Relative Velocity and Motion in a Plane

1. Introduction to Relative Velocity

When we study motion, we often describe it with respect to a stationary observer. However, in many real-world scenarios, objects are in motion relative to each other. The concept of relative velocity helps us analyze these situations. Relative velocity is the velocity of an object as observed from another object. It's not an absolute property but depends on the frame of reference of the observer.

Imagine you are on a train moving at 50 km/h. If someone on the ground watches you, your velocity is 50 km/h. But if another train is moving in the same direction at 60 km/h, from your perspective on the first train, the second train appears to be moving much slower. If it's moving in the opposite direction at 60 km/h, it appears to be moving much faster. This change in perceived velocity is what relative velocity describes.

2. Relative Velocity in One Dimension

Let's consider two objects, A and B, moving along a straight line (say, the x-axis). Let their velocities be $v_A$ and $v_B$, respectively. The velocity of object A relative to object B, denoted as $v_{A/B}$, is given by:

$v_{A/B} = v_A - v_B$

Similarly, the velocity of object B relative to object A, denoted as $v_{B/A}$, is:

$v_{B/A} = v_B - v_A$

Notice that $v_{A/B} = -v_{B/A}$. This means the velocity of A with respect to B is equal in magnitude but opposite in direction to the velocity of B with respect to A.

Example 1: Same Direction Motion

A car A is moving east at 60 km/h. Another car B is moving east at 40 km/h. What is the velocity of car A relative to car B?

Let east be the positive direction. $v_A = +60$ km/h $v_B = +40$ km/h $v_{A/B} = v_A - v_B = 60 - 40 = +20$ km/h. So, car A is moving east at 20 km/h relative to car B. From car B's perspective, car A is moving away from it towards the east.

Example 2: Opposite Direction Motion

A car A is moving east at 60 km/h. A truck B is moving west at 40 km/h. What is the velocity of car A relative to the truck B?

Let east be the positive direction. $v_A = +60$ km/h $v_B = -40$ km/h (since west is opposite to east) $v_{A/B} = v_A - v_B = 60 - (-40) = 60 + 40 = +100$ km/h. So, car A is moving east at 100 km/h relative to the truck B. From the truck's perspective, car A is approaching rapidly from the east.

3. Relative Velocity in Two Dimensions (Motion in a Plane)

When objects move in a plane, their velocities are vectors. Let the velocity of object A be $\vec{v}_A$ and the velocity of object B be $\vec{v}_B$. The velocity of A relative to B is given by:

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

This subtraction of vectors can be done using their components or graphically. If $\vec{v}_A = v_{Ax} \hat{i} + v_{Ay} \hat{j}$ and $\vec{v}_B = v_{Bx} \hat{i} + v_{By} \hat{j}$, then:

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

The magnitude and direction of $\vec{v}_{A/B}$ can then be calculated from its components.

3.1. Graphical Method for Vector Subtraction

To find $\vec{v}_A - \vec{v}_B$, we can reverse the vector $\vec{v}_B$ to get $-\vec{v}_B$ and then add $\vec{v}_A$ and $-\vec{v}_B$ using the parallelogram law or triangle law of vector addition.

Example 3: Boat Crossing a River

A boat can travel at a speed of 10 m/s relative to the water in still water. The river is flowing with a velocity of 2 m/s towards the east. The boat wants to travel due north. What is the velocity of the boat relative to the ground?

Let the velocity of the boat relative to the water be $\vec{v}_{BW}$. The magnitude is $|\vec{v}_{BW}| = 10$ m/s and its direction is due north. So, $\vec{v}_{BW} = 10 \hat{j}$ m/s. The velocity of the water relative to the ground is $\vec{v}_{WG} = 2 \hat{i}$ m/s (east is the positive x-direction). We want to find the velocity of the boat relative to the ground, $\vec{v}_{BG}$. The relation is $\vec{v}_{BG} = \vec{v}_{BW} + \vec{v}_{WG}$. Wait, this is not correct. The boat's speed of 10 m/s is its speed *relative to the water*. So, this is $\vec{v}_{BW}$. The river's speed is $\vec{v}_{WG}$. We want to find the boat's velocity relative to the ground, $\vec{v}_{BG}$. The fundamental relation is: $\vec{v}_{BG} = \vec{v}_{BW} + \vec{v}_{WG}$. However, the boat *intends* to go due north. This means the *resultant velocity relative to the ground* ($\vec{v}_{BG}$) must be in the north direction. The boat must steer itself slightly upstream to counteract the river's flow. Let the boat's velocity relative to water be $\vec{v}_{BW}$. Its magnitude is 10 m/s. Let it be directed at an angle $\theta$ upstream (west of north). $\vec{v}_{BW} = -10 \sin\theta \hat{i} + 10 \cos\theta \hat{j}$ The velocity of water relative to ground is $\vec{v}_{WG} = 2 \hat{i}$ m/s. The velocity of the boat relative to ground is $\vec{v}_{BG} = \vec{v}_{BW} + \vec{v}_{WG}$. $\vec{v}_{BG} = (-10 \sin\theta + 2) \hat{i} + (10 \cos\theta) \hat{j}$ For the boat to travel due north relative to the ground, the horizontal component of $\vec{v}_{BG}$ must be zero. $-10 \sin\theta + 2 = 0$ $10 \sin\theta = 2$ $\sin\theta = \frac{2}{10} = 0.2$ $\theta = \arcsin(0.2) \approx 11.54^\circ$. So, the boat must steer at an angle of approximately $11.54^\circ$ west of north. Now, the resultant velocity relative to the ground is: $\vec{v}_{BG} = (0) \hat{i} + (10 \cos\theta) \hat{j}$ We need to find $\cos\theta$. Since $\sin\theta = 0.2$, we can use $\sin^2\theta + \cos^2\theta = 1$. $\cos^2\theta = 1 - \sin^2\theta = 1 - (0.2)^2 = 1 - 0.04 = 0.96$. $\cos\theta = \sqrt{0.96} = \sqrt{\frac{96}{100}} = \frac{\sqrt{16 \times 6}}{10} = \frac{4\sqrt{6}}{10} = \frac{2\sqrt{6}}{5}$. So, $\vec{v}_{BG} = 10 \times \frac{2\sqrt{6}}{5} \hat{j} = 4\sqrt{6} \hat{j}$ m/s. The speed of the boat relative to the ground is $4\sqrt{6}$ m/s, which is approximately $4 \times 2.45 = 9.8$ m/s.

4. Relative Motion in Projectile Problems

Relative velocity is particularly useful in projectile motion problems, especially when dealing with multiple projectiles or when observing motion from a moving frame.

Consider two projectiles, A and B, launched simultaneously. If we want to know if they will collide, we can analyze their relative motion. If their initial positions are $\vec{r}_{A0}$ and $\vec{r}_{B0}$, and their velocities are $\vec{v}_A$ and $\vec{v}_B$, their positions at time $t$ are $\vec{r}_A(t) = \vec{r}_{A0} + \vec{v}_A t + \frac{1}{2}\vec{g} t^2$ and $\vec{r}_B(t) = \vec{r}_{B0} + \vec{v}_B t + \frac{1}{2}\vec{g} t^2$.

The relative position vector is $\vec{r}_{A/B}(t) = \vec{r}_A(t) - \vec{r}_B(t)$. $\vec{r}_{A/B}(t) = (\vec{r}_{A0} - \vec{r}_{B0}) + (\vec{v}_A - \vec{v}_B) t + \frac{1}{2}\vec{g} t^2 - \frac{1}{2}\vec{g} t^2$ $\vec{r}_{A/B}(t) = \vec{r}_{A0/B0} + \vec{v}_{A/B} t$

This equation shows that the relative motion of two projectiles under the same gravitational acceleration is equivalent to the motion of a single particle with initial position $\vec{r}_{A0/B0}$ and constant velocity $\vec{v}_{A/B}$. This significantly simplifies collision problems. If the initial relative position vector $\vec{r}_{A0/B0}$ and the relative velocity vector $\vec{v}_{A/B}$ are collinear (i.e., point in the same or opposite directions), the objects will collide (assuming no other constraints).

Example 4: Collision of Two Projectiles

Projectile A is fired from $(0,0)$ with velocity $v_A = 10 \hat{i} + 20 \hat{j}$ m/s. Projectile B is fired from $(10,0)$ with velocity $v_B = 0 \hat{i} + 20 \hat{j}$ m/s. Will they collide? (Assume $g=10$ m/s$^2$ downwards).

Initial position of A: $\vec{r}_{A0} = 0 \hat{i} + 0 \hat{j}$. Initial position of B: $\vec{r}_{B0} = 10 \hat{i} + 0 \hat{j}$. Initial relative position: $\vec{r}_{A0/B0} = \vec{r}_{A0} - \vec{r}_{B0} = (0-10)\hat{i} + (0-0)\hat{j} = -10 \hat{i}$ m.

Velocity of A: $\vec{v}_A = 10 \hat{i} + 20 \hat{j}$ m/s. Velocity of B: $\vec{v}_B = 0 \hat{i} + 20 \hat{j}$ m/s. Relative velocity: $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B = (10-0)\hat{i} + (20-20)\hat{j} = 10 \hat{i}$ m/s.

The relative motion is described by $\vec{r}_{A/B}(t) = \vec{r}_{A0/B0} + \vec{v}_{A/B} t$. $\vec{r}_{A/B}(t) = (-10 \hat{i}) + (10 \hat{i}) t = (-10 + 10t) \hat{i}$.

For collision, the relative position vector must become zero. $(-10 + 10t) \hat{i} = 0 \hat{i}$ $-10 + 10t = 0$ $10t = 10$ $t = 1$ second.

Since we found a positive time $t=1$ second for which the relative position is zero, the projectiles will collide.

Let's check their positions at $t=1$ s. $\vec{r}_A(1) = (0 \hat{i} + 0 \hat{j}) + (10 \hat{i} + 20 \hat{j})(1) + \frac{1}{2}(-10 \hat{j})(1)^2$ $\vec{r}_A(1) = 10 \hat{i} + 20 \hat{j} - 5 \hat{j} = 10 \hat{i} + 15 \hat{j}$. $\vec{r}_B(1) = (10 \hat{i} + 0 \hat{j}) + (0 \hat{i} + 20 \hat{j})(1) + \frac{1}{2}(-10 \hat{j})(1)^2$ $\vec{r}_B(1) = 10 \hat{i} + 20 \hat{j} - 5 \hat{j} = 10 \hat{i} + 15 \hat{j}$. Indeed, their positions are the same at $t=1$ s, confirming the collision.

5. Relative Velocity in Aircraft Navigation

Aircraft navigation provides a classic example of relative velocity in two dimensions. An airplane's velocity relative to the air (its airspeed and heading) is affected by the wind, which is the velocity of the air relative to the ground. The airplane's actual velocity relative to the ground (its ground speed and track) is the vector sum of its velocity relative to the air and the wind velocity.

Let $\vec{v}_{P/A}$ be the velocity of the plane relative to the air. Let $\vec{v}_{A/G}$ be the velocity of the air relative to the ground (the wind). Then, the velocity of the plane relative to the ground is $\vec{v}_{P/G} = \vec{v}_{P/A} + \vec{v}_{A/G}$.

Example 5: Flight Path Adjustment

An airplane has an airspeed of 500 km/h. It wants to fly due north. The wind is blowing from the west at 100 km/h. What should be the heading of the airplane, and what will be its ground speed?

We are given: Airspeed $|\vec{v}_{P/A}| = 500$ km/h. Wind velocity $\vec{v}_{A/G} = 100 \hat{i}$ km/h (blowing from the west means blowing towards the east). Desired ground velocity direction is due north, so $\vec{v}_{P/G} = v_{P/G} \hat{j}$ for some magnitude $v_{P/G}$.

The equation is $\vec{v}_{P/G} = \vec{v}_{P/A} + \vec{v}_{A/G}$. Let the heading of the plane be at an angle $\theta$ west of north. $\vec{v}_{P/A} = -500 \sin\theta \hat{i} + 500 \cos\theta \hat{j}$.

Substituting into the equation: $v_{P/G} \hat{j} = (-500 \sin\theta \hat{i} + 500 \cos\theta \hat{j}) + (100 \hat{i})$ $v_{P/G} \hat{j} = (-500 \sin\theta + 100) \hat{i} + (500 \cos\theta) \hat{j}$

For this equality to hold, the components must match. Equating the $\hat{i}$ components: $0 = -500 \sin\theta + 100$ $500 \sin\theta = 100$ $\sin\theta = \frac{100}{500} = 0.2$ $\theta = \arcsin(0.2) \approx 11.54^\circ$. So, the airplane must head approximately $11.54^\circ$ west of north.

Equating the $\hat{j}$ components: $v_{P/G} = 500 \cos\theta$. Using $\sin\theta = 0.2$, we find $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - (0.2)^2} = \sqrt{1 - 0.04} = \sqrt{0.96}$. $\cos\theta = \sqrt{\frac{96}{100}} = \frac{\sqrt{16 \times 6}}{10} = \frac{4\sqrt{6}}{10} = \frac{2\sqrt{6}}{5}$. $v_{P/G} = 500 \times \frac{2\sqrt{6}}{5} = 100 \times 2\sqrt{6} = 200\sqrt{6}$ km/h.

The ground speed is $200\sqrt{6}$ km/h, which is approximately $200 \times 2.449 = 489.8$ km/h.

6. Relative Motion of Frames of Reference

The concept of relative velocity extends to frames of reference. If frame S' moves with a constant velocity $\vec{v}_{S'/S}$ relative to an inertial frame S, then the position vector $\vec{r}'$ of a particle in S' is related to its position vector $\vec{r}$ in S by:

$\vec{r} = \vec{r}' + \vec{v}_{S'/S} t$ (assuming origins coincide at $t=0$)

Differentiating with respect to time, we get the velocities:

$\vec{v} = \vec{v}' + \vec{v}_{S'/S}$

This is precisely the relative velocity equation we've been using, where $\vec{v}$ is the velocity in frame S, $\vec{v}'$ is the velocity in frame S', and $\vec{v}_{S'/S}$ is the velocity of frame S' relative to frame S.

This principle, known as Galilean Transformation, is valid for speeds much less than the speed of light.

Example 6: Motion Observed from a Moving Train

A ball is thrown vertically upwards with a velocity of 20 m/s from a train moving horizontally at 15 m/s. What is the initial velocity of the ball as observed by a person standing on the ground?

Let the ground frame be S, and the train frame be S'. The velocity of the train relative to the ground is $\vec{v}_{S'/S} = 15 \hat{i}$ m/s (assuming motion is along the x-axis). The velocity of the ball relative to the train (frame S') is $\vec{v}' = 20 \hat{j}$ m/s (vertically upwards). We want to find the velocity of the ball relative to the ground (frame S), which is $\vec{v}$.

Using the Galilean transformation for velocities: $\vec{v} = \vec{v}' + \vec{v}_{S'/S}$ $\vec{v} = (20 \hat{j}) + (15 \hat{i})$ $\vec{v} = 15 \hat{i} + 20 \hat{j}$ m/s.

So, the person on the ground observes the ball having an initial horizontal velocity of 15 m/s (due to the train's motion) and an initial vertical velocity of 20 m/s. This means the ball will follow a parabolic path relative to the ground.

7. Summary of Key Concepts

Relative velocity is the velocity of an object as measured from the frame of reference of another object.

• In 1D: $v_{A/B} = v_A - v_B$.
• In 2D (or 3D): $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$.
• The relative motion of two objects under constant acceleration (like gravity) is equivalent to the motion of a single object with constant velocity equal to their relative velocity.
• Applications include river boat problems, aircraft navigation, and collision analysis.