When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first 20 s is extremely well modeled by the simple equation vx2 = (2P/m)t, where P = 3.6 ✕ 10⁴ watts is the car's power output, m = 1200 kg is its mass, and vx is in m/s. That is, the square of the car's velocity increases linearly with time. Find an algebraic expression in terms of P, m, and t for the car's acceleration at time t.
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Problem 82b
Textbook Question
Careful measurements have been made of Olympic sprinters in the 100 meter dash. A quite realistic model is that the sprinter's velocity is given by v𝓍 = a ( 1 - e⁻ᵇᵗ ) where t is in s, v𝓍 is in m/s, and the constants a and b are characteristic of the sprinter. Sprinter Carl Lewis's run at the 1987 World Championships is modeled with a = 11.81 m/s and b = 0.6887 s⁻¹. Find an expression for the distance traveled at time t.

1
Start by recalling the relationship between velocity and displacement. Velocity is the derivative of displacement with respect to time: \( v_x = \frac{dx}{dt} \). To find the distance traveled \( x(t) \), we need to integrate the velocity function with respect to time.
Substitute the given velocity function \( v_x = a (1 - e^{-bt}) \) into the integral: \( x(t) = \int v_x \, dt = \int a (1 - e^{-bt}) \, dt \).
Break the integral into two parts: \( x(t) = \int a \, dt - \int a e^{-bt} \, dt \). The first term is straightforward, and the second term requires the use of integration techniques for exponential functions.
Evaluate the first term: \( \int a \, dt = at \). For the second term, use the formula for the integral of an exponential function: \( \int e^{-bt} \, dt = -\frac{1}{b} e^{-bt} \). Thus, \( \int a e^{-bt} \, dt = -\frac{a}{b} e^{-bt} \).
Combine the results to get the final expression for \( x(t) \): \( x(t) = at + \frac{a}{b} e^{-bt} + C \), where \( C \) is the constant of integration. Since the sprinter starts at \( x = 0 \) when \( t = 0 \), substitute these values to solve for \( C \). This gives \( C = -\frac{a}{b} \). The final expression for the distance traveled is \( x(t) = at - \frac{a}{b} (1 - e^{-bt}) \).

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Exponential Functions
Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. In the context of the sprinter's velocity model, the term e⁻ᵇᵗ represents a decay factor that influences how quickly the sprinter reaches their maximum velocity. Understanding exponential growth and decay is crucial for analyzing how the sprinter's speed changes over time.
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Integration
Integration is a fundamental concept in calculus that allows us to find the area under a curve, which in this case represents the distance traveled over time. To find the distance traveled by the sprinter, we need to integrate the velocity function vₓ with respect to time t. This process will yield a function that describes the total distance covered as a function of time.
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Kinematic Equations
Kinematic equations describe the motion of objects under constant acceleration. Although the sprinter's motion is modeled with a velocity function that changes over time, the principles of kinematics still apply. By understanding how velocity relates to distance and time, we can derive the expression for distance traveled by integrating the velocity function.
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