Introduction: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…….
This scenario presents an interesting application of kinematics in classical physics. To solve the problem step-by-step, we need to consider the equations of motion for Ayesha and Fatima in two cases: when Fatima doesn’t accelerate and when Fatima accelerates with a constant acceleration of 0.1 m/s².
of motion for uniformly accelerated bodies.
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.
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….
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…….
Scenario 1: Fatima Does Not Accelerate
Initial Conditions:
- Speed of Ayesha (v_A) = 5 m/s
- Speed of Fatima (v_F) = 5 m/s
- Initial distance between Ayesha and Fatima = 10 m
- Fatima’s distance from the finish line = 50 m
- Ayesha accelerates with unknown acceleration (a_A) from her current speed (5 m/s) to overtake Fatima.
Since both runners are initially running at the same speed (5 m/s), Ayesha must accelerate to catch up to and overtake Fatima before Fatima reaches the finish line. Let’s first determine how long it will take Fatima to reach the finish line at her current speed.
Time for Fatima to Reach the Finish Line
Since Fatima is running at a constant speed and not accelerating, we can use the basic formula of motion:
tF=dvt_F = \frac{d}{v}tF =vd
where:
- ddd is the distance to the finish line (50 m),
- vvv is the speed of Fatima (5 m/s).
Substituting the values:
tF=505=10 secondst_F = \frac{50}{5} = 10 \text{ seconds}tF =550 =10 seconds
Thus, it will take Fatima 10 seconds to reach the finish line if she maintains her speed.
Distance Ayesha Must Cover
Ayesha, on the other hand, is 10 meters behind Fatima. This means that she must cover a total distance of:
dA=50+10=60 metersd_A = 50 + 10 = 60 \text{ meters}dA =50+10=60 meters
Ayesha needs to cover this distance in the same 10 seconds it takes Fatima to reach the finish line.
Equation for Ayesha’s Motion
Ayesha starts with an initial velocity of 5 m/s and accelerates at a constant rate aAa_AaA . The equation of motion that describes Ayesha’s displacement as a function of time is:
dA=vA⋅t+12aAt2d_A = v_A \cdot t + \frac{1}{2} a_A t^2dA =vA ⋅t+21 aA t2
We already know that:
- dA=60 metersd_A = 60 \, \text{meters}dA =60meters,
- vA=5 m/sv_A = 5 \, \text{m/s}vA =5m/s,
- t=10 secondst = 10 \, \text{seconds}t=10seconds.
Substituting these values into the equation:
60=5⋅10+12aA⋅10260 = 5 \cdot 10 + \frac{1}{2} a_A \cdot 10^260=5⋅10+21 aA ⋅102
This simplifies to:
60=50+50aA60 = 50 + 50 a_A60=50+50aA
Solving for aAa_AaA :
60−50=50aA60 – 50 = 50 a_A60−50=50aA 10=50aA10 = 50 a_A10=50aA aA=1050=0.2 m/s2a_A = \frac{10}{50} = 0.2 \, \text{m/s}^2aA =5010 =0.2m/s2
Thus, the least acceleration Ayesha must produce to overtake Fatima when Fatima is not accelerating is 0.2 m/s².
Scenario 2: Fatima Accelerates at 0.1 m/s²
Now let’s analyze the case where Fatima accelerates at 0.1 m/s².
Fatima’s Motion with Acceleration
In this case, Fatima is accelerating from an initial velocity of 5 m/s to some final velocity by the time she reaches the finish line. The equation of motion that describes Fatima’s displacement is:
dF=vF⋅t+12aFt2d_F = v_F \cdot t + \frac{1}{2} a_F t^2dF =vF ⋅t+21 aF t2
where:
- dF=50 metersd_F = 50 \, \text{meters}dF =50meters,
- vF=5 m/sv_F = 5 \, \text{m/s}vF =5m/s,
- aF=0.1 m/s2a_F = 0.1 \, \text{m/s}^2aF =0.1m/s2,
- ttt is the time it takes for Fatima to reach the finish line.
Substituting the known values:
50=5⋅t+12⋅0.1⋅t250 = 5 \cdot t + \frac{1}{2} \cdot 0.1 \cdot t^250=5⋅t+21 ⋅0.1⋅t2
This simplifies to:
50=5t+0.05t250 = 5t + 0.05t^250=5t+0.05t2
Multiplying through by 20 to eliminate the decimal:
1000=100t+t21000 = 100t + t^21000=100t+t2
Rearranging the equation:
t2+100t−1000=0t^2 + 100t – 1000 = 0t2+100t−1000=0
This is a quadratic equation. To solve for ttt, we use the quadratic formula:
t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}t=2a−b±b2−4ac
where:
- a=1a = 1a=1,
- b=100b = 100b=100,
- c=−1000c = -1000c=−1000.
Substituting these values into the formula:
t=−100±1002−4⋅1⋅(−1000)2⋅1t = \frac{-100 \pm \sqrt{100^2 – 4 \cdot 1 \cdot (-1000)}}{2 \cdot 1}t=2⋅1−100±1002−4⋅1⋅(−1000) t=−100±10000+40002t = \frac{-100 \pm \sqrt{10000 + 4000}}{2}t=2−100±10000+4000 t=−100±140002t = \frac{-100 \pm \sqrt{14000}}{2}t=2−100±14000 t=−100±118.322t = \frac{-100 \pm 118.32}{2}t=2−100±118.32
This gives two possible solutions for ttt:
t=−100+118.322=9.16 secondst = \frac{-100 + 118.32}{2} = 9.16 \, \text{seconds}t=2−100+118.32 =9.16seconds
or
t=−100−118.322=−109.16 secondst = \frac{-100 – 118.32}{2} = -109.16 \, \text{seconds}t=2−100−118.32 =−109.16seconds
Since time cannot be negative, we take t=9.16 secondst = 9.16 \, \text{seconds}t=9.16seconds.
Thus, Fatima reaches the finish line in 9.16 seconds when she accelerates at 0.1 m/s².
Ayesha’s Motion
Now, Ayesha needs to cover the 60 meters in 9.16 seconds instead of 10 seconds. Again, Ayesha’s displacement is described by the equation:
dA=vA⋅t+12aAt2d_A = v_A \cdot t + \frac{1}{2} a_A t^2dA =vA ⋅t+21 aA t2
Substituting the known values:
60=5⋅9.16+12aA⋅(9.16)260 = 5 \cdot 9.16 + \frac{1}{2} a_A \cdot (9.16)^260=5⋅9.16+21 aA ⋅(9.16)2
This simplifies to:
60=45.8+12aA⋅83.8860 = 45.8 + \frac{1}{2} a_A \cdot 83.8860=45.8+21 aA ⋅83.88 60−45.8=41.94aA60 – 45.8 = 41.94 a_A60−45.8=41.94aA 14.2=41.94aA14.2 = 41.94 a_A14.2=41.94aA aA=14.241.94=0.338 m/s2a_A = \frac{14.2}{41.94} = 0.338 \, \text{m/s}^2aA =41.9414.2 =0.338m/s2
Thus, the least acceleration Ayesha must produce to overtake Fatima when Fatima accelerates at 0.1 m/s² is approximately 0.338 m/s².
Conclusion: 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…….
- If Fatima does not accelerate, the least acceleration Ayesha must produce to overtake her is 0.2 m/s².
- If Fatima accelerates at 0.1 m/s², Ayesha must produce an acceleration of at least 0.338 m/s² to overtake Fatima before the finish line.
These results are derived using the fundamental principles of kinematics, specifically the equations
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