Problem 18 - Fundamentals of Physics

Question

The position of a particle moving along an \(x\) axis is given by \(x=12 t^{2}-2 t^{3}\), where \(x\) is in meters and \(t\) is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at \(t=3.0 \mathrm{~s}\). (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and \(\underline{(g)}\) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at \(t=0\) )? (i) Determine the average velocity of the particle between \(t=0\) and \(t=\overline{3} \mathrm{~s}\)

Step-by-Step Solution

Answer

The particle's position at \( t=3 \ s \) is 54 m, velocity is 18 m/s, and acceleration is -12 m/s². Max position is 64 m at \( t=4 \ s \); max velocity is 24 m/s at \( t=2 \ s \). Acceleration when particle stops is -24 m/s², and average velocity is 18 m/s.

Step 1: Determine the Position at t=3.0 s

Insert the given time into the position formula: \( x = 12t^2 - 2t^3 \). Calculate for \( t = 3.0 \ s \).\[ x = 12(3.0)^2 - 2(3.0)^3 \]\[ x = 108 - 54 = 54 \ \text{meters} \]

Step 2: Derive and Evaluate Velocity Formula

Velocity is the first derivative of the position with respect to time. Differentiate the position equation:\[ v(t) = \frac{dx}{dt} = 24t - 6t^2 \]Evaluate for \( t = 3.0 \ s \):\[ v(3.0) = 24(3.0) - 6(3.0)^2 = 72 - 54 = 18 \ \text{m/s} \]

Step 3: Derive and Evaluate Acceleration Formula

Acceleration is the derivative of velocity, or the second derivative of position with respect to time. Differentiate the velocity equation:\[ a(t) = \frac{dv}{dt} = 24 - 12t \]Evaluate for \( t = 3.0 \ s \):\[ a(3.0) = 24 - 12(3.0) = 24 - 36 = -12 \ \text{m/s}^2 \]

Step 4: Find Maximum Positive Position

To find the maximum positive position, set the derivative of the position (velocity) equal to zero and solve for \( t \).\[ 24t - 6t^2 = 0 \]Factor the equation:\[ 6t(4 - t) = 0 \]So, \( t = 0 \) or \( t = 4 \ s \). Evaluate the position at \( t = 4 \ s \):\[ x(4) = 12(4)^2 - 2(4)^3 = 192 - 128 = 64 \ \text{meters} \]

Step 5: Find Maximum Positive Velocity and Time

Velocity changes sign when the acceleration is zero. Set the acceleration equation to zero:\[ 24 - 12t = 0 \]Solve for \( t \):\[ t = 2 \ s \]Evaluate the velocity at \( t = 2 \ s \):\[ v(2) = 24(2) - 6(2)^2 = 48 - 24 = 24 \ \text{m/s} \]

Step 6: Calculate Acceleration at the Instant Particle Stops

A particle stops when its velocity is zero. We've previously calculated that this occurs at \( t = 4 \ s \).Evaluate the acceleration at \( t = 4 \ s \):\[ a(4) = 24 - 12(4) = 24 - 48 = -24 \ \text{m/s}^2 \]

Step 7: Determine Average Velocity

The average velocity over a time interval is given by the change in position over the change in time.\[ \text{Average Velocity} = \frac{x(3) - x(0)}{3 - 0} = \frac{54 - 0}{3} = 18 \ \text{m/s} \]

Key Concepts

Particle Motion

In kinematics, understanding particle motion is crucial for analyzing how objects move in space and time. Here, we are observing a particle moving along the x-axis, described by the position function \( x = 12t^2 - 2t^3 \). This equation tells us how the particle's position changes with respect to time \( t \). The polynomial form indicates a non-linear path, suggesting that the particle's movement involves varying velocities and accelerations as time progresses.
To study this motion, we examine its position at specific time intervals by substituting values of \( t \) into the equation. This approach helps us track the particle's location along the x-axis at any given moment.

• Position equation: Shows how position changes over time.
• Non-linear motion: Implies varying speed and acceleration.

Velocity Calculation

Velocity is the rate at which the position of the particle changes over time, making it an essential concept in kinematics. To obtain the velocity function, we differentiate the given position function \( x = 12t^2 - 2t^3 \) with respect to time \( t \).
The result, \( v(t) = 24t - 6t^2 \), is the velocity function, indicating that velocity is also a function of time, changing as the particle moves. To find the particle's instantaneous velocity at any point, simply substitute the desired time into this equation.
For example, at \( t = 3.0 \, ext{s} \), the calculation yields a velocity of 18 m/s.

• Differentiation: Used to find the velocity from the position function.
• Instantaneous velocity: Velocity at a specific point in time.

Acceleration Calculation

Acceleration refers to the change in velocity over time, showing how quickly the particle's speed is altering. To determine the acceleration, differentiate the velocity function \( v(t) = 24t - 6t^2 \). This yields the acceleration function \( a(t) = 24 - 12t \).
This function helps identify how the particle's velocity is increasing or decreasing at any point in time. By substituting specific values of \( t \), we obtain the instantaneous acceleration. At \( t = 3.0 \, ext{s} \), the particle's acceleration is calculated to be \(-12 \, ext{m/s}^2\), indicating a deceleration.

• Differentiation again: From velocity to get acceleration.
• Instantaneous acceleration: Acceleration at a specific moment.

Maximum Position

To identify the maximum position the particle achieves on the x-axis, we need to find when the velocity is zero because this indicates a turning point in motion. Setting the derived velocity equation \( 24t - 6t^2 \) equal to zero helps find these points.
Solving \( 24t - 6t^2 = 0 \) results in \( t = 0 \) and \( t = 4 \, ext{s} \). The particle reaches its highest position at \( t = 4 \, ext{s} \), which is 64 meters.
Analyzing these turning points provides insights into the particle's path, helping predict its farthest reach along the x-axis.

• Velocity set to zero: Finds turning points or extreme positions.
• Maximum position: The highest point reached, here at 64 meters.

Maximum Velocity

Maximum velocity occurs when a particle's rate of change of velocity, or acceleration, is zero. Thus, setting the acceleration equation \( a(t) = 24 - 12t \) to zero allows us to find the time when the velocity is at its peak.
Solving \( 24 - 12t = 0 \) gives \( t = 2 \, ext{s} \). Substituting this into the velocity function \( v(t) = 24t - 6t^2 \) yields a maximum velocity of 24 m/s. Understanding these dynamics helps check how fast the particle can go before it begins to slow down again.

• Acceleration set to zero: Indicates maximum velocity.
• Maximum velocity: Achieved at 2 seconds with 24 m/s.

Average Velocity

Average velocity provides insight into the overall motion of the particle over a specific time span. It is calculated by dividing the difference in position by the time interval.
Using the given position function, the position at \( t = 0 \) and \( t = 3 \, ext{s} \) are used to evaluate the average velocity as follows: \[ \text{Average Velocity} = \frac{x(3) - x(0)}{3 - 0} = \frac{54 - 0}{3} = 18 \, ext{m/s} \] This calculation shows how quickly the particle, on average, travels between these two points.

• Total distance over time: Method for finding average velocity.
• Average velocity: 18 m/s over the interval from 0 to 3 seconds.