Two friends, Barbara and Neil, are out rollerblading. With respect to the ground, Barbara is skating due south at a speed of 5.2 m/s. Neil is in front of her. With respect to the ground, Neil is skating due west at a speed of 3.2 m/s. Find Neil's velocity ((a) magnitude and (b) direction relative to due west, as seen by Barbara.

With vector addition in regards to the first part of the question, would (a) = 8.4 m/s? I may not be understanding the phrasing...

No, that is incorrect.

The phrasing asks for Neil's velocity as seen by Barbara, so we need to find the relative velocity of Neil with respect to Barbara. We will use the formula V_n/b = V_n - V_b where V_n/b is Neil's velocity relative to Barbara, V_n is Neil's velocity, and V_b is Barbara's velocity.

V_n = 3.2 m/s west
V_b = 5.2 m/s south

To find the relative velocity, we subtract the two vectors:

V_n/b = V_n - V_b = 3.2 m/s west - 5.2 m/s south

This gives us a vector pointing southwest.

(a) To find the magnitude we use the Pythagorean theorem:

|V_n/b| = √((3.2 m/s)² + (5.2 m/s)²) ≈ 6.1 m/s

So, the magnitude of the relative velocity is 6.1 m/s.

(b) To find the direction relative to due west, we can find the angle formed by the x-component (west) and the resulting vector.

θ = arctan(|V_south|/|V_west|) = arctan(5.2/3.2) = arctan(1.625) ≈ 58.6°

So, the direction of Neil's velocity relative to Barbara is 58.6° south of west.

To find Neil's velocity relative to Barbara, we need to use vector addition. We can add the two velocities using the Pythagorean theorem to find the magnitude and trigonometry to find the direction.

(a) To find the magnitude of Neil's velocity relative to Barbara, we can use the Pythagorean theorem:

Magnitude of Neil's velocity = sqrt((magnitude of Barbara's velocity)^2 + (magnitude of Neil's velocity)^2)

Magnitude of Neil's velocity = sqrt((5.2 m/s)^2 + (3.2 m/s)^2)

Magnitude of Neil's velocity = sqrt(27.04 + 10.24)

Magnitude of Neil's velocity = sqrt(37.28)

Magnitude of Neil's velocity ≈ 6.1 m/s

So, the magnitude of Neil's velocity relative to Barbara is approximately 6.1 m/s.

(b) To find the direction of Neil's velocity relative to due west, as seen by Barbara, we can use trigonometry. We can consider the angle between Barbara's velocity and the resultant velocity of the two velocities as the angle "θ".

θ = arctan((magnitude of Barbara's velocity) / (magnitude of Neil's velocity))

θ = arctan(5.2 m/s / 3.2 m/s)

θ ≈ arctan(1.625)

θ ≈ 58.6°

Since Neil's velocity is due west and Barbara's velocity is due south, the resultant velocity direction is in the southwest quadrant. Therefore, the direction of Neil's velocity relative to due west, as seen by Barbara, is approximately 58.6° southwest.

To find Neil's velocity relative to Barbara, we need to add their velocities as vectors.

First, we need to understand that velocities are vector quantities, which means they have both magnitude and direction. In this case, we have Barbara's velocity, which is due south at 5.2 m/s, and Neil's velocity, which is due west at 3.2 m/s.

To add these velocities, we need to consider their directions. Since Barbara is skating due south, her velocity can be represented as a vector pointing downwards. Neil, on the other hand, is skating due west, so his velocity can be represented as a vector pointing to the left.

To add these vectors, draw them on a coordinate system where the x-axis represents east-west direction, and the y-axis represents north-south direction. Place Barbara's velocity vector pointing downwards, and Neil's velocity vector pointing to the left.

To perform vector addition, we place the vectors head-to-tail. In this case, we place the tail of Neil's vector at the head of Barbara's vector. The resultant vector, which represents their combined velocity, is the vector drawn from the tail of Barbara's vector to the head of Neil's vector.

To find the magnitude of the resultant velocity, we can use the Pythagorean theorem. The magnitude, or length, of the resultant vector is the square root of the sum of the squares of the individual components' magnitudes. In this case, it would be:

Magnitude = sqrt((5.2)^2 + (3.2)^2)
= sqrt(27.04 + 10.24)
= sqrt(37.28)
≈ 6.1 m/s

So the magnitude of Neil's velocity relative to Barbara is approximately 6.1 m/s.

To find the direction of Neil's velocity relative to due west, as seen by Barbara, we can use trigonometry. We can find the angle between the resultant vector and the horizontal (west) axis. This can be calculated as:

Angle = arctan((5.2)/(3.2))
= arctan(1.625)
≈ 58.47 degrees

So Neil's velocity, relative to due west as seen by Barbara, has a magnitude of approximately 6.1 m/s and a direction of approximately 58.47 degrees north of west.