Relative velocity in two dimensions Knowledge-Based Study

Study notes

1. Frames of Reference and Vector Foundations

   Motion is inherently relative; the measured velocity of an object depends entirely on the observer's frame of reference. In one-dimensional motion, relative velocity is handled using simple positive or negative signs along a single line. In two dimensions, velocity must be treated as a full vector possessing both magnitude and direction.

   Consider two moving objects, A and B, observed from a common stationary reference frame G (typically the ground). The velocity of object A relative to object B, denoted as $\vec{v}_{A/B}$, describes how object A moves as seen by an observer traveling with object B.

   The fundamental vector relationship is expressed as:

   $$\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$$

   Equivalently, this relation can be written in terms of vector addition:

   $$\vec{v}_A = \vec{v}_{A/B} + \vec{v}_B$$

   Here, $\vec{v}_A$ and $\vec{v}_B$ represent the velocities of objects A and B relative to the ground.

   Resolving Vectors into Orthogonal Components

   To analyze two-dimensional relative motion systematically, velocities are resolved into perpendicular components along Cartesian coordinates (usually horizontal $x$ and vertical $y$, or East-West and North-South).

   Given velocity vectors in component form:

   • $\vec{v}_A = v_{Ax}\hat{i} + v_{Ay}\hat{j}$
   • $\vec{v}_B = v_{Bx}\hat{i} + v_{By}\hat{j}$

   The relative velocity vector is obtained by subtracting corresponding components:

   $$\vec{v}_{A/B} = (v_{Ax} - v_{Bx})\hat{i} + (v_{Ay} - v_{By})\hat{j}$$

   The magnitude and direction of the relative velocity are calculated as:

   • Magnitude: $$|\vec{v}_{A/B}| = \sqrt{(v_{Ax} - v_{Bx})^2 + (v_{Ay} - v_{By})^2}$$
   • Direction: $$\theta = \arctan\left(\frac{v_{Ay} - v_{By}}{v_{Ax} - v_{Bx}}\right)$$

   This component method avoids ambiguities that can arise when constructing scale vector diagrams by hand.

   Motion in a Moving Medium

   A classic application of two-dimensional relative velocity involves an object moving within a moving fluid, such as a boat crossing a flowing river or an aircraft flying through wind.

   Let:

   • $\vec{v}_{b/w}$ = velocity of the boat relative to the water (the heading and engine speed)
   • $\vec{v}_{w/g}$ = velocity of the water relative to the ground (river current)
   • $\vec{v}_{b/g}$ = velocity of the boat relative to the ground (the actual path traversed)

   By vector addition:

   $$\vec{v}_{b/g} = \vec{v}_{b/w} + \vec{v}_{w/g}$$

   Key Scenarios:

   1. Crossing in Minimum Time: The boat aims perpendicular to the riverbank. The crossing time depends strictly on the river width $d$ and the perpendicular component of speed: $$t = \frac{d}{v_{b/w}}$$ The river current causes downstream drift without altering this crossing time.
   2. Crossing Along the Shortest Path: To travel straight across perpendicular to the banks, the resultant velocity $\vec{v}_{b/g}$ must point directly across. The boat must steer upstream at an angle $\alpha$ such that: $$\sin(\alpha) = \frac{v_{w/g}}{v_{b/w}}$$ This is only possible if $v_{b/w} > v_{w/g}$.

   Apparent Velocity and Apparent Direction

   Another classic two-dimensional problem is determining the apparent direction of falling rain observed by a moving person or vehicle.

   Let $\vec{v}_{r/g}$ be the velocity of the rain relative to the ground (often directed vertically downward) and $\vec{v}_{m/g}$ be the velocity of the moving observer relative to the ground (directed horizontally).

   The velocity of the rain relative to the moving observer, $\vec{v}_{r/m}$, is:

   $$\vec{v}_{r/m} = \vec{v}_{r/g} - \vec{v}_{m/g}$$

   If rain falls vertically with speed $v_r$ and an observer walks horizontally with speed $v_m$:

   • $\vec{v}_{r/g} = -v_r\hat{j}$
   • $\vec{v}_{m/g} = v_m\hat{i}$
   • $\vec{v}_{r/m} = -v_m\hat{i} - v_r\hat{j}$

   The apparent rain appears to come from an angle $\theta$ relative to the vertical, requiring the observer to tilt an umbrella forward:

   $$\tan(\theta) = \frac{v_m}{v_r}$$

   As the observer's forward speed increases, the rain strikes at an increasingly tilted angle.

   Structured Problem Analysis

   Example: Car A travels north at 20 m/s relative to the ground. Car B travels east at 15 m/s relative to the ground. Determine the velocity of Car A as observed from Car B.

   Step-by-Step Solution:

   1. Define coordinate axes: Let East be $+\hat{i}$ and North be $+\hat{j}$.
   2. State given vectors:$$\vec{v}_A = 20\hat{j}\text{ m/s}$$$$\vec{v}_B = 15\hat{i}\text{ m/s}$$
   3. Apply the relative velocity formula:$$\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B = -15\hat{i} + 20\hat{j}\text{ m/s}$$
   4. Calculate the magnitude:$$|\vec{v}_{A/B}| = \sqrt{(-15)^2 + (20)^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ m/s}$$
   5. Determine the direction: The vector points into the second quadrant (west and north). The angle $\theta$ north of west is:$$\tan(\theta) = \frac{20}{15} \implies \theta = \arctan\left(\frac{4}{3}\right) \approx 53.1^\circ\text{ North of West}$$

   Summary:

   Relative velocity in two dimensions requires consistent reference frame labeling and accurate vector subtraction: $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$. Resolving into components reduces two-dimensional vector problems into two independent one-dimensional calculations.