RGBY Drag Race
Four cars -- Red, Green, Blue, and Yellow -- race a standard 1320-foot drag strip. They all start at the same time, each at constant speed.
You are told the following finishing gaps: - When Red finishes, Green is 330 feet behind. - When Green finishes, Blue is 330 feet behind. - When Blue finishes, Yellow is 330 feet behind.
What is Yellow's position (in feet from the start) when Red crosses the finish line?
Hints
- Each finishing gap tells you something about the ratio of two cars' speeds -- when car A covers 1320 ft, how far does car B travel?
- If Green covers 990 ft while Red covers 1320 ft, their speed ratio is $v_G/v_R = 990/1320 = 3/4$. The same logic applies to each consecutive pair.
- Chain the three speed ratios: $v_Y/v_R = (v_G/v_R)(v_B/v_G)(v_Y/v_B) = (3/4)^3$. Multiply by 1320 to get Yellow's position.
Worked Solution
How to Think About It: The key is to extract speed ratios from the finishing gaps. "When car A finishes, car B is 330 feet behind" is a statement about how far B travels in the time it takes A to cover 1320 feet. Since the speeds are constant, this pins down the ratio $v_B / v_A$ exactly. Chain those ratios and you get Yellow's speed as a fraction of Red's -- then it's trivial to find Yellow's position when Red hits 1320.
Quick Estimate: Green covers $990/1320 = 3/4$ of the track while Red covers the full track. So $v_G / v_R = 3/4$. Similarly $v_B / v_G = 3/4$ and $v_Y / v_B = 3/4$. So $v_Y / v_R = (3/4)^3 = 27/64 \approx 0.42$. Yellow covers about 42% of 1320 feet, which is around 556 feet. The exact answer will be slightly above that.
Approach: Use the constant-speed ratio to derive a closed-form answer.
Formal Solution:
Step 1: Find $v_G / v_R$.
When Red finishes (traveling 1320 ft), Green has traveled
Step 2: Find $v_B / v_G$ and $v_Y / v_B$ by the same logic: $\frac{v_B}{v_G} = \frac{990}{1320} = \frac{3}{4}, \quad \frac{v_Y}{v_B} = \frac{3}{4}$
Step 3: Chain the ratios: $\frac{v_Y}{v_R} = \left(\frac{3}{4}\right)^3 = \frac{27}{64}$
Step 4: Yellow's distance when Red finishes: $d_Y = \frac{27}{64} \times 1320 = \frac{27 \times 1320}{64} = \frac{35640}{64} = \frac{4455}{8} = 556.875 \text{ feet}$
Answer: Yellow is at $\dfrac{4455}{8} = 556.875$ feet when Red crosses the finish line.
Intuition
This problem is really about proportional reasoning -- extracting speed ratios from position snapshots. The elegant part is that the gaps compound multiplicatively, not additively. Even though each car is exactly 330 feet behind the previous one at their respective finish times, Yellow ends up nearly 760 feet behind Red when Red finishes. The gaps look symmetric in presentation but the underlying physics compounds.
This pattern appears in real quant contexts whenever you chain multiplicative ratios: compounded returns, multi-period decay rates, sequential probability products. Recognizing that three equal-looking 'gaps' compound geometrically rather than stack arithmetically is exactly the kind of insight that distinguishes careful quantitative reasoning from casual observation.