Problem 15 - 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? $$f(t)=2 t^{3}-21 t^{2}+60 t ; 0 \leq t \leq 6$$

Step-by-Step Solution

Answer

Question: Based on the given information, determine the times when the object is stationary, moving right, moving left, the velocity and acceleration at \(t=1\), the acceleration when the velocity is zero, and the intervals when the speed is increasing. Answer: 1. The object is stationary when \(v(t)=0\). Solve the quadratic equation \(6t^2 - 42t + 60=0\) to find the times when the object is stationary. 2. The object is moving to the right when the velocity function is positive, and moving to the left when it's negative. Use the roots of the equation from part 1 to determine the intervals for each case. 3. The velocity at \(t=1\) is 24 feet/sec, and the acceleration at \(t=1\) is -30 feet/sec^2. 4. Calculate the acceleration for the times when the object is stationary (from part 1) using the acceleration function \(a(t)=12t - 42\). 5. Speed is increasing when the object's velocity and acceleration have the same sign. Determine the intervals for t where both the velocity function (\(v(t)=6t^2 - 42t + 60\)) and the acceleration function (\(a(t)=12t - 42\)) have the same sign (both positive or both negative).

Step 1: Plot the position function for the given range

Since the function \(f(t)=2t^3-21t^2+60t\) is provided and the range of t is from 0 to 6, plot the function within this range using a graphing tool or calculator. This will represent the object’s position with respect to time. #b. Find and graph the velocity function and determine times when the object is stationary, moving to the right, and moving to the left#

Step 2: Find the first derivative of the position function

To find the velocity function, calculate the first derivative of the position function (\(f(t)\)) with respect to time (t): \(v(t) = \frac{d}{dt}f(t)= \frac{d}{dt}(2t^3-21t^2+60t)\) Applying the power rule: \(v(t)= (3 \cdot 2t^2) - (2 \cdot 21 t) + 60\) \(v(t)=6t^2 - 42t + 60\) This is the velocity function.

Step 3: Plot the velocity function for the given range

Plot the velocity function with time t ranging from 0 to 6, using a graphing tool or calculator.

Step 4: Determine the times when the object is stationary, moving right, and moving left

An object is stationary when its velocity is equal to zero, so find all values of t where \(v(t)=0\). Solve the quadratic equation \(6t^2 - 42t + 60=0\). When the velocity function is positive, the object will be moving to the right, and when it's negative, it will be moving to the left. Using the roots of the equation, we can determine the intervals for each case. #c. Determine the velocity and acceleration of the object at \(t=1\)#

Step 5: Find the velocity at \(t=1\)

To find the velocity at 1 second, plug \(t=1\) into the velocity function we found in step 2: \(v(1)=6(1)^2 - 42(1) + 60 = 6 - 42 + 60 = 24\) The velocity at \(t=1\) is 24 feet/sec.

Step 6: Find the acceleration function

To find the acceleration function, calculate the first derivative of the velocity function (v(t)) with respect to time (t): \(a(t) = \frac{d}{dt}v(t)= \frac{d}{dt}(6t^2-42t+60)\) Applying the power rule: \(a(t)= (2 \cdot 6t) - (1 \cdot 42) \) \(a(t)=12t - 42\) This is the acceleration function.

Step 7: Find the acceleration at \(t=1\)

To find the acceleration at 1 second, plug \(t=1\) into the acceleration function we found in step 6: \(a(1)=12(1)-42=-30\) The acceleration at \(t=1\) is -30 feet/sec^2. #d. Determine the acceleration of the object when its velocity is zero#

Step 8: Calculate the acceleration for times when v(t) = 0

Using the values of t you found in step 4, when the object is stationary, calculate the corresponding acceleration values at these times using the acceleration function found in step 6. #e. On what intervals is the speed increasing?#

Step 9: Determine intervals when speed is increasing

Speed is increasing when the object's velocity and acceleration have the same sign. Find the intervals for t where both the velocity function and acceleration function have the same sign (both positive or both negative).

Key Concepts

Position Function

The position function describes the location of an object at any given time as it moves in a specified direction, typically measured in a linear path. Mathematically, it is defined by the function \(s = f(t)\), where \(s\) is the position and \(t\) is time. In our exercise, the position function is given as \(f(t)=2t^3-21t^2+60t\) and is valid for the time interval from 0 to 6 seconds. This function allows us to determine where the object is at any moment within the specified timeframe.

In practice, you can visualize this function by plotting it on a graph, which gives a representation of how the object’s position changes over time. This visualization can help in understanding the object's motion pattern - whether it is speeding up, slowing down, or changing direction. Graphing the position function within the specified range can reveal the critical points where the behavior of the object changes, such as turning points and intervals of increasing or decreasing movement, all of which are essential to fully understand the object's motion.

Acceleration Function

Acceleration gives information about how quickly the velocity of an object changes over time, which is crucial for understanding the dynamics of motion. It is defined as the derivative of the velocity function \(v(t)\) with respect to time \(t\). From our solution, we know that the velocity function is \(v(t) = 6t^2 - 42t + 60\). By taking the derivative of this function, we obtain the acceleration function:
\[a(t) = 12t - 42\]
This means that for any given time \(t\), you can calculate how fast the velocity is increasing or decreasing. Knowing the acceleration is important for determining forces acting on the object, predicting future motion, and even identifying when the object reaches maximum or minimum speeds.

Acceleration can have a profound impact on the movement. When the acceleration and velocity signs match (either both positive or both negative), the object speeds up. When the signs are opposite, the object slows down. This foundational concept helps break down complex equations into simple relations that describe real-world movements.

Derivatives

Derivatives are a fundamental concept in calculus representing the rate of change of a function. In the context of motion, derivatives can help us transition from one motion-related function to another, providing deeper insights into the nature and dynamics of the motion.

The primary use of derivatives in our exercise involves finding the velocity and acceleration functions from the given position function. The first derivative of the position function \(f(t)\) gives us the velocity function \(v(t)\), which expresses how quickly the position changes:

• \(v(t) = \frac{d}{dt}(2t^3-21t^2+60t) = 6t^2 - 42t + 60\)

The second derivative of the position function, or the first derivative of the velocity function, yields the acceleration function \(a(t)\):

• \(a(t) = \frac{d}{dt}(6t^2-42t+60) = 12t - 42\)

Understanding derivatives allows us to make predictions about the object’s behavior and provides critical insights into the mechanics of continuous motion.

Object Motion

Object motion encompasses the overall changes in position, velocity, and acceleration as an object moves through space over time. It is the culmination of the concepts of position, velocity, and acceleration that are interconnected through derivatives.

An object is in motion when it continuously changes position. By plotting the position function \(f(t)\) over time, we gain a tangible insight into how the object proceeds.

• When the velocity function \(v(t)\) is plotted, it helps identify specific moments when the object is moving towards the target (moving right), stationary, or retreating (moving left).
• It’s also possible to detail the forces acting on the object by examining the changes in acceleration collected from the acceleration function \(a(t)\).

During analysis, the critical moments often involve examining when an object's velocity is zero, as this indicates a change in direction or a resting state. These insights highlight where motion occurs and help determine intervals where the speed is increasing or decreasing. Such a comprehensive look at object motion is paramount in fields as diverse as physics, engineering, and even computer graphics to predict movement accurately.