As two boats approach the marina, the velocity of boat 1 relative to boat 2 is 2.20m/s in a direction 47.0 ∘ east of north.
If boat 1 has a velocity that is 0.745m/s due north, what is the velocity (magnitude and direction) of boat 2?
Boat 1 north velocity as seen from boat 2 = boat 1 north speed - boat 2 north speed
2.2 cos 47 = .745 - boat 2 north
Boat 1 east as seen from boat 2 = boat 1 east - boat 2 east
2.2 sin 47 = 0 - boat 2 east
To find the velocity of boat 2, we can use vector addition. The velocity of boat 2 relative to boat 1 is the negative of the velocity of boat 1 relative to boat 2.
Given:
Velocity of boat 1 relative to boat 2 = 2.20 m/s at 47.0 degrees east of north
Velocity of boat 1 = 0.745 m/s due north
Step 1: Convert the velocity of boat 1 relative to boat 2 into its x and y components.
The x-component can be found by using the cosine function:
x1 = 2.20 m/s * cos(47.0 degrees) = 1.50 m/s (eastward)
The y-component can be found by using the sine function:
y1 = 2.20 m/s * sin(47.0 degrees) = 1.66 m/s (northward)
Step 2: Since boat 2 is moving in the opposite direction of boat 1 relative to boat 2, the velocity of boat 2 relative to boat 1 is the negative of the velocity of boat 1 relative to boat 2.
So, the velocity of boat 2 relative to boat 1 = -1.50 m/s (in the opposite direction of boat 1) in the x-direction and -1.66 m/s (in the opposite direction of boat 1) in the y-direction.
Step 3: Add the velocity of boat 1 to the velocity of boat 2 relative to boat 1 to find the velocity of boat 2.
The x-component of the velocity of boat 2 = (velocity of boat 1) + (velocity of boat 2 relative to boat 1) = 0.745 m/s + (-1.50 m/s) = -0.755 m/s
The y-component of the velocity of boat 2 = (velocity of boat 1) + (velocity of boat 2 relative to boat 1) = 0.745 m/s + (-1.66 m/s) = -0.915 m/s
Step 4: Calculate the magnitude and direction of the velocity of boat 2.
The magnitude of the velocity of boat 2 = sqrt((-0.755 m/s)^2 + (-0.915 m/s)^2) = 1.18 m/s
The direction of the velocity of boat 2 is given by the angle between the negative x-axis and the velocity vector, which can be found using the arctan function:
angle = arctan((-0.915 m/s) / (-0.755 m/s)) = 49.7 degrees east of south
Therefore, the velocity of boat 2 is 1.18 m/s at 49.7 degrees east of south.
To find the velocity of boat 2, we need to determine the magnitude and direction of its velocity relative to the marina.
Let's break down the given information:
- The velocity of boat 1 relative to boat 2 is 2.20 m/s.
- The direction is 47.0° east of north.
To find the velocity of boat 2 relative to boat 1, we need to subtract the velocity of boat 1 from the velocity of boat 2.
Velocity (boat 2 relative to boat 1) = Velocity (boat 2) - Velocity (boat 1)
Since Velocity (boat 1) is given as 0.745 m/s due north, we can represent it as a vector:
Velocity (boat 1) = 0.745 m/s [up]
Now, let's find the velocity of boat 2 relative to boat 1:
Velocity (boat 2 relative to boat 1) = 2.20 m/s [47.0° east of north] - 0.745 m/s [up]
To subtract these vectors, we need to resolve them into their north and east (or up and down) components.
The north component of Velocity (boat 2 relative to boat 1) = 2.20 m/s [47.0° east of north] × sin(47.0°)
The east component of Velocity (boat 2 relative to boat 1) = 2.20 m/s [47.0° east of north] × cos(47.0°)
Now, we can calculate the values:
North component = 2.20 m/s × sin(47.0°) ≈ 1.60 m/s [up]
East component = 2.20 m/s × cos(47.0°) ≈ 1.45 m/s [east]
Therefore, the velocity of boat 2 relative to boat 1 is:
Velocity (boat 2 relative to boat 1) = 1.45 m/s [east] + 1.60 m/s [up]
Now, to find the magnitude and direction of the velocity of boat 2 relative to the marina, we'll use the Pythagorean theorem and trigonometry.
Magnitude = √(1.45 m/s)^2 + (1.60 m/s)^2
Direction = arctan(1.45 m/s ÷ 1.60 m/s)
Solving these equations will give us the final answer.