Problem 98 - Thomas Calculus

Question

Constant acceleration Suppose that the velocity of a falling body is \(v=k \sqrt{s} \mathrm{m} / \mathrm{sec}(k \text { a constant) at the instant the body has }\) fallen \(s \mathrm{m}\) from its starting point. Show that the body's acceleration is constant.

Step-by-Step Solution

Answer

The body's acceleration is constant at \(\frac{k^2}{2}\).

Step 1: Express Velocity in Terms of Distance

The velocity of the falling body is given by the formula: \(v = k\sqrt{s}\). Here, \(v\) is the velocity, \(k\) is a constant, and \(s\) is the distance fallen.

Step 2: Find the Expression for Acceleration

Acceleration \(a\) is related to velocity \(v\) via the derivative of velocity with respect to time, \(\frac{dv}{dt}\). We can also express velocity \(v\) in terms of \(s\) as \(v = \frac{ds}{dt} = k\sqrt{s}\).

Step 3: Use Chain Rule to Relate Acceleration and Distance

Using the chain rule, we know that \(\frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}\). Thus, acceleration \(a = \frac{dv}{dt} = \frac{dv}{ds} \cdot v\).

Step 4: Differentiate Velocity with Respect to Distance

Differentiate \(v = k\sqrt{s}\) with respect to \(s\): \(\frac{dv}{ds} = \frac{d}{ds}(k\sqrt{s}) = \frac{k}{2\sqrt{s}}\).

Step 5: Substitute Derivative into Acceleration Expression

Substitute \(\frac{dv}{ds} = \frac{k}{2\sqrt{s}}\) and \(v = k\sqrt{s}\) into \(a = \frac{dv}{ds} \cdot v\): \(a = \frac{k}{2\sqrt{s}} \cdot k\sqrt{s} = \frac{k^2}{2}\).

Step 6: Conclude with Constant Acceleration

The expression for acceleration \(a = \frac{k^2}{2}\) is constant because it does not depend on \(s\) or any other variable. Therefore, the body's acceleration is constant.

Key Concepts

Understanding Velocity

Velocity is a fundamental concept in physics, particularly in the study of motion. It describes how quickly an object is moving in a specific direction. In simpler terms, it's the speed that also considers direction. For a falling body, velocity is crucial as it determines how fast the body descends through the air.

The initial problem describes the velocity of a falling body as \(v = k\sqrt{s}\). This equation tells us that velocity increases as the body falls further. Here, \(k\) is a constant, and \(s\) is the distance fallen.

• This means as time progresses and \(s\) increases, the velocity will increase as well.
• Velocity is directly proportional to the square root of the distance fallen.

This relationship is key to further exploring how acceleration behaves with respect to distance.

Insights into Derivatives

In calculus, derivatives are used to determine how a function changes as its input changes. It's a way to measure the 'rate of change'.

For motion, the derivative of velocity with respect to time, \(\frac{dv}{dt}\), gives us the acceleration. However, since our given velocity depends on the distance \(s\), we use the derivative to express the change in velocity concerning distance.

• The differentiation of the equation \(v = k\sqrt{s}\) with respect to \(s\) uses the derivative \(\frac{dv}{ds}\).
• The derivative \(\frac{dv}{ds} = \frac{k}{2\sqrt{s}}\) shows how velocity changes as the distance increases.

Understanding derivatives in this context helps connect velocity changes to physical movement through space.

Applying the Chain Rule

The chain rule is a calculus technique used to differentiate composite functions. It's useful when the variable you are differentiating with respect to depends on another variable.

In this context, we applied the chain rule to find acceleration:

• We start with the known relation \(\frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt}\).
• Using \(\frac{ds}{dt} = v\), the body’s velocity concerning time, helps relate back to known quantities.
• This leads to the equation \(a = \frac{dv}{ds} \cdot v\).

The chain rule thus simplifies our ability to find acceleration, creating a bridge between changes in distance and velocity.

Understanding Falling Body Motion

A falling body is often used to illustrate constant acceleration and velocity concepts. When a body falls, gravity is the force dictating this motion.

The acceleration we calculated, \(a = \frac{k^2}{2}\), turns out to be constant. Here's why:

• The calculation shows that acceleration does not depend on \(s\), the distance, or any temporal variable.
• This ties to physical uniformity in gravity's pull, assuming other forces like air resistance are negligible.

Hence, a falling body under constant acceleration is a great way to grasp these fundamental physics principles, demonstrating how mathematical equations depict real-world phenomena.