Problem 36 - The Calculus with Analytic Geometry

Question

An object is moving along a straight line according to the equation of motion \(s=3 t /\left(t^{2}+9\right)\), with \(t \geq 0\), where \(s\) ft is the directed distance of the object from the starting point at \(t \mathrm{sec}\). (a) What is the instantaneous velocity of the object at \(t_{1}\) sec? (b) What is the instantaneous velocity at \(1 \mathrm{sec}\) ? (c) At what time is the instantaneous velocity zero?

Step-by-Step Solution

Answer

The instantaneous velocity function is \( \frac{-3t^2 + 27}{(t^2 + 9)^2} \). The instantaneous velocity at \(t = 1\) sec is 0.24 ft/sec. The instantaneous velocity is zero at \(t = 3\) seconds.

Step 1: Find the Instantaneous Velocity

The instantaneous velocity is the derivative of the position function. Differentiate the position function with respect to time:\[ s(t) = \frac{3t}{t^2 + 9} \]Use the quotient rule for differentiation \(\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\), where \(u = 3t\) and \(v = t^2 + 9\). First, find the derivatives of \(u\) and \(v\):\[ u' = 3 \]\[ v' = 2t \]Now, apply the quotient rule:\[ s'(t) = \frac{(3)(t^2 + 9) - (3t)(2t)}{(t^2 + 9)^2} = \frac{3t^2 + 27 - 6t^2}{(t^2 + 9)^2} = \frac{-3t^2 + 27}{(t^2 + 9)^2} \]So, the instantaneous velocity function is:\[ v(t) = \frac{-3t^2 + 27}{(t^2 + 9)^2} \]

Step 2: Find the Instantaneous Velocity at t = 1 sec

Substitute \(t = 1\) into the velocity function:\[ v(1) = \frac{-3(1)^2 + 27}{(1^2 + 9)^2} = \frac{-3 + 27}{10^2} = \frac{24}{100} = 0.24 \]So, the instantaneous velocity at \(t = 1 \) sec is 0.24 ft/sec.

Step 3: Find When the Instantaneous Velocity is Zero

Set the velocity function equal to zero and solve for \(t\):\[ \frac{-3t^2 + 27}{(t^2 + 9)^2} = 0 \]For a fraction to be zero, the numerator must be zero (since the denominator is never zero):\[ -3t^2 + 27 = 0 \]Solve for \(t\):\[ -3t^2 = -27 \]\[ t^2 = 9 \]\[ t = 3 \]So, the instantaneous velocity is zero at \(t = 3\) seconds.

Key Concepts

Equation of Motion

The equation of motion describes the position of an object as it changes over time. In our exercise, the equation of motion given is equation of motion: \( s = \frac{3t}{t^2 + 9} \) where \( s \) represents the distance from the starting point in feet, and \( t \) is time in seconds. This function tells us precisely where the object is at any given second. Understanding this equation is crucial because it allows us to analyze the object's movement and predict its future position.

Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of a function that is the ratio of two differentiable functions. It's expressed as \( \frac{u}{v}' = \frac{u'v - uv'}{v^2} \) where \( u \) and \( v \) are functions of \( t \). In our problem, we need to apply the quotient rule to find the instantaneous velocity. Let's decompose:

• \( u = 3t \)
• \( v = t^2 + 9 \)

We first differentiate each function separately before combining them:

• \( u' = 3 \)
• \( v' = 2t \)

With these derivatives, we now apply the quotient rule: \( \frac{3(t^2 + 9) - 3t(2t)}{(t^2 + 9)^2} = \frac{3t^2 + 27 - 6t^2}{(t^2 + 9)^2} = \frac{-3t^2 + 27}{(t^2 + 9)^2} \). Now we have the instantaneous velocity function: \( v(t) = \frac{-3t^2 + 27}{(t^2 + 9)^2} \).

Calculus Differentiation

Calculus differentiation is a fundamental concept that deals with how a function changes as its input changes. In terms of motion, differentiation provides us with the velocity of an object. Consider the position function \( s(t) = \frac{3t}{t^2 + 9} \). To find how fast the object is moving at any point in time (its instantaneous velocity), we differentiate this function with respect to \( t \). We use the quotient rule since our original function is a ratio.By differentiating, we get the velocity function: \( v(t) = \frac{-3t^2 + 27}{(t^2 + 9)^2} \). From here, we can answer specific questions:

• At \( t = 1 \): Plug in \( t = 1 \) into the velocity function: \( v(1) = \frac{-3(1)^2 + 27}{(1^2 + 9)^2} = \frac{24}{100} = 0.24 \text{ ft/sec} \).
• Finding when velocity is zero: Set the numerator of the velocity function to zero and solve: \( -3t^2 + 27 = 0 \), we find that \( t = 3 \) seconds.

Differentiation allows us to turn the abstract position function into tangible insights about movement and speed.