Three particles A, B and C start moving simultaneously with constant velocities from three places. T

Q: Why do we assume $x_A < x_B < x_C$?

The problem states the starting places are collinear and B is between A and C. We can set up our coordinate axis such that the positive direction points from A to C. This naturally gives us $x_A < x_B < x_C$.

Q: Why does A colliding with B imply $v_A > v_B$?

Since B starts ahead of A ($x_B > x_A$), the only way A can catch up to B is if A is moving faster than B. If $v_A \le v_B$, the distance between them would either remain constant or increase, making a collision impossible.

Q: How did we get the equation $v_B - v_C = (v_A - v_C) - (v_A - v_B)$?

This is a simple algebraic manipulation. If you expand the right side, you get $v_A - v_C - v_A + v_B$, which simplifies exactly to $v_B - v_C$. We do this because we already know the values of $(v_A - v_C)$ and $(v_A - v_B)$ from the given collision times.

Q: Why does $t_2 \le t_1$ guarantee a collision between B and C?

We know the initial distance $d_2$ (between A and C) is greater than $d_1$ (between A and B). If the time $t_2$ is also smaller than or equal to $t_1$, then the fraction $d_2/t_2$ must be strictly greater than $d_1/t_1$. This mathematically proves that $v_B > v_C$, which is the exact condition required for B to catch up to C.

Q: In Case 2, why might they not collide?

If $t_2 > t_1$, the denominator of $d_2/t_2$ is larger than the denominator of $d_1/t_1$. Even though the numerator $d_2$ is larger than $d_1$, the overall fraction $d_2/t_2$ could end up being smaller than $d_1/t_1$. If that happens, $v_B < v_C$, meaning C is faster than B, and B will never catch it.

Question

Three particles A, B and C start moving simultaneously with constant velocities from three places. The starting places are collinear and that of B is somewhere in between those of A and C. In the absence of C, particles A and B would have collided $t_1$ time after they started and in the absence of B, particles A and C would have collided $t_2$ time after they started. What would have happened, if A were not present?

Options:

(A) If $t_2 > t_1$, particles B and C may or may not collide.

(B) If $t_2 > t_1$, particles B and C collide in the interval $(t_1, t_2)$.

(C) If $t_2 \le t_1$, particles B and C must collide at an instant $t \le t_2$.

(D) If $t_2 \le t_1$, particles B and C must collide in the interval $[t_1, t_2]$.

Sigma Percentile · Accepted Answer

Imagine a straight highway where three particles—A, B, and C—are lined up and ready to race. Particle A is at the back, B is in the middle, and C is at the front. They all start moving simultaneously with their own constant velocities. The problem gives us two hypothetical scenarios: 1. If C wasn't there, A would catch up and collide with B at time $t_1$. 2. If B wasn't there, A would catch up and collide with C at time $t_2$. The ultimate question is: **If A wasn't there, what would happen between B and C?** This might seem like a puzzle with missing pieces. We don't know their exact positions or speeds! But physics is beautiful because it allows us to use algebra to uncover hidden truths. Let's break this down step-by-step. ### Setting Up the Mathematical Stage Let's place our particles on an x-axis. Their initial positions are $x_A$, $x_B$, and $x_C$. Because B is between A and C, we know for a fact that: $$x_A < x_B < x_C$$ Let their constant velocities be $v_A$, $v_B$, and $v_C$. Since they move with constant velocity, their position at any time $t$ is given by the standard kinematic equation: $$x(t) = x_0 + vt$$ ### Decoding the Collisions Let's look at the first clue. A and B collide at $t_1$. For A to catch B, A must be faster. At the moment of collision, their positions are equal: $$x_A + v_A t_1 = x_B + v_B t_1$$ We can rearrange this to find the relative velocity of A with respect to B: $$(v_A - v_B)t_1 = x_B - x_A$$ Let's call the initial distance between A and B as $d_1 = x_B - x_A$. So, we get: $$v_A - v_B = \frac{d_1}{t_1}$$ Now, let's apply the exact same logic to the second clue. A and C collide at $t_2$. $$x_A + v_A t_2 = x_C + v_C t_2$$ Let the initial distance between A and C be $d_2 = x_C - x_A$. This gives us: $$v_A - v_C = \frac{d_2}{t_2}$$ **Crucial Observation:** Because C starts further ahead than B, the distance $d_2$ is strictly greater than $d_1$ ($d_2 > d_1$). ### The Ultimate Showdown: B vs. C Now, we remove A from the picture. Will B and C collide? For B to catch up to C, the trailing particle (B) must be moving faster than the leading particle (C). The mathematical condition for their collision is simply: $$v_B > v_C$$ But how do we compare $v_B$ and $v_C$ when we only know their velocities relative to A? We use a clever algebraic trick! We can express the relative velocity of B and C using the equations we already derived: $$v_B - v_C = (v_A - v_C) - (v_A - v_B)$$ Substitute the fractions we found earlier: $$v_B - v_C = \frac{d_2}{t_2} - \frac{d_1}{t_1}$$ For B and C to collide, this difference must be positive. So, we need: $$\frac{d_2}{t_2} > \frac{d_1}{t_1}$$ ### Analyzing the Cases The options in the question ask us to consider two cases based on the times $t_1$ and $t_2$. **Case 1: What if $t_2 \le t_1$?** Let's look at our fractions. We already know that the numerator $d_2$ is greater than $d_1$. If the denominator $t_2$ is smaller than or equal to $t_1$, then the fraction $\frac{d_2}{t_2}$ is getting a "double boost"—it has a larger numerator AND a smaller denominator! Mathematically, it is guaranteed that: $$\frac{d_2}{t_2} > \frac{d_1}{t_1}$$ This means $v_B$ is definitely greater than $v_C$. **Particles B and C MUST collide.** But when? Let's find the collision time $t$. $$t = \frac{d_2 - d_1}{v_B - v_C}$$ If we subtract $t_2$ from $t$ and simplify, we find that $t - t_2 \le 0$, which means $t \le t_2$. This perfectly matches option (c). **Case 2: What if $t_2 > t_1$?** Now things get murky. The numerator $d_2$ is larger, which tries to make the fraction bigger. But the denominator $t_2$ is also larger, which tries to make the fraction smaller. Depending on the exact values, $\frac{d_2}{t_2}$ could be greater than, equal to, or less than $\frac{d_1}{t_1}$. - If it's greater, they collide. - If it's equal, they move at the exact same speed and stay parallel forever. - If it's less, C is faster and pulls away from B. Therefore, if $t_2 > t_1$, particles B and C **may or may not collide**. This matches option (a). ### The Takeaway This problem is a brilliant exercise in abstract kinematic reasoning. By translating physical constraints into algebraic inequalities, we can definitively predict the outcome of a race without knowing a single number!

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Visualized Solution

Initial Setup of Particles $A$ , $B$ , and $C$

Equations of Motion $x (t)$

Collision Condition for $A$ and $B$

Collision Condition for $A$ and $C$

Condition for $B$ and $C$ to Collide

Relative Velocity $v_{B} - v_{C}$

Evaluating Case 1: $t_{2} \leq t_{1}$

Time of Collision $t$ for Case 1

Evaluating Case 2: $t_{2} > t_{1}$

The Sigma Insight: Motion in a Straight Line

Setting Up the Mathematical Stage

Decoding the Collisions

The Ultimate Showdown

Analyzing the Cases

The Takeaway This problem is a brilliant exercise in abstract kinematic reasoning

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Visualized Solution

Equations of Motion .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x(t)

Collision Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

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Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C to Collide

Relative Velocity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vB​−vC​

Evaluating Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2​≤t1​

Time of Collision .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t for Case 1

Evaluating Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t2​>t1​