Problem 27 - 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. $$ s=\frac{4}{9} t^{3 / 2}+2 t^{1 / 2}, t \geq 0 $$

Step-by-Step Solution

Answer

Acceleration is zero at t = 1.5 sec. Distance is 3.27 ft. Velocity is 1.633 ft/sec.

Step 1 - Find the Velocity

To find the velocity, take the first derivative of the position function with respect to time. The position function given is: \[ s = \frac{4}{9} t^{3/2} + 2 t^{1/2} \]. The velocity function, v(t), is: \[ v(t) = \frac{d}{dt}\left( \frac{4}{9} t^{3/2} + 2 t^{1/2} \right) \]. Apply the power rule: \[ v(t) = \frac{4}{9} \cdot \frac{3}{2} t^{1/2} + 2 \cdot \frac{1}{2} t^{-1/2} \]. Simplify: \[ v(t) = \frac{2}{3} t^{1/2} + t^{-1/2} \].

Step 2 - Find the Acceleration

To find the acceleration, take the derivative of the velocity function with respect to time. The velocity function is: \[ v(t) = \frac{2}{3} t^{1/2} + t^{-1/2} \]. The acceleration function, a(t), is: \[ a(t) = \frac{d}{dt}\left( \frac{2}{3} t^{1/2} + t^{-1/2} \right) \]. Apply the power rule: \[ a(t) = \frac{2}{3} \cdot \frac{1}{2} t^{-1/2} + (-1/2) t^{-3/2} \]. Simplify: \[ a(t) = \frac{1}{3} t^{-1/2} - \frac{1}{2} t^{-3/2} \].

Step 3 - Set Acceleration to Zero

To find when the instantaneous acceleration is zero, set the acceleration function equal to zero and solve for t: \[ 0 = \frac{1}{3} t^{-1/2} - \frac{1}{2} t^{-3/2} \]. Multiply through by \( t^{3/2} \) to clear the fractions: \[ 0 = \frac{1}{3} t - \frac{1}{2} \]. Rearrange and solve for t: \[ \frac{1}{3} t = \frac{1}{2} \], \[ t = \frac{3}{2} = 1.5 \].

Step 4 - Find the Directed Distance at t = 1.5

Substitute \( t = 1.5 \) back into the original position function to find the directed distance: \[ s(1.5) = \frac{4}{9} (1.5)^{3/2} + 2 (1.5)^{1/2} \]. Calculate the values: \[ (1.5)^{3/2} = \sqrt{1.5^3} = \sqrt{3.375} \approx 1.837 \], \[ (1.5)^{1/2} = \sqrt{1.5} \approx 1.225 \]. So: \[ s(1.5) = \frac{4}{9} \cdot 1.837 + 2 \cdot 1.225 \], \[ s(1.5) \approx \frac{4}{9} \cdot 1.837 + 2.45 \], \[ s(1.5) \approx 0.82 + 2.45 = 3.27 \].

Step 5 - Find the Instantaneous Velocity at t = 1.5

Substitute \( t = 1.5 \) back into the velocity function to find the instantaneous velocity: \[ v(1.5) = \frac{2}{3} (1.5)^{1/2} + (1.5)^{-1/2} \]. Calculate the values: \[ (1.5)^{1/2} = \sqrt{1.5} \approx 1.225 \], \[ (1.5)^{-1/2} = \frac{1}{\sqrt{1.5}} \approx 0.816 \]. So: \[ v(1.5) = \frac{2}{3} \cdot 1.225 + 0.816 \], \[ v(1.5) \approx 0.817 + 0.816 = 1.633 \].

Key Concepts

Calculus

Calculus is an essential branch of mathematics that helps us understand changes between quantities. For instantaneous acceleration in particle motion, we utilize derivatives, a fundamental concept in calculus. Derivatives allow us to determine rates of change, such as velocity and acceleration, from position functions.

The position function tells us how far the particle is from a reference point at any given time. Taking the derivative of this function with respect to time gives us the velocity, showing how the position changes over time. If we take the derivative again, we get the acceleration, showing how the velocity changes over time.

Derivative of Position Function

To find the instantaneous velocity, we first need the derivative of the position function, which is given by:
\ s = \frac{4}{9} t^{3/2} + 2 t^{1/2}. \

Using the power rule, we can derive the velocity function:
\ v(t) = \frac{2}{3} t^{1/2} + t^{-1/2}. \

For acceleration, we take another derivative of the velocity function. It becomes slightly more complex:
\ a(t) = \frac{1}{3} t^{-1/2} - \frac{1}{2} t^{-3/2}. \

Understanding the process of getting these derivatives is crucial. The power rule states that for any term of the form \( t^n \), the derivative is \( n t^{n-1} \). This rule is repeatedly applied to obtain both velocity and acceleration functions.

Velocity and Acceleration in Motion

Velocity and acceleration are central concepts in analyzing particle motion. Velocity tells us how fast and in what direction the particle is moving. Acceleration shows the rate of change of velocity.

To find when the instantaneous acceleration is zero, set the acceleration function to zero and solve for \( t \). We get:
\ 0 = \frac{1}{3} t^{-1/2} - \frac{1}{2} t^{-3/2}, \ which simplifies to \( t = 1.5 \).

Given \( t = 1.5 \), we substitute back into the original position and velocity functions:

Position: \( s(1.5) = 3.27 \),
Velocity: \( v(1.5) = 1.633 \).

This means that at \( t = 1.5 \ seconds \), the particle's acceleration is zero, it's 3.27 feet from the origin, and it has a velocity of 1.633 feet per second.