Problem 15
Question
The function \(s(t)\) describes the position of a particle moving along a coordinate line, where \(s\) is in meters and \(t\) is in seconds. (a) Make a table showing the position, velocity, and acceleration to two decimal places at times \(t=1,2,3,4,5\). (b) At each of the times in part (a), determine whether the particle is stopped; if it is not, state its direction of motion. (c) At each of the times in part (a), determine whether the particle is speeding up, slowing down, or neither. $$ s(t)=\sin \frac{\pi t}{4} $$
Step-by-Step Solution
Verified Answer
(a) Position: \(0.71, 1.00, 0.71, 0.00, -0.71\).
Velocity: \(0.55, 0, -0.55, -0.79, -0.55\).
Acceleration: \(-0.88, -1.94, -0.88, 0, 0.88\).
(b) Stopped at \(t=2\), direction: Right, Stopped, Left, Left, Left.
(c) Speed: Slows, Neither, Speeds, Neither, Slows.
1Step 1: Position Calculation
To find the position of the particle at given times, we need to evaluate the function \( s(t) = \sin \left(\frac{\pi t}{4}\right) \). We do this for each \( t = 1, 2, 3, 4, 5\) and round to two decimal places. - For \( t = 1 \), \( s(1) = \sin \left(\frac{\pi}{4}\right) \approx 0.71 \)- For \( t = 2 \), \( s(2) = \sin \left(\frac{\pi}{2}\right) = 1.00 \)- For \( t = 3 \), \( s(3) = \sin \left(\frac{3\pi}{4}\right) \approx 0.71 \)- For \( t = 4 \), \( s(4) = \sin \left(\pi\right) = 0.00 \)- For \( t = 5 \), \( s(5) = \sin \left(\frac{5\pi}{4}\right) \approx -0.71 \)
2Step 2: Velocity Calculation
The velocity of the particle is the derivative of the position function, \( v(t) = s'(t) = \frac{d}{dt} \left(\sin \frac{\pi t}{4}\right) \). Using the chain rule, we find \( v(t) = \frac{\pi}{4} \cos \frac{\pi t}{4} \). Evaluate this at \( t = 1, 2, 3, 4, 5\):- \( v(1) \approx 0.55 \)- \( v(2) = 0 \)- \( v(3) \approx -0.55 \)- \( v(4) \approx -0.79 \)- \( v(5) \approx -0.55 \)
3Step 3: Acceleration Calculation
The acceleration is the derivative of the velocity, \( a(t) = v'(t) = \frac{d}{dt} \left(\frac{\pi}{4} \cos \frac{\pi t}{4}\right) \). This gives \( a(t) = -\left(\frac{\pi}{4}\right)^2 \sin \frac{\pi t}{4} \). Calculate for \( t = 1, 2, 3, 4, 5\):- \( a(1) \approx -0.88 \)- \( a(2) \approx -1.94 \)- \( a(3) \approx -0.88 \)- \( a(4) = 0 \)- \( a(5) \approx 0.88 \)
4Step 4: Determine Stop and Direction
A particle stops when its velocity is zero. Thus, check the velocity values from Step 2:- At \( t = 2 \), the velocity is 0, so the particle is stopped. - At other times, the particle is not stopped. Check the sign of velocity to determine direction: positive for right/forwards and negative for left/backwards. - \( t = 1 \): moves right (\( v > 0 \))- \( t = 3 \): moves left (\( v < 0 \))- \( t = 4, 5 \): moves left (\( v < 0 \))
5Step 5: Determine Speeding Up or Slowing Down
The particle speeds up if velocity and acceleration have the same sign, slows down if they have opposite signs, and neither if velocity is zero.- \( t = 1 \): Slowing down (\( v > 0, a < 0 \))- \( t = 2 \): Neither (\( v = 0 \))- \( t = 3 \): Speeding up (\( v < 0, a < 0 \))- \( t = 4 \): Neither (\( a = 0 \))- \( t = 5 \): Slowing down (\( v < 0, a > 0 \))
Key Concepts
Concept of CalculusUnderstanding VelocityDecoding AccelerationPosition Function Dynamics
Concept of Calculus
Calculus is a branch of mathematics that focuses on the study of change and motion. It relies on two fundamental concepts: differentiation and integration. Differentiation helps us understand how a function changes, while integration is used to calculate areas under curves, among other things. In the context of particle motion, differentiation is used to determine both velocity and acceleration from the position function. This allows us to analyze how a particle's motion evolves over time and to predict future behavior.
When dealing with particle motion, calculus provides the tools to extract useful information. By differentiating a position function, you can find the velocity function. Differentiating again, you can then obtain the acceleration function. This step-by-step transformation, made possible by calculus, is essential for understanding how forces and various factors affect an object's movement.
When dealing with particle motion, calculus provides the tools to extract useful information. By differentiating a position function, you can find the velocity function. Differentiating again, you can then obtain the acceleration function. This step-by-step transformation, made possible by calculus, is essential for understanding how forces and various factors affect an object's movement.
Understanding Velocity
Velocity is a vector quantity that refers to the speed of an object in a specific direction. In calculus terms, it is the first derivative of the position function with respect to time. This derivative gives us a velocity function that illustrates how quickly and in which direction the position of a particle is changing over time.
For the given position function, the velocity is expressed as:\[v(t) = \frac{d}{dt} \left(\sin \frac{\pi t}{4}\right)\]Using the chain rule, this becomes:\[v(t) = \frac{\pi}{4} \cos \frac{\pi t}{4}\]
Now, by evaluating the velocity function at various times, we can determine how fast and in which direction the particle is moving. If velocity is positive, the particle moves in the forward direction; if negative, it moves backward. Zero velocity indicates that the particle is momentarily stopped.
For the given position function, the velocity is expressed as:\[v(t) = \frac{d}{dt} \left(\sin \frac{\pi t}{4}\right)\]Using the chain rule, this becomes:\[v(t) = \frac{\pi}{4} \cos \frac{\pi t}{4}\]
Now, by evaluating the velocity function at various times, we can determine how fast and in which direction the particle is moving. If velocity is positive, the particle moves in the forward direction; if negative, it moves backward. Zero velocity indicates that the particle is momentarily stopped.
Decoding Acceleration
Acceleration is the rate of change of velocity with respect to time. Similar to velocity being the derivative of position, acceleration is the derivative of velocity. It provides insight into how the velocity of a particle changes, thus affecting the particle's speed and direction. In mathematical terms, acceleration is the second derivative of the position function.
For our function, the acceleration is derived as follows:\[a(t) = v'(t) = \frac{d}{dt} \left(\frac{\pi}{4} \cos \frac{\pi t}{4}\right)\]This results in:\[a(t) = -\left(\frac{\pi}{4}\right)^2 \sin \frac{\pi t}{4}\]
Evaluating acceleration at different points helps to identify whether a particle is speeding up or slowing down. If both velocity and acceleration are of the same sign (both positive or both negative), the particle speeds up. If they are opposite, the particle slows down.
For our function, the acceleration is derived as follows:\[a(t) = v'(t) = \frac{d}{dt} \left(\frac{\pi}{4} \cos \frac{\pi t}{4}\right)\]This results in:\[a(t) = -\left(\frac{\pi}{4}\right)^2 \sin \frac{\pi t}{4}\]
Evaluating acceleration at different points helps to identify whether a particle is speeding up or slowing down. If both velocity and acceleration are of the same sign (both positive or both negative), the particle speeds up. If they are opposite, the particle slows down.
Position Function Dynamics
The position function, denoted as \(s(t)\), represents the location of a particle along a coordinate line at any given time \(t\). In our exercise, the position is given by the function \(s(t) = \sin \frac{\pi t}{4}\). This periodic function provides oscillating values that allow the particle to move forward and backward along the line.
To determine the position of the particle at specific times, the position function is evaluated directly at those times. For instance:- At \( t = 1 \), \( s(t) \approx 0.71 \)
- At \( t = 2 \), \( s(t) = 1.00 \)
Each value gives the exact position in meters, showing the particle's path at incremental time periods.
Understanding the position function is crucial, as it sets the foundation for calculating other quantities like velocity and acceleration. This function shows the exact path and span of the motion, allowing for comprehensive analyses of how the particle traverses over time.
To determine the position of the particle at specific times, the position function is evaluated directly at those times. For instance:- At \( t = 1 \), \( s(t) \approx 0.71 \)
- At \( t = 2 \), \( s(t) = 1.00 \)
Each value gives the exact position in meters, showing the particle's path at incremental time periods.
Understanding the position function is crucial, as it sets the foundation for calculating other quantities like velocity and acceleration. This function shows the exact path and span of the motion, allowing for comprehensive analyses of how the particle traverses over time.
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