Question

Two vehicles approach an intersection: a truck moving eastbound at 16.0 m/s and an SUV moving southbound at 20.0 m/s. Suppose that the +x-axis is directed eastward and the +y-axis is directed northward. Find the magnitude of the velocity vector of the truck relative to the SUV. Express your answer with the appropriate units.

A. Find the direction angle of the velocity vector of the truck relative to the SUV. The angle is measured from the +x-axis toward the +y-axis. Express your answer in degrees.
B. Find the direction angle of the velocity vector of the SUV relative to the truck. The angle is measured from the +x-axis toward the +y-axis. Express your answer in degrees.
C. Find the magnitude of the velocity vector of the SUV relative to the truck. Express your answer with the appropriate units.

Answer 1

The magnitude of the truck's velocity relative to the SUV is 25.6 m/s. The direction angle from the +x-axis toward the +y-axis for the truck relative to the SUV is -51.34 degrees, and for the SUV relative to the truck is 38.66 degrees.

Step-by-step explanation:

Finding Relative Velocities

To find the magnitude of the velocity vector of the truck relative to the SUV, we use the Pythagorean theorem where we consider the truck's velocity as the horizontal component (Vx = 16.0 m/s eastbound) and the SUV's velocity as the vertical component (Vy = -20.0 m/s southbound, which can be considered as negative because it's in the opposite direction of the positive y-axis).

Magnitude of the truck's velocity relative to the SUV (Vtr-suv) is calculated as:
|Vtr-suv| = √(Vx² + Vy²)
= √(16.0 m/s)² + (-20.0 m/s)²
= √(256 + 400) m²/s²
= √(656) m²/s²
= 25.6 m/s

To find the direction angle of the velocity vector (θ) from the +x-axis toward the +y-axis:

θ = tan⁻¹(Vy / Vx)
= tan⁻¹(-20.0 / 16.0)
= -51.34°
The negative angle indicates that the direction is below the +x-axis, consistent with south of east.

For the SUV relative to the truck, we change the reference frame. In this case, the direction angle is the complement of the previous angle because the reference frame has effectively been rotated 90 degrees.

Direction angle of the SUV relative to the truck (θsuv-tr) is thus:

θsuv-tr = 90° - |θ|
= 90° - 51.34°
= 38.66°

The magnitude of the velocity of the SUV relative to the truck is same as the truck relative to the SUV because it's a relative measure between the two, hence also 25.6 m/s.