Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 23-28, 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. $$ s=t^{2}+\frac{16}{t}, t>0 $$

Step-by-Step Solution

Answer

(a) \(v(t) = 2t - \frac{16}{t^2}\), \(a(t) = 2 + \frac{32}{t^3}\); (b) \(t > 2\); (c) \(0 < t < 2\); (d) Never.

Step 1: Find the Velocity Function

To find the velocity function, we need to take the derivative of the position function \(s(t) = t^2 + \frac{16}{t}\). The derivative of the first term \(t^2\) is \(2t\), and the derivative of \(\frac{16}{t}\) is \(-\frac{16}{t^2}\). Therefore, the velocity function is \(v(t) = 2t - \frac{16}{t^2}\).

Step 2: Find the Acceleration Function

The acceleration function is the derivative of the velocity function. Taking the derivative of \(v(t) = 2t - \frac{16}{t^2}\), we get \(a(t) = 2 + \frac{32}{t^3}\). Therefore, the acceleration function is \(a(t) = 2 + \frac{32}{t^3}\).

Step 3: Determine When the Object is Moving Right

The object moves to the right when the velocity is positive \(v(t) > 0\). Solve \(2t - \frac{16}{t^2} > 0\). Simplifying, we multiply by \(t^2\) to get \(2t^3 - 16 > 0\). Thus, \(t^3 > 8\), and \(t > 2\). So, the object moves to the right when \(t > 2\).

Step 4: Determine When the Object is Moving Left

The object moves to the left when the velocity is negative \(v(t) < 0\). Solve \(2t - \frac{16}{t^2} < 0\). With similar calculations from the previous step, \(t^3 < 8\), and \(t < 2\). So, the object moves to the left when \(0 < t < 2\).

Step 5: Determine When the Acceleration is Negative

Acceleration is negative when \(a(t) < 0\). Solve \(2 + \frac{32}{t^3} < 0\). Isolate \(\frac{32}{t^3} < -2\) or \(t^3 < -16\). However, since \(t > 0\), there is no real value of \(t\) that satisfies this inequality, meaning the acceleration is never negative in the domain of \(t > 0\).

Step 6: Draw a Schematic Diagram

The schematic diagram should depict the object moving from left to right starting at the origin, turning back to the left until \(t = 2\), and then moving to the right when \(t > 2\). An important feature to highlight is at \(t = 2\) where the velocity changes sign, indicating a change in direction from left to right.

Key Concepts

Velocity Function

In calculus, the velocity function is a crucial concept when studying motion. It tells us the rate at which an object's position changes over time. If we know the position function, denoted as \( s(t) \), we can find the velocity by taking its derivative. Essentially, the velocity function \( v(t) \) gives us the first derivative of the position function with respect to time.

• For example, if the position is given as \( s(t) = t^2 + \frac{16}{t} \), then the velocity is the derivative of this function, which comes out to \( v(t) = 2t - \frac{16}{t^2} \).

This tells us how fast and in which direction the object is moving at any time \( t \). When \( v(t) > 0 \), the object moves to the right, and when \( v(t) < 0 \), it moves to the left.
Knowing how to derive these functions helps you understand the object's real-time motion behavior.

Acceleration Function

The acceleration function describes how an object's velocity changes over time. It’s the derivative of the velocity function, and represents the rate of change of velocity. This is helpful for understanding whether the object is speeding up or slowing down.

• Given the velocity function \( v(t) = 2t - \frac{16}{t^2} \), the acceleration function is calculated by finding the derivative of this, resulting in \( a(t) = 2 + \frac{32}{t^3} \).

The acceleration function tells us when and how quickly an object’s velocity is changing:
- If \( a(t) > 0 \), velocity is increasing.
- If \( a(t) < 0 \), velocity is decreasing.
In this exercise, the calculation shows that acceleration is never negative for \( t > 0 \). This means the object does not slow down in this time frame.

Derivative Applications

The concept of derivatives plays a central role in calculus, especially when it comes to understanding motion. By taking derivatives, we’re not only calculating instantaneous rates of change, but also deciphering the behavior of moving objects.

• The first derivative is used to find velocity from the position function. This tells us the object's speed and direction of movement.
• The second derivative gives us the acceleration, indicating changes in velocity.

In practical terms, derivatives are tools that predict how objects in motion behave at any given point in time. They’re pivotal in determining when speed increases or decreases, as well as identifying critical points of change, such as when an object switches from moving left to right or vice versa.

Motion Along a Line

Motion along a line is a simple yet fundamental concept in physics and calculus that describes how an object moves on a single straight path. To fully describe this motion, we analyze various properties, such as position, velocity, and acceleration.

• The position function \( s(t) \) provides the location of the object at a time \( t \).
• The velocity function \( v(t) \), derived from the position function, indicates the speed and direction.
• The acceleration function \( a(t) \), the derivative of velocity, shows how velocity changes.

Understanding motion along a line involves finding when the object moves in different directions (to the right or left) and identifying any changes in speed or reversals in direction. In this particular exercise, we find that:- The object moves to the right when \( t > 2 \).- It moves to the left when \( 0 < t < 2 \).- There are no intervals where the acceleration is negative for \( t > 0 \).
This comprehensive analysis is crucial for designing models that predict actual physical movement.