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Races and Relative Speeds

Introduction

The concept of Races and Relative Speeds builds on time-speed-distance fundamentals, where two or more participants move simultaneously. This pattern is crucial because it helps you compare their speeds, calculate head starts (leads), and determine who wins or when they meet.

Such problems often appear in aptitude tests and require clear understanding of relative speed when competitors move in the same or opposite directions.

Pattern: Races and Relative Speeds

Pattern

Key concept: Compare distances or times using the formula Speed = Distance ÷ Time and the idea of relative speed.

  • When moving in the same direction: Relative speed = (Speed₁ - Speed₂)
  • When moving in opposite directions: Relative speed = (Speed₁ + Speed₂)
  • If both complete the same distance: Time is inversely proportional to speed → T₁ : T₂ = S₂ : S₁
  • If one gets a head start (lead): Use difference in times or distances to find winner or margin.

Step-by-Step Example

Question

In a 500 m race, A runs at 5 m/s and B at 4 m/s. How much lead does A have when A finishes the race?

Solution

  1. Step 1: Find A’s time to finish

    Time = Distance ÷ Speed = 500 ÷ 5 = 100 seconds.
  2. Step 2: Find B’s distance in same time

    Distance = Speed × Time = 4 × 100 = 400 m.
  3. Step 3: Find the lead

    Lead = 500 - 400 = 100 m.
  4. Final Answer:

    A beats B by 100 m.
  5. Quick Check:

    Ratio of speeds = 5 : 4 → difference = 1/5 of 500 = 100 ✅

Quick Variations

1. Finding lead (difference in distances).

2. Finding head start required for a tie.

3. Races with time difference instead of distance.

4. Ratio-based questions (Speed ratio ↔ Distance ratio ↔ Time ratio).

5. Multiple participants (A, B, C) comparisons.

Trick to Always Use

  • Step 1: Always find who finishes first using Speed × Time or Distance ÷ Speed.
  • Step 2: For lead, compute difference of distances covered in the same time.
  • Step 3: For head start, set distances equal to find required lead.
  • Step 4: Keep units consistent (m/s or km/h).

Summary

Summary

  • When speeds differ, faster person gains lead = (Speed diff) × Time.
  • For equal distances, Time ∝ 1/Speed.
  • Head start = Distance required to make times equal.
  • Use consistent units throughout the calculation.

Practice

(1/5)
1. In a 400 m race, A runs at 8 m/s and B at 6 m/s. How much lead does A have when A finishes the race?
easy
A. 50 m
B. 150 m
C. 200 m
D. 100 m

Solution

  1. Step 1: Time for A to finish

    Time = Distance ÷ Speed = 400 ÷ 8 = 50 s.
  2. Step 2: Distance B covers in same time

    Distance = Speed × Time = 6 × 50 = 300 m.
  3. Step 3: Lead = Remaining distance

    Lead = 400 - 300 = 100 m.
  4. Final Answer:

    A beats B by 100 m → Option D.
  5. Quick Check:

    Speed ratio 8:6 = 4:3 → 1/4 of 400 = 100 m ✅
Hint: Lead = Distance × (1 - slower/faster).
Common Mistakes: Mixing up time and speed ratios.
2. A runs at 10 m/s and B at 8 m/s in a 500 m race. If B is given a 50 m head start, who wins and by how much?
easy
A. A wins by 50 m
B. A wins by 25 m
C. B wins by 25 m
D. B wins by 50 m

Solution

  1. Step 1: Time for A to finish

    Time = 500 ÷ 10 = 50 s.
  2. Step 2: Distance B covers in same time

    Distance = 8 × 50 = 400 m; plus head start = 400 + 50 = 450 m.
  3. Step 3: Difference

    A = 500 m, B = 450 m → A wins by 50 m.
  4. Final Answer:

    A wins by 50 m → Option A.
  5. Quick Check:

    Without head start A would be 100 m ahead (speed ratio), head start reduces it to 50 m ✅
Hint: Compute both distances in the same time, include head start before comparing.
Common Mistakes: Forgetting to add the head start for B.
3. A and B start together in a 600 m race. A’s speed is 9 m/s and B’s is 8 m/s. When A finishes, how far is B from the finish line?
medium
A. 50.67 m
B. 66.67 m
C. 75.67 m
D. 80.67 m

Solution

  1. Step 1: Time for A to finish

    Time = 600 ÷ 9 = 66.666... s.
  2. Step 2: Distance B covers in same time

    Distance = 8 × 66.666... = 533.333... m.
  3. Step 3: Remaining distance to finish

    600 - 533.333... = 66.666... m (≈ 66.67 m).
  4. Final Answer:

    B is about 66.67 m from the finish → Option B.
  5. Quick Check:

    Speed ratio 9:8 → (1/9) of 600 = 66.666... m ✅
Hint: Remaining gap = Distance × (1 - slower/faster).
Common Mistakes: Rounding too early; keep fractions until final step.
4. In a 1 km race, A can run 1 km in 180 seconds and B in 200 seconds. How much start can A give B to finish together?
medium
A. 100 m
B. 80 m
C. 90 m
D. 120 m

Solution

  1. Step 1: Speeds

    A's speed = 1000 ÷ 180 ≈ 5.555... m/s; B's speed = 1000 ÷ 200 = 5 m/s.
  2. Step 2: Let head start = x (m) for B

    Then B runs (1000 - x) m in same time A runs 1000 m:
    (1000 - x) ÷ 5 = 1000 ÷ 5.555... → 1000 - x = 900 → x = 100 m.
  3. Final Answer:

    A can give B a 100 m head start → Option A.
  4. Quick Check:

    If B starts 100 m ahead, B runs 900 m at 5 m/s = 180 s same as A ✅
Hint: Set times equal and solve for head start: (D - x)/v_slow = D/v_fast.
Common Mistakes: Comparing times directly without forming equation for head start.
5. In an 800 m race, A beats B by 80 m and B beats C by 40 m. By how much does A beat C?
medium
A. 100 m
B. 110 m
C. 116 m
D. 120 m

Solution

  1. Step 1: Derive speed ratios

    When A runs 800 m, B runs 720 m → A:B = 800:720 = 10:9.
    When B runs 800 m, C runs 760 m → B:C = 800:760 = 20:19.
  2. Step 2: Combine ratios to get A:C

    A:C = (10/9) × (20/19) = 200:171.
  3. Step 3: When A runs 800 m, C runs (171/200)×800 = 684 m → Lead = 800 - 684 = 116 m.

  4. Final Answer:

    A beats C by 116 m → Option C.
  5. Quick Check:

    Chaining ratios is equivalent to composing relative distances → 116 m is consistent with given pairwise leads ✅
Hint: Multiply A:B and B:C ratios to get A:C; then compute remaining distance.
Common Mistakes: Adding pairwise leads instead of using speed/distance ratios.

Mock Test

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