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Ch 02: Kinematics in One Dimension
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 2, Problem 81a

A good model for the acceleration of a car trying to reach top speed in the least amount of time is ax = a0 ─ kvx, where a₀ is the initial acceleration and k is a constant. Find an expression for k in terms of a0 and the car's top speed vmax.

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Step 1: Start by analyzing the given equation for acceleration, a𝓍 = a₀ ─ kv𝓍. At the car's top speed vₘₐₓ, the acceleration a𝓍 becomes zero because the car can no longer increase its speed.
Step 2: Set a𝓍 = 0 in the equation, which gives 0 = a₀ ─ k * vₘₐₓ. This represents the condition at the car's maximum speed.
Step 3: Rearrange the equation to solve for k. Add k * vₘₐₓ to both sides, resulting in k * vₘₐₓ = a₀.
Step 4: Divide both sides of the equation by vₘₐₓ to isolate k. This gives k = a₀ / vₘₐₓ.
Step 5: Conclude that the expression for k in terms of a₀ and vₘₐₓ is k = a₀ / vₘₐₓ.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Acceleration

Acceleration is the rate of change of velocity of an object with respect to time. In this context, it describes how quickly the car can increase its speed. The initial acceleration (a₀) represents the car's acceleration at the start, while the model suggests that this acceleration decreases as the car's speed increases, influenced by the constant k.
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Top Speed

Top speed, denoted as vₘₐₓ, is the maximum velocity that a car can achieve under ideal conditions. It is a critical parameter in the model, as it defines the limit of acceleration. Understanding the relationship between initial acceleration and top speed is essential for deriving the constant k, which quantifies how acceleration diminishes as the car approaches its maximum speed.
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Exponential Decay

The equation a𝓍 = a₀ - kv𝓍 represents a form of exponential decay, where the acceleration decreases as the velocity increases. The constant k determines the rate of this decay, indicating how quickly the car's acceleration diminishes as it approaches its top speed. This concept is vital for understanding the dynamics of the car's motion and how forces interact as speed increases.
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Related Practice
Textbook Question

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 s later. What was the rocket's acceleration?

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Textbook Question

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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Textbook Question

A sprinter can accelerate with constant acceleration for 4.0 s before reaching top speed. He can run the 100 meter dash in 10.0 s. What is his speed as he crosses the finish line?

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Textbook Question

A good model for the acceleration of a car trying to reach top speed in the least amount of time is a𝓍 = a ─ kv𝓍, where a is the initial acceleration and k is a constant. Find an expression for the car's velocity as a function of time.

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Textbook Question

A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500m/s . It is loaded with 600 kg of fuel, which it burns in 30 s. What is the rocket's speed 10 s, 20 s, and 30 s after launch?

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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.

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