Problem 35 - University Calculus: Early Transcendentals

Question

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t .\) Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$s=t^{3}-6 t^{2}+7 t, \quad 0 \leq t \leq 4$$

Step-by-Step Solution

Answer

The object is at rest at \( t \approx 1.55 \) and \( t \approx 2.78 \), changes direction there, speeds up when \( v \) and \( a \) share signs, slows otherwise, is fastest or slowest at critical points, and farthest at max \( |s(t)| \).

Step 1: Calculate the Position Function

The given position function is \( s = f(t) = t^3 - 6t^2 + 7t \). We will use this function to graph the position versus time for the interval \( 0 \leq t \leq 4 \).

Step 2: Derive the Velocity Function

To find the velocity function, take the first derivative of the position function with respect to \( t \):\[ v(t) = f'(t) = \frac{d}{dt}(t^3 - 6t^2 + 7t) = 3t^2 - 12t + 7 \].

Step 3: Derive the Acceleration Function

To find the acceleration function, take the second derivative of the position function with respect to \( t \):\[ a(t) = f''(t) = \frac{d^2}{dt^2}(t^3 - 6t^2 + 7t) = 6t - 12 \].

Step 4: Determine when the object is at rest

The object is at rest when its velocity is zero. Set the velocity function equal to zero and solve for \( t \):\[ 3t^2 - 12t + 7 = 0 \].Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -12 \), and \( c = 7 \), we get:\[ t = \frac{12 \pm \sqrt{144 - 84}}{6} = \frac{12 \pm \sqrt{60}}{6} \approx 1.55, 2.78 \]. Thus, the object is momentarily at rest at \( t \approx 1.55 \) and \( t \approx 2.78 \).

Step 5: Determine Change of Direction

An object changes direction at points where the velocity changes sign. The roots that solve the velocity function in Step 4, \( t \approx 1.55 \) and \( t \approx 2.78 \), are candidates for direction change. Check the signs of \( v(t) \) around these points to confirm direction changes.

Step 6: Determine Speeding Up or Slowing Down

The object speeds up when velocity and acceleration have the same signs, and slows down when they have opposite signs. Compute \( v(t) \) and \( a(t) \) over the interval \( 0 \leq t \leq 4 \) to see where they are both positive or both negative.

Step 7: Determine Fastest and Slowest Speeds

Velocity magnitude \( |v(t)| \) indicates speed. Calculate critical points within the interval using the derivative \( v'(t) \) or by evaluating key points \( t = 0, 1.55, 2.78, 4 \) to find highest and lowest values of \( |v(t)| \).

Step 8: Determine Farthest from Origin

The object is farthest from the origin at maximum value of \( |s(t)| \). Evaluate \( |s(t)| \) at endpoints and critical points \( t = 0, 1.55, 2.78, 4 \) and compare to determine when the distance is maximized.

Key Concepts

Velocity Function

The velocity function is key for understanding how fast an object moves and in which direction. It is the first derivative of the position function, which means it tells us how the position changes over time. For the given exercise, the velocity function is derived from the position function, resulting in:

• \( v(t) = 3t^2 - 12t + 7 \)

This equation helps us pinpoint when the object is stationary or changing direction. By setting the velocity equal to zero, we find the times when the object is at rest, which are \( t \approx 1.55 \) and \( t \approx 2.78 \).
By testing the intervals around these points, we can see if the velocity function changes its sign, indicating a change in direction. If \( v(t) \) transitions from positive to negative or vice versa, it means the object is switching its path.

Acceleration Function

Acceleration is the rate at which velocity changes over time. It's obtained by taking the derivative of the velocity function, leading to our understanding of how an object's speed increases or decreases. In this example, the acceleration function is:

• \( a(t) = 6t - 12 \)

The sign of this function tells us whether the object is speeding up or slowing down:

• When both \( v(t) \) and \( a(t) \) are positive, the object speeds up.
• If they are both negative, it also speeds up but in the opposite direction.
• When the signs differ, the object is slowing down.

By evaluating these functions over the interval \( 0 \leq t \leq 4 \), we learn more about the object's motion dynamics.

Kinematics

Kinematics deals with the motion of objects, characterized by parameters like position, velocity, and acceleration without getting into the forces causing the motion. Each function—position, velocity, and acceleration—gives us a piece of the puzzle:

• The position function describes where the object is on the axis.
• The velocity function shows how quickly and in what direction it moves.
• The acceleration function indicates changes in that velocity, telling us whether the pace of movement is increasing or lessening.

By graphing and analyzing these functions, one can map the object's motion through time, capturing moments when it rests, the sectors where it speeds up, or those in which it slows down.

Motion Analysis

Conducting a motion analysis means observing everything about how an object moves through time. It involves checking different aspects, such as:

• Rest points: Where the velocity function equals zero.
• Direction changes: Identified by sign changes in the velocity function.
• Speed alterations: Determined by how the velocity function and acceleration function interact.
• Critical movements: Finding maximum or minimum values in these functions can expose times of fastest, slowest, or farthest distances from a point.

By carefully observing these aspects, students can get a detailed picture of the object's behavior with respect to time, aiding comprehension of the object's full motion pattern.