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Chapter 3: Problem 17

An object moves so that its velocity, \(v,\) is related to its position, \(s,\) according to \(v=\sqrt{b^{2}+2 g s},\) where \(b\) and \(g\) are constants. Show that the acceleration of the object is constant.

Short Answer

Expert verified
The acceleration of the object is constant and equal to \(g\).

Step by step solution

01

Understanding the velocity equation

The velocity of the object is given by the equation \(v = \sqrt{b^2 + 2gs}\), where \(b\) and \(g\) are constants, and \(v\) is the velocity while \(s\) is the position. Our task is to deduce whether the acceleration is constant.
02

Express velocity as a function of position

Since \(v = \frac{ds}{dt}\), we can relate velocity thus: \(\frac{ds}{dt} = \sqrt{b^2 + 2gs}\). We'll manipulate this to find the acceleration.
03

Find acceleration using chain rule

Acceleration \(a\) can be written as \(\frac{dv}{dt}\), and using chain rule, it can also be expressed as \(a = \frac{d}{dt}(\frac{ds}{dt}) = \frac{dv}{ds} \cdot \frac{ds}{dt}\). We need to find \(\frac{dv}{ds}\).
04

Differentiate the velocity equation with respect to position

Differentiating \(v = \sqrt{b^2 + 2gs}\) with respect to \(s\), we get \(\frac{dv}{ds} = \frac{d}{ds}\left(\sqrt{b^2 + 2gs}\right) = \frac{1}{2\sqrt{b^2 + 2gs}} \cdot 2g = \frac{g}{\sqrt{b^2 + 2gs}}\).
05

Combine expressions to find acceleration

Substituting \(\frac{dv}{ds}\) and \(v\) in the chain rule expression for acceleration: \(a = \frac{dv}{ds} \cdot v = \frac{g}{\sqrt{b^2 + 2gs}} \cdot \sqrt{b^2 + 2gs}\).
06

Simplify acceleration expression

Simplify the acceleration: \(a = g\). This shows that the acceleration is indeed constant and equal to \(g\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Velocity-Position Relationship
In the realm of calculus, understanding how an object's velocity relates to its position is crucial for problem-solving. When we're given a formula that ties these two together, like in the equation \(v = \sqrt{b^2 + 2gs}\), it sets the stage for analyzing motion in a more nuanced way. Here:
  • \(v\) is the velocity, which is how fast the object positions itself over time.
  • \(s\) is the position of the object at any given time.
  • \(b\) and \(g\) are constants that shape the graph of the velocity function.
By rearranging this equation in terms of \(s\), students can analyze how changes in position affect velocity.
Differentiation: The Key Calculus Tool
Differentiation is like a magnifying glass for mathematical expressions. It allows us to see how one quantity changes with respect to another. Given the formula \(v = \sqrt{b^2 + 2gs}\), we differentiate with respect to \(s\) to find the rate of change of the velocity, or \(\frac{dv}{ds}\). This gives us the expression \(\frac{g}{\sqrt{b^2 + 2gs}}\).

Using differentiation, we can extend our understanding to find acceleration by applying the chain rule. The chain rule tells us that:\[a = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \times v\]
Differentiation therefore uncovers the relationship between velocity and acceleration by analyzing the position.
Acceleration and Its Constancy
Acceleration is all about the change in velocity over time. It's vital in understanding how an object's speed varies as it moves. Once we have the equation for acceleration:\[a = \frac{g}{\sqrt{b^2 + 2gs}} \cdot \sqrt{b^2 + 2gs}\]

Simplifying this expression results in \(a = g\). This tells us something powerful 鈥� the acceleration is constant and equals the constant \(g\), irrespective of the position or velocity of the object. This constancy implies that the object's movement, influenced by \(g\), shows steady change in velocity, allowing predictions and calculations over time without variability. Understanding acceleration's constancy in this way is foundational in physics and calculus, highlighting how calculus can predict real-world phenomena.

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