Problem 17 - Calculus - AP Edition

Question

The function \(s(t)\) describes the position of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, speed, and acceleration at time \(t=1\) (c) At what times is the particle stopped? (d) When is the particle speeding up? Slowing down? (e) Find the total distance traveled by the particle from time \(t=0\) to time \(t=5\) $$s(t)=t^{3}-3 t^{2}, \quad t \geq 0$$

Step-by-Step Solution

Answer

(a) \(v(t) = 3t^2 - 6t\), \(a(t) = 6t - 6\). (b) At \(t=1\), \(s = -2\) ft, \(v = -3\) ft/s, \(a = 0\) ft/s². (c) Stopped at \(t=0, 2\). (d) Speeding up for \(t<1\) and \(t>2\), slowing down for \(1

Step 1: Differentiate to Find Velocity

The velocity function is the derivative of the position function. Given \(s(t) = t^3 - 3t^2\), differentiate it with respect to \(t\): \[ v(t) = \frac{d}{dt}(t^3 - 3t^2) = 3t^2 - 6t \]

Step 2: Differentiate to Find Acceleration

The acceleration function is the derivative of the velocity function. Differentiate \(v(t) = 3t^2 - 6t\) with respect to \(t\): \[ a(t) = \frac{d}{dt}(3t^2 - 6t) = 6t - 6 \]

Step 3: Evaluate Functions at t=1

Calculate the position, velocity, and acceleration at \(t=1\): For position, \(s(1) = 1^3 - 3(1)^2 = 1 - 3 = -2\) feet.For velocity, \(v(1) = 3(1)^2 - 6(1) = 3 - 6 = -3\) ft/s.For acceleration, \(a(1) = 6(1) - 6 = 0\) ft/s².For speed, which is the absolute value of velocity: \(\text{speed} = |v(1)| = |-3| = 3\) ft/s.

Step 4: Determine When Particle is Stopped

A particle is stopped when its velocity is zero. Set \(v(t) = 0\) and solve:\[ 3t^2 - 6t = 0 \]Factoring out \(3t\), \[ 3t(t - 2) = 0 \] gives two solutions: \(t = 0\) and \(t = 2\). The particle is stopped at \(t = 0\) seconds and \(t = 2\) seconds.

Step 5: Analyze Speeding Up and Slowing Down

A particle is speeding up when velocity and acceleration have the same sign and slowing down when they have opposite signs. For \(a(t) = 6t - 6\):- When \(t < 1\): \(a(t) < 0\), \(v(t) < 0\) \(\Rightarrow\) Speeding up as both are negative.- When \(1 < t < 2\): \(a(t) > 0\), \(v(t) < 0\) \(\Rightarrow\) Slowing down as signs differ.- When \(t > 2\): \(a(t) > 0\), \(v(t) > 0\) \(\Rightarrow\) Speeding up as both are positive.

Step 6: Calculate Total Distance Traveled from t=0 to t=5

First, recognize that total distance requires the integral of the absolute value of velocity. We consider the intervals \(([0, 2]) \cup ([2, 5])\) as particle changes direction at charge in sign of \(v(t)\).- Integrate from \(t=0\) to \(t=2\): \(\int_0^2 |3t^2 - 6t| \, dt = \int_0^2 -(3t^2 - 6t) \, dt\) because \(v(t) < 0\).- Integrate from \(t=2\) to \(t=5\): \(\int_2^5 |3t^2 - 6t| \, dt = \int_2^5 (3t^2 - 6t) \, dt\) because \(v(t) > 0\).Solve these integrals:- \(\int -(3t^2 - 6t) \, dt = [-t^3 + 3t^2]_0^2 = -8\).- \(\int (3t^2 - 6t) \, dt = [t^3 - 3t^2]_2^5\). Calculate: - At \(t=5\): \((5)^3 - 3(5)^2 = 125 - 75 = 50\). - At \(t=2\): \((2)^3 - 3(2)^2 = 8 - 12 = -4\). - Distance from \(2\) to \(5\): \(50 - (-4) = 54\).Adding these:Total distance = \(8 + 54 = 62\) feet.

Key Concepts

Differentiation

Differentiation plays a critical role in calculus, especially when determining the rate of change for functions. In this exercise, differentiation is used to find both the velocity and acceleration functions from the given position function, \(s(t)\). Here, the position function \(s(t) = t^3 - 3t^2\) describes how far a particle has moved over time along a coordinate line. To explore changes over time, we differentiate this function with respect to time \(t\). This process involves applying standard rules of differentiation, such as the power rule, to calculate the derivative, which reflects the function's rate of change.
Understanding differentiation helps in determining how positions change over time into other physical quantities such as velocity and acceleration. Without this foundational skill, analyzing particle motion and other real-world phenomena would be much more challenging. Differentiation effectively allows us to translate position into movement-related functions critical for further analysis.

Velocity Function

The velocity function gives us insight into the speed and direction of a particle's motion at any given time. In our exercise, we differentiated the position function to derive the velocity function: \(v(t) = 3t^2 - 6t\). This function indicates how fast the particle's position is changing per unit of time, and it can be positive or negative based on the particle's direction.
When you solve for velocity, it's crucial to check its value at specific times to understand the particle's motion better. For instance, setting \(v(t)\) to zero helps identify when the particle stops. Additionally, analyzing the velocity function over specific intervals helps determine when the particle is moving forward or backward. Understanding the velocity function equips you to comprehend speed and direction, foundational concepts in physics and engineering.

• Derivative of position function = velocity function
• Tracks how quickly position changes over time
• Indicates particle stopping or direction change when \(v(t) = 0\)

Acceleration Function

Acceleration describes how the velocity of a particle changes over time. it's a crucial concept in understanding motion dynamics. From the velocity function we previously derived, \(v(t) = 3t^2 - 6t\), we further differentiate to find the acceleration function: \(a(t) = 6t - 6\). Essentially, acceleration gives the rate at which velocity is changing with respect to time.
Acceleration helps us understand whether a particle is picking up speed or slowing down. Positive acceleration means the velocity is increasing, while negative acceleration indicates a decrease. It's important to examine not just if acceleration is positive or negative, but also the relationship between velocity and acceleration:

• Both positive – speeding up
• Both negative – speeding up in opposite direction
• Opposite signs – slowing down

Mastering acceleration is vital for analyzing complex systems where motion changes dynamically, such as in automotive engineering or physics.

Position Function

The position function \(s(t) = t^3 - 3t^2\) tells us the exact location of the particle along the coordinate line at any time \(t\). This foundational function is our starting point for deriving other functions related to motion, such as velocity and acceleration.
Calculating the position at a specific \(t\) gives us spatial coordinates – or how far the particle has traveled from the starting point. This calculation is generally straightforward: plug the time value into the position function to find the particle's exact location. For instance, evaluating \(s(t)\) at \(t = 1\) would tell us where the particle is at exactly one second.
Position functions are essential in scenarios ranging from simple physics problems to tracking the trajectories of spacecraft. Understanding how position functions operate allows us to model and control movement in multiple dimensions or complex systems.

Distance Traveled

Distance traveled is a measure of the total length of the path taken by a particle over a given time interval. Unlike displacement, which is the shortest path between two points, distance considers the actual path traveled irrespective of direction. In this exercise, calculating the total distance traveled involves integrating the absolute value of the velocity function \(|v(t)|\).
To determine distance from \(t = 0\) to \(t = 5\), we evaluate the integral of velocity over intervals where it changes sign, capturing areas under the velocity curve accurately. It involves:

• Identifying intervals from when velocity is negative \(t = 0\) to when it turns positive \(t = 2\)
• Calculating separate integrals for each velocity direction
• Summing results to ensure all motion is accounted

The result is the cumulative travel without concern for direction, giving the total path length covered, crucial in scenarios such as fuel efficiency calculations or motion analysis in robotics.

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Key Concepts

Differentiation

Differentiation plays a critical role in calculus, especially when determining the rate of change for functions. In this exercise, differentiation is used to find both the velocity and acceleration functions from the given position function, \(s(t)\). Here, the position function \(s(t) = t^3 - 3t^2\) describes how far a particle has moved over time along a coordinate line. To explore changes over time, we differentiate this function with respect to time \(t\). This process involves applying standard rules of differentiation, such as the power rule, to calculate the derivative, which reflects the function's rate of change.
Understanding differentiation helps in determining how positions change over time into other physical quantities such as velocity and acceleration. Without this foundational skill, analyzing particle motion and other real-world phenomena would be much more challenging. Differentiation effectively allows us to translate position into movement-related functions critical for further analysis.

Velocity Function

The velocity function gives us insight into the speed and direction of a particle's motion at any given time. In our exercise, we differentiated the position function to derive the velocity function: \(v(t) = 3t^2 - 6t\). This function indicates how fast the particle's position is changing per unit of time, and it can be positive or negative based on the particle's direction.
When you solve for velocity, it's crucial to check its value at specific times to understand the particle's motion better. For instance, setting \(v(t)\) to zero helps identify when the particle stops. Additionally, analyzing the velocity function over specific intervals helps determine when the particle is moving forward or backward. Understanding the velocity function equips you to comprehend speed and direction, foundational concepts in physics and engineering.

• Derivative of position function = velocity function
• Tracks how quickly position changes over time
• Indicates particle stopping or direction change when \(v(t) = 0\)

Acceleration Function

Acceleration describes how the velocity of a particle changes over time. it's a crucial concept in understanding motion dynamics. From the velocity function we previously derived, \(v(t) = 3t^2 - 6t\), we further differentiate to find the acceleration function: \(a(t) = 6t - 6\). Essentially, acceleration gives the rate at which velocity is changing with respect to time.
Acceleration helps us understand whether a particle is picking up speed or slowing down. Positive acceleration means the velocity is increasing, while negative acceleration indicates a decrease. It's important to examine not just if acceleration is positive or negative, but also the relationship between velocity and acceleration:

• Both positive – speeding up
• Both negative – speeding up in opposite direction
• Opposite signs – slowing down

Mastering acceleration is vital for analyzing complex systems where motion changes dynamically, such as in automotive engineering or physics.

Position Function

The position function \(s(t) = t^3 - 3t^2\) tells us the exact location of the particle along the coordinate line at any time \(t\). This foundational function is our starting point for deriving other functions related to motion, such as velocity and acceleration.
Calculating the position at a specific \(t\) gives us spatial coordinates – or how far the particle has traveled from the starting point. This calculation is generally straightforward: plug the time value into the position function to find the particle's exact location. For instance, evaluating \(s(t)\) at \(t = 1\) would tell us where the particle is at exactly one second.
Position functions are essential in scenarios ranging from simple physics problems to tracking the trajectories of spacecraft. Understanding how position functions operate allows us to model and control movement in multiple dimensions or complex systems.

Distance Traveled

Distance traveled is a measure of the total length of the path taken by a particle over a given time interval. Unlike displacement, which is the shortest path between two points, distance considers the actual path traveled irrespective of direction. In this exercise, calculating the total distance traveled involves integrating the absolute value of the velocity function \(|v(t)|\).
To determine distance from \(t = 0\) to \(t = 5\), we evaluate the integral of velocity over intervals where it changes sign, capturing areas under the velocity curve accurately. It involves:

• Identifying intervals from when velocity is negative \(t = 0\) to when it turns positive \(t = 2\)
• Calculating separate integrals for each velocity direction
• Summing results to ensure all motion is accounted

The result is the cumulative travel without concern for direction, giving the total path length covered, crucial in scenarios such as fuel efficiency calculations or motion analysis in robotics.