As two boats approach the marina, the velocity of boat 1 relative to boat 2 is 2.29 m/s in a direction 37.8° east of north. If boat 1 has a velocity that is 0.745 m/s due north, what is the magnitude of the velocity of boat 2?
To find the magnitude of the velocity of boat 2, we can use vector addition. The velocity of boat 1 relative to boat 2 is 2.29 m/s at an angle of 37.8° east of north. The velocity of boat 1 is 0.745 m/s due north.
First, let's break down the velocity of boat 1 relative to boat 2 into its north and east components. The north component can be calculated using the formula:
Velocity north = Velocity relative × sin(θ)
where θ is the angle of 37.8° east of north. Plugging in the values:
Velocity north = 2.29 m/s × sin(37.8°) = 1.39 m/s
The east component can be calculated using the formula:
Velocity east = Velocity relative × cos(θ)
Velocity east = 2.29 m/s × cos(37.8°) = 1.82 m/s
Now, let's add the north component of boat 1's velocity to the velocity of boat 1 to find the north component of boat 2's velocity:
Velocity north2 = Velocity north1 + Velocity north
Velocity north2 = 0.745 m/s + 1.39 m/s = 2.135 m/s
Finally, we can find the magnitude of boat 2's velocity by using the Pythagorean theorem:
Velocity2 = √(Velocity north2^2 + Velocity east^2)
Velocity2 = √(2.135 m/s)^2 + (1.82 m/s)^2) = √(4.56 m^2/s^2 + 3.3124 m^2/s^2) = √7.8724 m^2/s^2 = 2.8 m/s
Therefore, the magnitude of the velocity of boat 2 is 2.8 m/s.