Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

An object is moving along a horizontal coordinate line according to the formula \(s=f(t)\), where \(s\), the directed distance from the origin, is in feet and \(t\) is in seconds. In each case, answer the following questions (see Examples 2 and 3\()\). (a) What are \(v(t)\) and \(a(t)\), the velocity and acceleration, at time \(t\) ? (b) When is the object moving to the right? (c) When is it moving to the left? (d) When is its acceleration negative? (e) Draw a schematic diagram that shows the motion of the object. $$ s=t+\frac{4}{t}, t>0 $$

Step-by-Step Solution

Answer

(a) \( v(t) = 1 - \frac{4}{t^2} \), \( a(t) = \frac{8}{t^3} \). (b) Right for \( t > 2 \). (c) Left for \( 0 < t < 2 \). (d) Never negative.

Step 1: Find the Velocity Function

The velocity function, \( v(t) \), is the first derivative of the position function \( s(t) \). Given \( s(t) = t + \frac{4}{t} \), calculate \( v(t) = \frac{d}{dt}\left(t + \frac{4}{t}\right)\). Using the power rule and the quotient rule, we find that: \[ v(t) = 1 - \frac{4}{t^2} \]

Step 2: Find the Acceleration Function

The acceleration function, \( a(t) \), is the derivative of the velocity function \( v(t) \). So, calculate \( a(t) = \frac{d}{dt}(1 - \frac{4}{t^2}) \). Using the power rule again, we find: \[ a(t) = \frac{8}{t^3} \]

Step 3: Determine Motion to the Right

The object moves to the right when its velocity \( v(t) > 0 \). Set the inequality: \[ 1 - \frac{4}{t^2} > 0 \]. Solve for \( t \): \[ 1 > \frac{4}{t^2} \] leads to \[ t^2 > 4 \], thus \( t > 2 \).

Step 4: Determine Motion to the Left

The object moves to the left when its velocity \( v(t) < 0 \). Using the same critical inequality:\[ 1 - \frac{4}{t^2} < 0 \] Solve for \( t \): \[ 1 < \frac{4}{t^2} \] which gives \( t^2 < 4 \), thus, \( 0 < t < 2 \).

Step 5: Determine Negative Acceleration

The acceleration \( a(t) \) is negative when \( a(t) < 0 \). From Step 2, \( a(t) = \frac{8}{t^3} \).Since \( \frac{8}{t^3} > 0 \) for \( t > 0 \), the acceleration is never negative in this domain.

Step 6: Schematic Diagram

The object transitions from moving left to right at \( t = 2 \), with a positive acceleration throughout. Draw a line representing \( t \) increasing on the horizontal axis with marks indicating \( t = 2 \) as a transition point. Indicate a trajectory that moves left for \( t < 2 \) and right for \( t > 2 \). The line should curve upwards as acceleration is positive throughout.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how quantities change. It involves finding the rate at which one variable changes with respect to another. This is incredibly useful when studying motion, as we can determine how quickly an object is moving or accelerating at any point in time. In our exercise, differentiation is used to find two crucial functions: velocity and acceleration.

• The velocity function, noted as \(v(t)\), is the derivative of the position function \(s(t)\). This shows how fast the position is changing over time.
• The acceleration function, \(a(t)\), is the derivative of the velocity function, revealing how fast the velocity changes.

By applying differentiation rules such as the power rule and the quotient rule, we can efficiently calculate the derivatives needed in real-world applications like this one.

Velocity

Velocity is a vector quantity that represents the speed of an object in a specific direction. Calculating velocity involves taking the derivative of the position function with respect to time. In simpler terms, it tells us how fast something is moving and in which direction.

• In this exercise, the velocity function \(v(t) = 1 - \frac{4}{t^2}\) was derived using differentiation methods.
• By analyzing this function, we determine when the object is moving to the right or left by checking when \(v(t)\) is positive or negative, respectively.

Remember, a positive velocity implies movement to the right, while a negative velocity means the object is moving left. Also, if velocity is zero, the object is momentarily stationary before changing its direction.

Acceleration

Acceleration is another vector quantity that indicates how fast the velocity of an object is changing. It can tell us whether an object is speeding up or slowing down, including the direction of that change. Acceleration provides insights into the forces affecting motion
. The exercise illustrates that calculating acceleration involves differentiating the velocity function:

• The derived function for acceleration is \(a(t) = \frac{8}{t^3}\).
• This function helps determine the nature of the motion. Since \(a(t)\) is always positive for \(t > 0\), the acceleration throughout this interval is consistently positive.

This means the object is constantly speeding up in its movement direction. Analyzing acceleration allows us to understand the dynamics of motion on a deeper level.

Motion

Understanding motion involves looking at how an object moves along a path over time. In this context, motion is not just about the position but also includes the velocity and acceleration that affect how the object behaves during its movement.
This exercise provides a comprehensive view of motion:

• The object moves to the left for \(0 < t < 2\), as indicated by a negative velocity.
• At \(t = 2\), the object reaches a pivotal point where it stops momentarily as it changes direction.
• For \(t > 2\), the object moves to the right with increasing speed due to the positive acceleration noted throughout this period.

Visualizing motion with a schematic diagram helps in comprehending these changes. Such diagrams plot time against the movement trajectory, displaying key transitions and the curved path indicative of positive acceleration. This holistic approach demystifies how objects travel within the physical world.