Problem 25 - The Calculus with Analytic Geometry

Question

A particle is moving along a straight line according to the given equation of motion, where \(s\) \(\mathrm{ft}\) is the directed distance of the particle from the origin at \(t \mathrm{sec}\). Find the time when the instantaneous acceleration is zero, and then find the directed distance of the particle from the origin and the instantaneous velocity at this instant. $$ s=\frac{125}{16 t+32}-\frac{2}{5} t^{5}, t \geq 0 $$

Step-by-Step Solution

Answer

Instantaneous acceleration is zero at \(t = 1\). Directed distance: \(s \approx 1.95 \text{ ft}\). Instantaneous velocity: \(v = -22 \text{ ft/sec}\).

Step 1: Find the Velocity Function

The velocity function, denoted as \(v(t)\), is the first derivative of the position function \(s(t)\). Compute the derivative of \(s(t) = \frac{125}{16t + 32} - \frac{2}{5} t^5\) with respect to \(t\) using the chain rule and the power rule.

Step 2: Find the Acceleration Function

The acceleration function, denoted as \(a(t)\), is the first derivative of the velocity function \(v(t)\). Compute the first derivative of the velocity function obtained in Step 1.

Step 3: Set the Acceleration to Zero

Set the acceleration function, \(a(t)\), equal to zero. Solve for \(t\) to determine when the instantaneous acceleration is zero.

Step 4: Compute the Directed Distance

Substitute the value of \(t\) obtained in Step 3 back into the original position function \(s(t)\) to find the directed distance of the particle from the origin at this instant.

Step 5: Compute the Instantaneous Velocity

Substitute the value of \(t\) obtained in Step 3 into the velocity function \(v(t)\) to find the instantaneous velocity at this instant.

Key Concepts

derivative

To start with the problem, you need to understand what a derivative is. In calculus, the derivative of a function represents the rate at which the function's value changes. If you have a function, say \(f(x)\), the derivative, \(f'(x)\), tells you the slope of the function at any point \(x\). For the position function \(s(t)\) given in the exercise, its derivative gives us the velocity function \(v(t)\). In this problem, you would compute the derivative of \(s(t) = \frac{125}{16t + 32} - \frac{2}{5} t^5\) with respect to \(t\). This involves applying both the chain rule and the power rule.

velocity function

The velocity function \(v(t)\) is essentially the first derivative of the position function \(s(t)\). It describes how fast the particle is moving along the straight line at any given time \(t\). By finding the derivative of \(s(t) = \frac{125}{16t + 32} - \frac{2}{5} t^5\), you'll get \(v(t)\). This gives you a function that shows the particle's velocity as a function of time. For the given function:

• Apply the chain rule to \(\frac{125}{16t + 32}\)
• Apply the power rule to \(-\frac{2}{5} t^5\)

When you combine these, you have the complete velocity function.

acceleration function

Next, you need the acceleration function \(a(t)\). This is obtained by differentiating the velocity function \(v(t)\). Differentiation again comes into play. By finding the derivative of \(v(t)\), you get the acceleration \(a(t)\), telling you how the velocity changes over time. It shows the rate of change of velocity, which is crucial for determining when the instantaneous acceleration is zero.

instantaneous acceleration

Instantaneous acceleration is the rate of change of velocity at a specific point in time. To find when the instantaneous acceleration is zero:

• Set \(a(t) = 0\)
• Solve for \(t\)

Doing this in the problem helps identify the specific time when the velocity stops changing. It's a key step that allows you to find the exact moment in time you're particularly interested in.

chain rule

The chain rule is a fundamental concept used when differentiating composite functions. A composite function is like \( \frac{125}{16t + 32} \) from the exercise, where function is inside another function. To apply the chain rule:

• Differentiate the outer function
• Multiply it by the derivative of the inner function

Using the chain rule properly simplifies finding the first derivative for complex functions.

power rule

The power rule is one of the simplest and most frequently used differentiation rules. It states that for any function \(f(t) = t^n\), the derivative is given by \(f'(t) = nt^{n-1}\). For example, in \(-\frac{2}{5} t^5\), using the power rule, takes less effort. This rule is applied when dealing with polynomial terms within the velocity and acceleration functions, making differentiation straightforward for those specific parts.