91Ó°ÊÓ

In physics, understanding the relationship between velocity and displacement is essential for analyzing motion. When an object is in motion, its velocity describes how quickly it moves, while displacement is the distance it covers in a particular direction from its starting point.

In this exercise, we have the velocity of a falling body represented as a function of its displacement: \( v = k \sqrt{s} \). Here, \( v \) represents velocity, \( k \) is a constant specific to the problem, and \( s \) indicates displacement in meters. This expression shows that as the displacement increases, the velocity changes consistently with the square root of the displacement.

Key takeaways include:

• Velocity depends on how far the object has fallen (\( s \)).
• The constant \( k \) plays a crucial role in determining the rate of increase in velocity with respect to displacement.
• Understanding this relation helps in determining other motion properties like acceleration.

To delve deeper into motion, we often need to link different rates of change. This is where the chain rule in calculus becomes vital. The chain rule allows us to differentiate composite functions, essentially letting us "chain" together the effect of different variables.

In this problem, the velocity \( v = k \sqrt{s} \) is differentiated with respect to time \( t \) to find acceleration, but it is only a function of displacement \( s \).

To solve this, we apply the chain rule formula:

• The formula is \( \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} \).
• \( \frac{dv}{ds} \) is computed directly from differentiating the given velocity expression with respect to \( s \).
• \( \frac{ds}{dt} \) is simply the velocity \( v \), giving us a way to incorporate the effect of time indirectly through displacement.

Once we have these pieces, they can be multiplied to find \( \frac{dv}{dt} \), showing the acceleration.

Differentiation is a fundamental concept in calculus that helps us find how functions change. It involves calculating derivatives, which gives us the rate at which a quantity changes with another quantity.

In this exercise, differentiating is used to convert the velocity function into an acceleration function. Let's break it down:

• We start with velocity: \( v = k \sqrt{s} \).
• When differentiating \( v \) with respect to \( s \), the formula \( \frac{dv}{ds} = \frac{k}{2\sqrt{s}} \) is obtained.
• This represents how velocity changes as displacement changes.

Next, apply this result using the chain rule, which includes combining it with \( \frac{ds}{dt} = v \). The simplifying process results in a constant acceleration \( \frac{dv}{dt} = \frac{k^2}{2} \). Differentiation here helps translate the changing displacement directly into the acceleration, showcasing its versatility in analyzing motion.