Problem 13 - 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? $$f(t)=2 t^{2}-9 t+12 ; 0 \leq t \leq 3$$

Step-by-Step Solution

Answer

Answer: The position function is \(f(t) = 2t^2 - 9t + 12\), and its graph is a parabola. The velocity function is \(v(t) = 4t - 9\), and it can be graphed on the interval \([0, 3]\). The object is: - stationary at \(t = \frac{9}{4}\) - moving to the right when \(t > \frac{9}{4}\) - moving to the left when \(t < \frac{9}{4}\) At \(t=1\), the object has a velocity of \(-5\) ft/s and an acceleration of \(4\) ft/s². The acceleration when the velocity is zero is \(4\) ft/s². The speed is increasing on the interval \((\frac{9}{4}, 3]\).

Step 1: a. Graph the position function

To graph the position function, plot \(f(t) = 2t^2 - 9t + 12\) on the interval \([0, 3]\). This is a quadratic function, and it will look like a parabola. Use a graphing calculator or software to draw the graph.

Step 2: b. Find and graph the velocity function

To find the velocity function, calculate the first derivative of the position function: $$v(t) = \frac{d}{dt}(2t^2 - 9t + 12) = 4t - 9.$$ The object is moving to the right when \(v(t) > 0\), to the left when \(v(t) < 0\), and is stationary when \(v(t) = 0\). Let's find when \(v(t)=0\): $$4t - 9 = 0 \Rightarrow t = \frac{9}{4}.$$ Now, test the intervals: - For \(t < \frac{9}{4}\): \(v(t) = 4t - 9 < 0\), the object is moving to the left. - For \(t > \frac{9}{4}\): \(v(t) = 4t - 9 > 0\), the object is moving to the right. - At \(t = \frac{9}{4}\), the object is stationary. Graph the velocity function \(v(t) = 4t - 9\) on the interval \([0, 3]\) using a graphing calculator or software.

Step 3: c. Determine the velocity and acceleration at \(t=1\)

We found the velocity function to be \(v(t) = 4t - 9\). Now, we can find the velocity at \(t=1\): $$v(1) = 4(1) - 9 = -5.$$ To find the acceleration function, take the derivative of the velocity function: $$a(t) = \frac{d}{dt}(4t - 9) = 4.$$ The acceleration does not depend on time. To calculate the acceleration at \(t=1\), we simply plug in \(t=1\): $$a(1) = 4.$$ So, at \(t=1\), the velocity is \(-5\) ft/s and the acceleration is \(4\) ft/s².

Step 4: d. Determine the acceleration when the velocity is zero

Since the acceleration does not depend on time, the acceleration when the velocity is zero remains the same at \(4\) ft/s².

Step 5: e. On what intervals is the speed increasing?

Speed is increasing when the velocity and acceleration have the same sign. Since the acceleration is always positive (\(4\) ft/s²), we need to find the interval where the velocity is also positive. As we found earlier, the object is moving to the right (positive velocity) when \(t > \frac{9}{4}\). Therefore, the speed is increasing on the interval \((\frac{9}{4}, 3]\).

Key Concepts

Position Function

The position function, denoted as \( s = f(t) \), describes the position of an object over time. It's a mathematical representation of where the object is located along a horizontal path at any given time \( t \). In this exercise, the position function is given as \( f(t) = 2t^2 - 9t + 12 \). This is a quadratic equation, which means its graph will be a parabola.

• The parabola's shape can help visualize the motion of the object. It gives insight into how the object moves over a period.
• For \( 0 \leq t \leq 3 \), the function tells us the position of the object ranging right from the origin, as \( s > 0 \) means it is moving in the positive direction.

To graph this function, you plot points by substituting different values of \( t \) within the specified range \([0, 3]\). Seeing the curve helps in analyzing changes in motion, such as when or if it reverses direction.

Velocity Function

The velocity function indicates how fast an object's position changes over time. It is essentially the derivative of the position function. In calculus, the derivative represents the rate of change. For this exercise, the velocity function is\[v(t) = \frac{d}{dt}(2t^2 - 9t + 12) = 4t - 9.\]This linear function illustrates how velocity changes over time, indicating both direction and speed.

• A positive velocity means the object moves to the right, while a negative velocity means it moves to the left.
• Finding when \( v(t) = 0 \) (at \( t = \frac{9}{4} \)) helps determine when the object is stationary.

This is essential for understanding motion, as it breaks down into intervals:

• For \( t < \frac{9}{4} \), \( v(t) < 0 \), meaning the object moves left.
• For \( t > \frac{9}{4} \), \( v(t) > 0 \), so the object moves right.

Graphing \( v(t) \) provides a straightforward visual of these changes.

Acceleration

Acceleration describes how the velocity of an object changes over time. It is found by taking the derivative of the velocity function. Here, the acceleration function is\[a(t) = \frac{d}{dt}(4t - 9) = 4.\]This result means that acceleration is constant at \( 4 \) ft/s² for all times \( t \), regardless of the object's current velocity or position.

• A constant acceleration indicates uniform change in velocity.
• Since the acceleration is always positive, it suggests a consistently increasing velocity when the object moves to the right \((t > \frac{9}{4})\).

Understanding acceleration is crucial because:

• It impacts how quickly an object can change direction or speed.
• Determining the acceleration when velocity is zero helps analyze momentary pauses in motion or changes in direction.

Even though acceleration is constant, it plays a huge role in understanding the dynamics of the object's motion over time.