Problem 12 - Calculus Early Transcendentals

Question

Suppose the position of an object moving horizontally after t seconds is given by the following functions \(s=f(t),\) where \(s\) is measured in feet, with \(s>0\) corresponding to positions right of the origin. a. Graph the position function. b. Find and graph the velocity function. When is the object stationary, moving to the right, and moving to the left? c. Determine the velocity and acceleration of the object at \(t=1\). d. Determine the acceleration of the object when its velocity is zero. e. On what intervals is the speed increasing? $$f(t)=-t^{2}+4 t-3 ; 0 \leq t \leq 5$$

Step-by-Step Solution

Answer

Answer: When the velocity of the object is zero, its acceleration is -2, and the direction changes from right to left.

Step 1: Graph the position function

To graph the position function \(f(t)=-t^2+4t-3\), we can plot the function on a graphing calculator or online graphing tool like Desmos or GeoGebra. The given interval is \(0 \leq t \leq 5\).

Step 2: Find and graph the velocity function

The velocity function is the derivative of the position function \(f(t)\). So, we differentiate \(f(t)\) with respect to \(t\), obtaining: $$ f'(t) = \frac{d}{dt}(-t^2 + 4t - 3) = -2t + 4 $$To graph the velocity function, we can plot this function on a graphing calculator or online graphing tool like Desmos or GeoGebra.

Step 3: Determine the stationary points and movement direction

To find when the object is stationary, we look for when the velocity function is equal to zero: $$ -2t + 4 = 0 $$Solving for \(t\), we get \(t = 2\). To determine the movement direction of the object, analyze the sign of the velocity function: - If \(f'(t) > 0\), the object is moving to the right - If \(f'(t) < 0\), the object is moving to the left The object moves to the right when \(0

Step 4: Determine the velocity and acceleration at \(t = 1\)

Evaluate the velocity (\(f'(t)\)) and acceleration (\(f''(t)\)) functions at \(t = 1\). First, calculate the acceleration function by taking the derivative of the velocity function: $$ f''(t) = \frac{d}{dt}(-2t + 4) = -2 $$Now, evaluate \(f'(1)\) and \(f''(1)\): $$ f'(1) = -2(1) + 4 = 2 \\ f''(1) = -2 $$At \(t = 1\), the velocity and acceleration of the object are: $$ \text{velocity} = 2 \\ \text{acceleration} = -2 $$

Step 5: Determine the acceleration when its velocity is zero

From step 3, the velocity is zero at \(t = 2\). Calculate the acceleration at \(t = 2\): $$ f''(2) = -2 $$The acceleration when its velocity is zero is \(-2\).

Step 6: Determine the intervals when the speed is increasing

The speed of an object is increasing when its velocity and acceleration have the same sign. - If \(f'(t) > 0\) and \(f''(t) > 0\), the speed is increasing. - If \(f'(t) < 0\) and \(f''(t) < 0\), the speed is increasing. The acceleration function is constant: \(f''(t) = -2\) (which is negative). Therefore, the speed is increasing when the velocity function is also negative. From Step 3, we found that the velocity was negative when \(2 < t < 5\). So, the speed is increasing on the interval \((2, 5)\).

Key Concepts

Position Function

The position function signifies the position of an object with respect to time. In our problem, it's expressed as \( f(t) = -t^2 + 4t - 3 \), where \( t \) is the time in seconds and \( s \) is the position in feet. This quadratic function represents how the object's position changes over time as it moves horizontally. To visualize it, we should plot this function over the interval \( 0 \leq t \leq 5 \).
The shape of the graph is a parabola opening downwards because of the negative coefficient of \( t^2 \). At the vertex, the position is maximized, and on either side of the vertex, the position decreases. This means that the object changes direction and starts returning back towards its starting point.
Graphing this function gives students an understanding of how the position changes, showing the peak and endpoints clearly. Tools like graphing calculators can help in plotting these kinds of functions.

Velocity Function

The velocity function tells us the rate at which the object's position changes over time. To find it, we differentiate the position function. In calculus terms, this means taking the derivative of \( f(t) \).

• The position function is \( f(t) = -t^2 + 4t - 3 \).
• By differentiating, we get the velocity function \( f'(t) = -2t + 4 \).

This linear function represents the object's speed and direction. Positive values indicate movement to the right, while negative values indicate movement to the left. At \( t = 2 \), the velocity is zero, meaning the object is momentarily stationary before changing direction.
Graphing the velocity function shows when the object speeds up or slows down, and when it switches direction. It's crucial to understand these concepts to predict movements accurately.

Acceleration Function

Acceleration measures how the velocity of the object changes over time. To find it, we differentiate the velocity function. This is the second derivative of the position function.

• The velocity function we found is \( f'(t) = -2t + 4 \).
• Taking its derivative gives us the acceleration function: \( f''(t) = -2 \).

Interestingly, this acceleration is constant at \( -2 \) for any value of \( t \). The negative value indicates that the velocity is decreasing over time.
At \( t = 1 \), the velocity is \( f'(1) = 2 \) and the acceleration is \( f''(1) = -2 \). This tells us that although the object is moving to the right initially, it is decelerating. Understanding acceleration is pivotal for grasping changes in motion dynamics effectively.

Graphing

Graphing is a powerful tool to visualize how functions behave over time. For position, velocity, and acceleration, graphing not only displays changes but also relationships and trends.

• Position graphs show curves (like parabolas) that display motion paths.
• Velocity graphs show linear trends that communicate changes in speed and direction.
• Acceleration is constant and depicted as a flat line on its own graph.

When graphing these functions, look for crucial points where velocity is zero, which indicate changes in direction, or peaks in position graphs indicating highest or lowest points in motion.
Using graphing technology simplifies understanding. It connects abstract concepts of calculus to tangible visual aids, making learning intuitive and engaging.

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Key Concepts

Position Function

The position function signifies the position of an object with respect to time. In our problem, it's expressed as \( f(t) = -t^2 + 4t - 3 \), where \( t \) is the time in seconds and \( s \) is the position in feet. This quadratic function represents how the object's position changes over time as it moves horizontally. To visualize it, we should plot this function over the interval \( 0 \leq t \leq 5 \).
The shape of the graph is a parabola opening downwards because of the negative coefficient of \( t^2 \). At the vertex, the position is maximized, and on either side of the vertex, the position decreases. This means that the object changes direction and starts returning back towards its starting point.
Graphing this function gives students an understanding of how the position changes, showing the peak and endpoints clearly. Tools like graphing calculators can help in plotting these kinds of functions.

Velocity Function

The velocity function tells us the rate at which the object's position changes over time. To find it, we differentiate the position function. In calculus terms, this means taking the derivative of \( f(t) \).

• The position function is \( f(t) = -t^2 + 4t - 3 \).
• By differentiating, we get the velocity function \( f'(t) = -2t + 4 \).

This linear function represents the object's speed and direction. Positive values indicate movement to the right, while negative values indicate movement to the left. At \( t = 2 \), the velocity is zero, meaning the object is momentarily stationary before changing direction.
Graphing the velocity function shows when the object speeds up or slows down, and when it switches direction. It's crucial to understand these concepts to predict movements accurately.

Acceleration Function

Acceleration measures how the velocity of the object changes over time. To find it, we differentiate the velocity function. This is the second derivative of the position function.

• The velocity function we found is \( f'(t) = -2t + 4 \).
• Taking its derivative gives us the acceleration function: \( f''(t) = -2 \).

Interestingly, this acceleration is constant at \( -2 \) for any value of \( t \). The negative value indicates that the velocity is decreasing over time.
At \( t = 1 \), the velocity is \( f'(1) = 2 \) and the acceleration is \( f''(1) = -2 \). This tells us that although the object is moving to the right initially, it is decelerating. Understanding acceleration is pivotal for grasping changes in motion dynamics effectively.

Graphing

Graphing is a powerful tool to visualize how functions behave over time. For position, velocity, and acceleration, graphing not only displays changes but also relationships and trends.

• Position graphs show curves (like parabolas) that display motion paths.
• Velocity graphs show linear trends that communicate changes in speed and direction.
• Acceleration is constant and depicted as a flat line on its own graph.

When graphing these functions, look for crucial points where velocity is zero, which indicate changes in direction, or peaks in position graphs indicating highest or lowest points in motion.
Using graphing technology simplifies understanding. It connects abstract concepts of calculus to tangible visual aids, making learning intuitive and engaging.