A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 15.0 m/s due north relative to the water.
V(boat/water) = V(boat/ground) - V(water/ground) when velocity is a vector.
V(water/ground) = 1.60i (m/s)
V(boat/water) = 15.0j (m/s)
Thus, V(boat/ground) = 1.60i + 15.0j = 15.09 m/s due 17 degrees east of north.
i is a unit vector along east and j is a unit vector along north.
Setup it up like a triangle and use the pythagorean theorem.
A^2 + B^2 = C^2
Thus,
1.6m/s^2 + 15.0m/s^2 = 227.5
sqrt(227.5) = 15.0748134
=15.1m/s.
You can also use tangent to calculate the angle, but your question doesn't ask for it.
Hope I helped, good luck.
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