Problem 12
Question
In Exercises, \(s(t)\) is the position function of a body moving along a coordinate line, where \(t \geq 0\), and \(s(t)\) is measured in feet and \(t\) in seconds. (a) Determine the times \((s)\) and the position \((s)\) when the body is stationary. (b) When is the body moving in the positive direction? In the negative direction? (c) Sketch a schematic showing the position of the body at any time t. $$ s(t)=\frac{1}{3} t^{3}-\frac{3}{2} t^{2}+1 $$
Step-by-Step Solution
Verified Answer
The body is stationary at times \(t = 0\) s and \(t = 3\) s with positions s(0) = 1 ft and s(3) = -1 ft. It moves in the positive direction for \(t < 0\) and \(t > 3\) and in the negative direction for \(0 < t < 3\). The sketch of the position vs. time graph is a cubic function representing these behaviors.
1Step 1: Find the first derivative of the position function s(t)
To find when the body is stationary and in which direction it is moving, we need the first derivative of the position function, which represents the velocity of the body.
Find the first derivative of s(t):
\(s(t)=\frac{1}{3}t^3 - \frac{3}{2}t^2 + 1\)
Using the power rule:
\(s'(t)=3(\frac{1}{3}t^2) - 2(\frac{3}{2}t^1) + 0\)
Simplifying:
\(s'(t)=t^2 - 3t\)
2Step 2: Determine when the body is stationary
The body is stationary when its velocity is zero. So we need to find when \(s'(t) = 0\).
Find the roots of the equation \(s'(t) = 0\):
\(t^2 - 3t = 0\)
\(t(t - 3) = 0\)
The roots are:
\(t = 0\) and \(t = 3\)
Now, we must plug in these values of t into the original position function, s(t), to find the position when the body is stationary.
For \(t = 0\):
\(s(0)=\frac{1}{3}(0^3)-\frac{3}{2}(0^2)+1 = 1\)
For \(t = 3\):
\(s(3)=\frac{1}{3}(3^3)-\frac{3}{2}(3^2)+1 = - 1\)
So, the body is stationary at times \(t = 0\) s and \(t = 3\) s, with positions s(0) = 1 ft and s(3) = -1 ft.
3Step 3: Determine when the body is moving in positive and negative directions.
The body is moving in a positive direction when \(s'(t) > 0\), and in a negative direction when \(s'(t) < 0\).
The first derivative \(s'(t) = t^2 - 3t\).
We already found the roots t = 0 and t = 3 when solving for when the body is stationary. Now, we must determine the signs of the first derivative in the intervals created by these roots.
Solve for \(s'(t)\) for different intervals.
- For \(t < 0\):
\(s'(-1) = (-1)^2 - 3(-1) = 1 + 3 = 4 > 0\)
Therefore, the body is moving in the positive direction for \(t < 0\).
- For \(0 < t < 3\):
\(s'(1) = (1)^2 - 3(1) = 1 - 3 = -2 < 0\)
Therefore, the body is moving in the negative direction for \(0 < t < 3\).
- For \(t > 3\):
\(s'(4) = (4)^2 - 3(4) = 16 - 12 = 4 > 0\)
Therefore, the body is moving in the positive direction for \(t > 3\).
4Step 4: Sketch the position vs time graph
Now, using the information we've collected, we can sketch a position vs time graph of the body:
1. The body starts at the position `s(0) = 1 ft`.
2. It is stationary at times \(t = 0\) and \(t = 3\) and positions `s(0) = 1 ft` and `s(3) = -1 ft`.
3. The body moves in a positive direction for \(t < 0\) and \(t > 3 \).
4. The body moves in a negative direction for \(0 < t < 3\).
With this information, you can sketch a cubic function showing that the body is initially in the positive direction, then it becomes stationary at the time \(t = 0\), then moves in the negative direction until the time \(t = 3\), becomes stationary again, and finally moves in the positive direction.
Key Concepts
Position functionVelocityStationary pointsDerivativeGraph sketching
Position function
The position function, often represented as \( s(t) \), describes the location of a moving object along a coordinate line as a function of time \( t \). In this exercise, the position function given is \( s(t) = \frac{1}{3}t^3 - \frac{3}{2}t^2 + 1 \), where \( s(t) \) is measured in feet and \( t \) in seconds. This function is crucial because it provides the formula needed to determine where the object is at any given time.
Understanding how the position changes over time helps predict the nature of the object's motion, such as when it is moving forward, backward, or standing still.
Understanding how the position changes over time helps predict the nature of the object's motion, such as when it is moving forward, backward, or standing still.
Velocity
Velocity is a measure of how fast an object is moving and the direction of its motion. It is calculated by finding the derivative of the position function, \( s(t) \). The first derivative, \( s'(t) \), indicates the velocity of the object at time \( t \).
In our exercise, the velocity is represented by \( s'(t) = t^2 - 3t \). This expression allows us to determine at what times the object is moving, the direction it is moving, and when it is stationary. For example:
In our exercise, the velocity is represented by \( s'(t) = t^2 - 3t \). This expression allows us to determine at what times the object is moving, the direction it is moving, and when it is stationary. For example:
- If \( s'(t) > 0 \), the body moves in a positive direction.
- If \( s'(t) < 0 \), the body moves in a negative direction.
- If \( s'(t) = 0 \), the body is stationary.
Stationary points
Stationary points occur when an object is not moving—that means its velocity is zero. Hence, they are often found by solving \( s'(t) = 0 \). For this exercise, the roots of the equation \( t^2 - 3t = 0 \) are \( t = 0 \) and \( t = 3 \).
Substituting these \( t \)-values back into the position function \( s(t) \) gives the position of the object at these stationary points. We find that:
Substituting these \( t \)-values back into the position function \( s(t) \) gives the position of the object at these stationary points. We find that:
- At \( t = 0 \), the position is \( s(0) = 1 \) ft.
- At \( t = 3 \), the position is \( s(3) = -1 \) ft.
Derivative
The derivative is a fundamental tool in calculus that tells us how a function changes at any given point and is essential for understanding motion dynamics, such as velocity and acceleration. In this context, deriving the position function \( s(t) \) as \( s'(t) = t^2 - 3t \) reveals the velocity of the object.
The derivative facilitates investigating how fast the position is changing over time, which assists in determining direction changes and intervals of motion. It allows for precise calculations of when an object speeds up, slows down, or halts.
The derivative facilitates investigating how fast the position is changing over time, which assists in determining direction changes and intervals of motion. It allows for precise calculations of when an object speeds up, slows down, or halts.
Graph sketching
Graph sketching visually represents the changes in the position of an object over time, using information derived from the position and velocity functions. For this exercise, sketching a graph requires:
- Plotting the stationary points at \( t = 0 \) and \( t = 3 \).
- Showing where the object moves positively \((\) on \( t < 0 \) and \( t > 3 \)).
- Indicating negative movement for \( 0 < t < 3 \).
Other exercises in this chapter
Problem 12
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