The Problem :
Two runners, Ayesha and Fatima, are leading the field in a long-distance race. They are both running at 5 ms, with Ayesha 10 m behind Fatima. When Fatima is 50 m from the tape, Ayesha accelerates but Fatima doesn’t. What is the least acceleration Ayesha must produce to overtake Fatima? If instead Fatima accelerates at 0.1 m s2 up to the tape, what is the least acceleration Ayesha must produce?
Solution:
Two runners, Ayesha and Fatima, are leading the field in a long-distance race. They are both running at 5 ms, with Ayesha 10 m behind Fatima….
Ayesha and Fatima Race Problem Solution
Ayesha and Fatima Race Problem – Solution
Given Data:
- Both Ayesha and Fatima run at 5 m/s.
- Ayesha is 10 meters behind Fatima.
- Fatima is 50 meters from the finish line.
- Ayesha accelerates to overtake Fatima.
Case 1: Fatima Runs at Constant Speed
Fatima’s motion is uniform at 5 m/s.
Time to finish: t = 50 / 5 = 10 seconds
Ayesha must cover (50 + 10) = 60 meters in 10 seconds.
Using the formula:
s = ut + (1/2) a t²
Substituting values:
60 = 5(10) + (1/2) a (10)²
60 = 50 + 50a
50a = 10
Acceleration, a = 0.2 m/s²
Case 2: Fatima Accelerates at 0.1 m/s²
Fatima’s motion follows the equation:
s = ut + (1/2) a t²
50 = 5t + (1/2) (0.1) t²
50 = 5t + 0.05 t²
Solving for t:
0.05t² + 5t – 50 = 0
Using the quadratic formula:
t = (-5 + sqrt(25 + 10)) / 0.1
t = (-5 + sqrt(35)) / 0.1
Approximating sqrt(35) ≈ 5.92:
t = (-5 + 5.92) / 0.1 = 9.2 seconds
Ayesha must cover 60 meters in 9.2 seconds:
60 = 5(9.2) + (1/2) a (9.2)²
60 = 46 + 42.32a
42.32a = 14
Acceleration, a = 0.33 m/s²
Final Answers:
- If Fatima does not accelerate: Ayesha needs 0.2 m/s² acceleration.
- If Fatima accelerates at 0.1 m/s²: Ayesha needs 0.33 m/s² acceleration.
