Two cars A and B are running in the same direction with constant speeds v_A = 25 m/s and v_B = 27 m/

Q: Why does car C have to be on the perpendicular bisector?

In geometry, the locus of all points equidistant from two fixed points is their perpendicular bisector. Since car C is always equidistant from A and B, it must continuously travel along this moving bisector.

Q: Why is the x-velocity of C the average of A and B's velocities?

The x-coordinate of the bisector is exactly halfway between A and B, meaning $x_C = (x_A + x_B)/2$. Differentiating this position equation with respect to time directly gives $v_{Cx} = (v_A + v_B)/2$.

Q: Can car C move in the negative y-direction?

Yes! The math gives us $v_{Cy}^2 = 224$, which means $v_{Cy}$ could be positive or negative. However, since we are asked for the *modulus* (magnitude) of the relative velocity, the sign of $v_{Cy}$ doesn't affect the final answer.

Q: Why doesn't the relative velocity depend on the distance between the roads?

The relative velocity depends only on the velocity vectors of the cars, not their absolute positions. As long as the velocity vectors remain constant and satisfy the equidistant constraint, the relative speed remains constant.

Q: Is there a shortcut formula for this specific type of problem?

Yes! As derived in the solution, the magnitude of the relative velocity simplifies beautifully to $\sqrt{v_C^2 - v_A v_B}$. It's a neat algebraic trick that saves time during exams.

Question

Two cars A and B are running in the same direction with constant speeds $v_A = 25 	ext{ m/s}$ and $v_B = 27 	ext{ m/s}$ on a straight road. Another car C is running with a constant speed $v_C = 30 	ext{ m/s}$ on another straight road. If the car C always remains equidistant from the cars A and B, find moduli of velocities of the car C relative to the car A and the car B.

Sigma Percentile · Accepted Answer

The problem of finding the relative velocity of a car that maintains an equal distance from two other moving cars is a classic test of geometric intuition and kinematic principles. It beautifully intertwines the abstract concept of a locus with the physical reality of moving bodies.

### Analyzing the Setup

Imagine a straight, endless highway. On this highway, two cars, **A** and **B**, are cruising in the same direction. Car A is moving at a steady $25 	ext{ m/s}$, while car B is slightly faster, moving at $27 	ext{ m/s}$.

Now, picture a third car, **C**, driving on a completely different straight road nearby. Car C is moving at a constant speed of $30 	ext{ m/s}$. The problem introduces a fascinating constraint: no matter how much time passes, car C is always exactly the same distance from car A as it is from car B.

### The Geometric Constraint

What does it mean for a point to be equidistant from two other moving points? In geometry, the locus of all points equidistant from two given points is their **perpendicular bisector**.

Therefore, car C must always lie on the perpendicular bisector of the line segment connecting cars A and B. Let's set up a coordinate system. Let the highway of cars A and B be the x-axis. The positions of A and B at any time $t$ are $x_A$ and $x_B$.

Because car C is on the perpendicular bisector, its x-coordinate must be exactly halfway between A and B:
$$x_C = \frac{x_A + x_B}{2}$$

### The Master Equation

This simple position equation is the key to unlocking the entire problem. By differentiating this equation with respect to time, we transition from geometry to kinematics. The rate of change of position is velocity, so:
$$v_{Cx} = \frac{v_A + v_B}{2}$$

This tells us that the horizontal velocity component of car C is simply the average of the velocities of cars A and B. Let's plug in the given values:
$$v_{Cx} = \frac{25 + 27}{2} = 26 	ext{ m/s}$$

So, car C is moving to the right at $26 	ext{ m/s}$ along the x-axis. But we know its total speed is $30 	ext{ m/s}$. Where is the rest of the speed coming from? It must be moving vertically (along the y-axis) as well.

Using the Pythagorean theorem, we can find the y-component of its velocity:
$$v_{Cy}^2 = v_C^2 - v_{Cx}^2$$
$$v_{Cy}^2 = 30^2 - 26^2 = 900 - 676 = 224$$

### Final Calculation

We are asked to find the modulus (magnitude) of the velocity of car C relative to car A. The relative velocity vector is defined as:
$$\vec{v}_{C/A} = \vec{v}_C - \vec{v}_A$$

Breaking this down into components, we subtract their x-velocities, while the y-velocity of C remains unchanged (since A has no vertical motion):
$$\vec{v}_{C/A} = (v_{Cx} - v_A)\hat{i} + v_{Cy}\hat{j}$$
$$\vec{v}_{C/A} = (26 - 25)\hat{i} + \sqrt{224}\hat{j} = 1\hat{i} + \sqrt{224}\hat{j}$$

Now, we calculate the magnitude of this relative velocity vector:
$$|\vec{v}_{C/A}| = \sqrt{1^2 + 224} = \sqrt{225} = 15 	ext{ m/s}$$

The relative speed of car C with respect to car A is exactly **$15 	ext{ m/s}$**.

If we were to calculate the relative velocity with respect to car B, the x-velocity difference would be $(26 - 27) = -1 	ext{ m/s}$. Squaring this still gives $1$, leading to the exact same magnitude of $15 	ext{ m/s}$.

### A Beautiful Generalization

If we look closely at the algebra, a stunning general formula emerges. Let's expand the magnitude equation before plugging in the numbers:
$$|\vec{v}_{C/A}|^2 = (v_{Cx} - v_A)^2 + v_{Cy}^2$$
$$|\vec{v}_{C/A}|^2 = v_{Cx}^2 - 2v_{Cx}v_A + v_A^2 + v_C^2 - v_{Cx}^2$$
$$|\vec{v}_{C/A}|^2 = v_C^2 - 2v_{Cx}v_A + v_A^2$$

Substitute $v_{Cx} = \frac{v_A + v_B}{2}$:
$$|\vec{v}_{C/A}|^2 = v_C^2 - 2\left(\frac{v_A + v_B}{2}ight)v_A + v_A^2$$
$$|\vec{v}_{C/A}|^2 = v_C^2 - (v_A + v_B)v_A + v_A^2$$
$$|\vec{v}_{C/A}|^2 = v_C^2 - v_A^2 - v_A v_B + v_A^2$$
$$|\vec{v}_{C/A}|^2 = v_C^2 - v_A v_B$$

Taking the square root gives us a powerful shortcut:
$$|\vec{v}_{C/A}| = \sqrt{v_C^2 - v_A v_B}$$

This elegant formula perfectly encapsulates the physics of the problem, proving that sometimes, the most complex constraints lead to the most beautiful mathematical symmetries.

Visualized Solution

Visualizing the Setup

The Equidistant Constraint

Kinematics of the Bisector

Velocity Constraint along X-axis

Calculating $v_{C x}$

Finding the Y-component of Velocity

Relative Velocity Formulation

Magnitude of Relative Velocity

The Elegant General Formula

The Sigma Insight: Relative Velocity

Analyzing the Setup

The Geometric Constraint

The Master Equation

Final Calculation

A Beautiful Generalization

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