Two particles start simultaneously from the same point and move along two straight lines, one with u

Q: Why do we minimize the square of the velocity instead of the velocity itself?

Great question! Minimizing the magnitude $v_{rel} = \sqrt{(v_{rel,x})^2 + (v_{rel,y})^2}$ involves differentiating a square root, which leads to a messy chain rule. Since the square of a positive quantity is also minimized at the same point as the quantity itself, minimizing $v_{rel}^2$ is a standard, elegant trick to keep the algebra clean and error-free.

Q: Can't we just use the relative acceleration and initial relative velocity?

You absolutely can! The relative acceleration $\vec{a}_{rel} = \vec{a}_2 - \vec{a}_1$ is constant. The relative velocity at time $t$ is $\vec{v}_{rel}(t) = \vec{v}_{rel}(0) + \vec{a}_{rel}t$. The magnitude is minimized when $\vec{v}_{rel}(t)$ is perpendicular to $\vec{a}_{rel}$. This is a more advanced vector approach, but it yields the same result. The calculus method we used is just a more explicit way to see the same physics.

Q: What if the angle $\alpha$ was 90 degrees? Does the formula still hold?

Yes, it does! If $\alpha = 90^\circ$, then $\cos 90^\circ = 0$. Our formula gives $t = \frac{u \cos 90^\circ}{f} = 0$. This means the relative velocity is already at its minimum at $t=0$, which makes sense because the particles are moving perpendicularly, and their separation velocity is dominated by the initial velocity $u$.

Question

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity $u$ and the other from rest with uniform acceleration $f$. Let $\alpha$ be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time

Options:

(A) $\frac{u \cos \alpha}{f}$

(B) $\frac{u \sin \alpha}{f}$

(C) $\frac{f \cos \alpha}{u}$

(D) $u \sin \alpha$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a vast, empty plane. Two particles, Particle 1 and Particle 2, start their journey from this exact point at the same moment.

Particle 1 moves with a constant, uniform velocity $u$ along the $x$-axis. Particle 2 starts from rest but is driven by a uniform acceleration $f$ at an angle $\alpha$ to the path of the first particle.

Our goal is to find the precise moment when these two particles are moving as 'slowly' as possible relative to each other. This is a fundamental study of relative motion in classical mechanics.

### The Language of Vectors

To solve this, we translate the physical motion into vector notation. For Particle 1, the velocity is constant:
$$\vec{v}_1 = u \hat{i}$$

For Particle 2, we resolve the acceleration vector $\vec{a}_2$ into its horizontal and vertical components:
$$\vec{a}_2 = (f \cos \alpha) \hat{i} + (f \sin \alpha) \hat{j}$$

Since Particle 2 starts from rest, its velocity at any time $t$ is the integral of its acceleration:
$$\vec{v}_2 = (ft \cos \alpha) \hat{i} + (ft \sin \alpha) \hat{j}$$

### The Relative Velocity Vector

We define the relative velocity as $\vec{v}_{rel} = \vec{v}_2 - \vec{v}_1$. By subtracting the velocity of the first particle from the second, we obtain:
$$\vec{v}_{rel} = (ft \cos \alpha - u) \hat{i} + (ft \sin \alpha) \hat{j}$$

To find when this relative velocity is 'least', we minimize the square of its magnitude, $v_{rel}^2$, to avoid the complexity of square roots:
$$v_{rel}^2 = (ft \cos \alpha - u)^2 + (ft \sin \alpha)^2$$

### The Calculus of Minimization

Expanding the expression for the square of the magnitude, we get:
$$v_{rel}^2 = f^2 t^2 \cos^2 \alpha + u^2 - 2uft \cos \alpha + f^2 t^2 \sin^2 \alpha$$

Using the trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$, the expression simplifies to:
$$v_{rel}^2 = f^2 t^2 + u^2 - 2uft \cos \alpha$$

To find the minimum, we differentiate this expression with respect to $t$ and set the derivative to zero:
$$\frac{d}{dt}(f^2 t^2 + u^2 - 2uft \cos \alpha) = 2f^2 t - 2uf \cos \alpha = 0$$

Solving for $t$, we find the time of minimum relative velocity:
$$t = \frac{u \cos \alpha}{f}$$

### The Physical Insight

This result indicates that the minimum relative velocity occurs when the component of the second particle's velocity along the first particle's path ($ft \cos \alpha$) matches the first particle's velocity ($u$).

At this moment, the relative motion in the $x$-direction vanishes, leaving only the perpendicular component. This is the **'sweet spot'** of their relative journey.

This derivation demonstrates how calculus and vector algebra work in harmony to reveal the hidden truths of motion. **The particles are at their closest relative velocity at time $t = \frac{u \cos \alpha}{f}$.**

Select Answer:

Visualized Solution

Visualizing the Setup

Defining Velocity Vectors

Velocity of Particle 2 at time $t$

Relative Velocity Vector Setup

Grouping Components

Magnitude of Relative Velocity

Expanding the Expression

Simplifying the Expression

Applying Calculus for Minimum

Differentiating the Expression

Solving for Time

Conclusion and Takeaway

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Language of Vectors

The Relative Velocity Vector

The Calculus of Minimization

The Physical Insight

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