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(magnitude and direction) of boat 2?
To find the magnitude and direction of boat 2's velocity, we can use vector addition.
Given:
Velocity of boat 1 relative to boat 2 = 2.20 m/s
Direction of boat 1's velocity relative to boat 2 = 47.0° East of North
Velocity of boat 1 in the North direction = 0.745 m/s
Let's break down the velocity of boat 1 relative to boat 2 into its North and East components.
The North component of boat 1's velocity relative to boat 2 can be found using trigonometry:
North component = velocity * sin(direction)
North component = 2.20 m/s * sin(47.0°)
Calculating the value of the North component:
North component = 2.20 m/s * 0.7313
North component = 1.609 m/s
The East component of boat 1's velocity relative to boat 2 can also be found using trigonometry:
East component = velocity * cos(direction)
East component = 2.20 m/s * cos(47.0°)
Calculating the value of the East component:
East component = 2.20 m/s * 0.6820
East component = 1.499 m/s
Now, let's add the North component of boat 1's velocity to the North component of boat 2's velocity to get the total North component:
Total North component = North component of boat 1 + North component of boat 2
Since boat 1's velocity in the North direction is given as 0.745 m/s, the North component of boat 2's velocity can be calculated:
Total North component = 0.745 m/s + North component of boat 2
Since the velocity in the North direction is conserved, the total North component must be equal to 0.745 m/s:
0.745 m/s + North component of boat 2 = 0.745 m/s
Therefore, the North component of boat 2's velocity is 0 m/s.
Now, let's add the East component of boat 1's velocity to the East component of boat 2's velocity to get the total East component:
Total East component = East component of boat 1 + East component of boat 2
The East component of boat 1's velocity is 1.499 m/s, and the West component of boat 2's velocity is unknown.
Total East component = 1.499 m/s + East component of boat 2
Since the velocity in the East direction is conserved, the total East component must be equal to 0 m/s:
1.499 m/s + East component of boat 2 = 0 m/s
Therefore, the East component of boat 2's velocity is -1.499 m/s.
Now, we have the magnitude and direction of boat 2's velocity. The magnitude can be found using the Pythagorean theorem:
Magnitude of boat 2's velocity = sqrt(North component^2 + East component^2)
Magnitude of boat 2's velocity = sqrt((0 m/s)^2 + (-1.499 m/s)^2)
Magnitude of boat 2's velocity = sqrt(0 + 2.246 m^2/s^2)
Magnitude of boat 2's velocity = sqrt(2.246) m/s
Magnitude of boat 2's velocity ≈ 1.498 m/s
To find the direction, we can use trigonometry:
Direction of boat 2's velocity = arctan(East component / North component)
Direction of boat 2's velocity = arctan(-1.499 m/s / 0 m/s)
Direction of boat 2's velocity ≈ -90°
Therefore, the magnitude of boat 2's velocity is approximately 1.498 m/s, and its direction is approximately 90° West of North.
To find the magnitude and direction of boat 2, we can break down the velocity of boat 1 into its North and East components.
Given:
Velocity of boat 1 (v1) = 0.745 m/s due North
Velocity of boat 1 relative to boat 2 (vrel) = 2.20 m/s at 47.0° East of North
First, let's break down the velocity of boat 1 into its North and East components using trigonometry.
North component of v1 (v1N) = v1 * cos θ1
v1N = 0.745 m/s * cos 90° (since it is due North)
v1N = 0.745 m/s * 1
v1N = 0.745 m/s
East component of v1 (v1E) = v1 * sin θ1
v1E = 0.745 m/s * sin 90°
v1E = 0.745 m/s * 0
v1E = 0 m/s
Now, let's find the North and East components of the relative velocity between boat 1 and boat 2.
North component of vrel (vrelN) = vrel * cos θ2
vrelN = 2.20 m/s * cos 47.0°
vrelN ≈ 2.20 m/s * 0.682 (rounded to three decimal places)
vrelN ≈ 1.502 m/s
East component of vrel (vrelE) = vrel * sin θ2
vrelE = 2.20 m/s * sin 47.0°
vrelE ≈ 2.20 m/s * 0.731 (rounded to three decimal places)
vrelE ≈ 1.609 m/s
Now, let's find the North (v2N) and East (v2E) components of the velocity of boat 2 by subtracting the components of boat 1 from the relative velocity components.
v2N = vrelN - v1N
v2N = 1.502 m/s - 0.745 m/s
v2N ≈ 0.757 m/s
v2E = vrelE - v1E
v2E = 1.609 m/s - 0 m/s
v2E = 1.609 m/s
Finally, we can find the magnitude and direction of boat 2's velocity using these components.
Magnitude of boat 2's velocity (|v2|) = sqrt(v2N^2 + v2E^2)
|v2| = sqrt((0.757 m/s)^2 + (1.609 m/s)^2)
|v2| ≈ sqrt(0.572 m^2/s^2 + 2.588 m^2/s^2)
|v2| ≈ sqrt(3.16 m^2/s^2)
|v2| ≈ 1.776 m/s
Direction of boat 2's velocity = arctan(v2E/v2N)
Direction of boat 2's velocity ≈ arctan(1.609 m/s / 0.757 m/s)
Direction of boat 2's velocity ≈ arctan(2.125)
Direction of boat 2's velocity ≈ 64.8°
Therefore, the magnitude of boat 2's velocity is approximately 1.776 m/s and its direction is approximately 64.8° (East of North).