In a relay race, runner A is carrying the baton and has a speed of 5.6 – Free 76A

In a relay race, runner A is carrying the baton and has a speed of 5.6 ????/????. When runner A is 96 ???? behind runner B, runner A starts slowing down with 0.02 ????/???? 2 . At the same moment, runner B starts from rest and accelerates rightward with 0.06 ????/???? 2 . a) How long afterwards will A catch up with B to pass the baton to B? b) What are the speed of runners at the meeting point? c) What is the distance travelled by runner B at the meeting time?

Answer

Relay Race Baton Passing Explained with Kinematics – Learnlyfly

Physics Behind Baton Passing in a Relay Race

Let’s analyze a real-world physics problem involving two runners during a relay race. One runner is decelerating while the other accelerates from rest. We’ll calculate:

  • Time when Runner A catches up with Runner B
  • Speeds of both runners at the meeting point
  • Distance traveled by Runner B before receiving the baton

📘 Given:

  • Runner A’s speed: vA = 5.6 m/s
  • Runner A’s acceleration (deceleration): aA = -0.02 m/s²
  • Initial gap between runners: 96 m
  • Runner B starts from rest: vB = 0
  • Runner B’s acceleration: aB = 0.06 m/s²

🧮 Step 1: Position Equations

Let’s define the origin where Runner A starts. Both runners are moving rightward.

xA(t) = 5.6t – 0.5 × 0.02t² = 5.6t – 0.01t²
xB(t) = 96 + 0.5 × 0.06t² = 96 + 0.03t²

🔄 Step 2: Set xA = xB to Find Time

5.6t – 0.01t² = 96 + 0.03t² ⇒ -0.04t² + 5.6t – 96 = 0

Using the quadratic formula:

t = [-5.6 ± √(5.6² – 4 × (-0.04) × (-96))] / (2 × -0.04) = [-5.6 ± √(16)] / -0.08 = [-5.6 ± 4] / -0.08
  • t₁ = (-5.6 + 4)/-0.08 = -1.6 / -0.08 = 20 seconds
  • t₂ = (-5.6 – 4)/-0.08 = -9.6 / -0.08 = 120 seconds ❌ (too far)

Runner A catches up to Runner B at t = 20 seconds

🏃‍♂️ Step 3: Find Speeds at Meeting Point

Runner A:

vA = 5.6 + (-0.02 × 20) = 5.6 – 0.4 = 5.2 m/s

Runner B:

vB = 0 + 0.06 × 20 = 1.2 m/s

Speeds at handoff: A = 5.2 m/s, B = 1.2 m/s

📏 Step 4: Distance Runner B Covers

xB = 0.5 × 0.06 × 20² = 0.03 × 400 = 12 meters

Runner B travels 12 meters before receiving the baton

🧠 Summary

  • ⏱️ Time to catch up: 20 seconds
  • 🏃 Speed of Runner A: 5.2 m/s
  • 🏃‍♂️ Speed of Runner B: 1.2 m/s
  • 📍 Distance B runs: 12 meters

🎯 Concepts Used:

  • ✔️ Position-time equations for uniformly accelerated motion
  • ✔️ Relative motion between two moving objects
  • ✔️ Quadratic formula for solving time

💬 Conclusion

This example shows how simple kinematic equations can be applied to real-world scenarios like sports. Relay races involve more than just speed — understanding acceleration and timing is key to a successful handoff!

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