Math Problem Statement

  1. Sprinters Justin Gatlin and Usain Bolt are in a race. When Gatlin is 25.0 m from the finish line, he has a speed of 11.3 m/s and is 2.00 m behind Bolt, who has a speed of 12.2 m/s at this moment. Bolt thinks he has an easy win and so during the remaining portion of the race he decelerates at a constant rate of magnitude 0.100 m/s2 to the finish line. Above what minimum acceleration does Gatlin now need during the remaining portion of the race, if he wishes to beat Bolt?

Solution

We can solve this problem using kinematics. Here's the step-by-step breakdown:

1. Define the given variables:

  • Gatlin:

    • Initial distance to the finish line: dG=25.0md_G = 25.0 \, \text{m}
    • Initial speed: vG0=11.3m/sv_{G0} = 11.3 \, \text{m/s}
    • Distance behind Bolt: 2.00 m
    • Gatlin needs to cover dG=25.0m+2.00m=27.0md_G = 25.0 \, \text{m} + 2.00 \, \text{m} = 27.0 \, \text{m} to beat Bolt.
  • Bolt:

    • Initial distance to the finish line: dB=25.0md_B = 25.0 \, \text{m}
    • Initial speed: vB0=12.2m/sv_{B0} = 12.2 \, \text{m/s}
    • Deceleration: aB=0.100m/s2a_B = -0.100 \, \text{m/s}^2

2. Find the time it takes Bolt to reach the finish line:

Bolt is decelerating, so we use the kinematic equation:

dB=vB0t+12aBt2d_B = v_{B0} t + \frac{1}{2} a_B t^2

Substitute the values:

25.0=12.2t+12(0.100)t225.0 = 12.2 t + \frac{1}{2} (-0.100) t^2

This simplifies to:

25.0=12.2t0.05t225.0 = 12.2 t - 0.05 t^2

Rearrange into a standard quadratic form:

0.05t212.2t+25.0=00.05 t^2 - 12.2 t + 25.0 = 0

Solve this quadratic equation to find tt, the time Bolt takes to reach the finish line.

3. Use the quadratic formula:

The quadratic formula is:

t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here, a=0.05a = 0.05, b=12.2b = -12.2, and c=25.0c = 25.0. Substituting these values:

t=(12.2)±(12.2)24(0.05)(25.0)2(0.05)t = \frac{-(-12.2) \pm \sqrt{(-12.2)^2 - 4(0.05)(25.0)}}{2(0.05)}

Now we compute this to find tt.The two solutions for time are:

  • t1=241.93secondst_1 = 241.93 \, \text{seconds} (which is not physically meaningful in this context)
  • t2=2.07secondst_2 = 2.07 \, \text{seconds} (this is the correct and relevant time)

So, Bolt takes approximately tB=2.07secondst_B = 2.07 \, \text{seconds} to reach the finish line.

4. Determine the acceleration Gatlin needs:

For Gatlin to beat Bolt, he must cover his remaining distance (27.0 m) in less than 2.07 seconds. We can use the following kinematic equation:

dG=vG0t+12aGt2d_G = v_{G0} t + \frac{1}{2} a_G t^2

Where:

  • dG=27.0md_G = 27.0 \, \text{m}
  • vG0=11.3m/sv_{G0} = 11.3 \, \text{m/s}
  • t=2.07secondst = 2.07 \, \text{seconds}

Substituting the known values:

27.0=11.3×2.07+12aG(2.07)227.0 = 11.3 \times 2.07 + \frac{1}{2} a_G (2.07)^2

Now, let's solve for aGa_G, the minimum acceleration Gatlin needs.It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to continue with any part of the explanation manually, or ask another question? Here are some related topics we can explore:

  1. How does constant deceleration affect the distance in kinematics?
  2. Why do we disregard unphysical solutions in quadratic equations?
  3. How does the quadratic formula help solve time-related problems in physics?
  4. What are the different types of acceleration in a race scenario?
  5. How can one optimize running speed and acceleration over a distance?

Tip: Always double-check units in kinematic equations to ensure consistent results!

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Math Problem Analysis

Mathematical Concepts

Kinematics
Quadratic Equations
Deceleration

Formulas

d = v₀t + (1/2)at²
Quadratic formula: t = (-b ± √(b² - 4ac)) / 2a

Theorems

Kinematic equations of motion

Suitable Grade Level

Grades 11-12