Problem 7 - Thomas Calculus

Question

Particle motion At time \(t\) , the position of a body moving along the \(s\) -axis is \(s=t^{3}-6 t^{2}+9 t \mathrm{m}\) . a. Find the body's acceleration each time the velocity is zero. b. Find the body's speed each time the acceleration is zero. c. Find the total distance traveled by the body from \(t=0\) to \(t=2\) .

Step-by-Step Solution

Answer

a. Acceleration: \(-6\) m/s² at \(t=1\), \(6\) m/s² at \(t=3\). b. Speed: 3 m/s when \(t=2\). c. Total distance: 8 meters.

Step 1: Find the Velocity Function

The position of the body is given by the equation \( s(t) = t^3 - 6t^2 + 9t \). To find the velocity, differentiate this function with respect to \( t \):\[ v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9. \]

Step 2: Find Times When Velocity is Zero

Set the velocity function equal to zero and solve for \( t \):\[ 3t^2 - 12t + 9 = 0. \]Divide the entire equation by 3:\[ t^2 - 4t + 3 = 0. \]Factor the quadratic equation:\[ (t-1)(t-3) = 0. \]Thus, the velocity is zero at \( t = 1 \) and \( t = 3 \).

Step 3: Find the Acceleration Function

Differentiate the velocity function to find acceleration:\[ a(t) = \frac{dv}{dt} = 6t - 12. \]

Step 4: Find Acceleration When Velocity is Zero

Using times from Step 2, substitute \( t = 1 \) and \( t = 3 \) into the acceleration function:- At \( t = 1 \), \( a(1) = 6(1) - 12 = -6 \).- At \( t = 3 \), \( a(3) = 6(3) - 12 = 6 \).

Step 5: Find Times When Acceleration is Zero

Set the acceleration function equal to zero and solve for \( t \):\[ 6t - 12 = 0. \]Solving gives:\[ t = 2. \]

Step 6: Find Speed When Acceleration is Zero

Substitute \( t = 2 \) into the velocity function to find the speed:\[ v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3. \]The speed is the absolute value of velocity, which is 3 m/s.

Step 7: Find the Total Distance Traveled

Calculate the position at \( t = 0 \), \( t = 1 \), \( t = 2 \), and \( t = 3 \):- \( s(0) = 0^3 - 6(0)^2 + 9(0) = 0 \)- \( s(1) = 1^3 - 6(1)^2 + 9(1) = 1 - 6 + 9 = 4 \)- \( s(2) = 2^3 - 6(2)^2 + 9(2) = 8 - 24 + 18 = 2 \)- \( s(3) = 3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0 \)The total distance is \(|s(1) - s(0)| + |s(2) - s(1)| + |s(3) - s(2)| = 4 + 2 + 2 = 8 \) meters.

Key Concepts

Acceleration

Acceleration measures how quickly the velocity of an object changes over time.
It is a fundamental concept in the study of particle motion.
To find acceleration, we differentiate the velocity function with respect to time. In this exercise, the velocity function is given as

• \[ v(t) = 3t^2 - 12t + 9. \]

The acceleration function is then found by differentiating the velocity function:

• \[ a(t) = \frac{dv}{dt} = 6t - 12. \]

This tells us how the velocity changes at any point in time.
For instance, at time \( t = 1 \), the acceleration is -6 m/s², indicating the velocity is decreasing.
At \( t = 3 \), the acceleration is 6 m/s², meaning the velocity is increasing.

Velocity

Velocity is the rate of change of the position of a body with respect to time.
It tells us how fast something is moving and in which direction.
In calculus, velocity is found by differentiating the position function.Consider the position function given by

• \[ s(t) = t^3 - 6t^2 + 9t. \]

By differentiating with respect to time, we obtain the velocity function:

• \[ v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9. \]

When the velocity is zero, it means the object is momentarily at rest.
Solving for \( t \) when \( v(t) = 0 \) gives us critical moments when these changes occur.

Distance Traveled

The distance traveled by an object is different from its displacement.
Distance considers the total path covered, regardless of direction.
In this exercise, to determine the total distance traveled, the positions at various times were calculated.Initially, the object's position is at \( t = 0 \):

• \( s(0) = 0 \)
• \( s(1) = 4 \)
• \( s(2) = 2 \)
• \( s(3) = 0 \)

To find the total distance, the differences between consecutive positions were summed:

• \( |4 - 0| + |2 - 4| + |0 - 2| = 4 + 2 + 2 = 8 \) meters.

This approach ensures the calculated distance accounts for changes in direction.

Differentiation

Differentiation is a mathematical process used to find the rate at which a quantity changes.
It's crucial in understanding particle motion because it helps us find velocity and acceleration from a position function.
To differentiate a function with respect to time, we apply the power rule, which helps derive these functions. In this exercise, we differentiated the position function:

• \[ s(t) = t^3 - 6t^2 + 9t. \]

First, we found the velocity by differentiating the position:
• \[ v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9. \]

Then, we derived the acceleration from the velocity:

• \[ a(t) = \frac{dv}{dt} = 6t - 12. \]

Differentiation provides insights into how motion variables change over time, thereby enhancing our understanding of the object's behavior.