Problem 25 - 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. $$ s=t^{3}-9 t^{2}+24 t $$

Step-by-Step Solution

Answer

(a) \(v(t) = 3t^2 - 18t + 24\), \(a(t) = 6t - 18\). (b) Right for \(t < 2\) or \(t > 4\). (c) Left for \(2 < t < 4\). (d) Negative for \(t < 3\).

Step 1: Find the Velocity Function

The velocity function, \(v(t)\), is the derivative of \(s(t)\) with respect to \(t\). We start by differentiating the position function \(s(t) = t^3 - 9t^2 + 24t\). The derivative is given by: \[ v(t) = \frac{d}{dt}(t^3 - 9t^2 + 24t) = 3t^2 - 18t + 24. \]

Step 2: Find the Acceleration Function

The acceleration function, \(a(t)\), is the derivative of the velocity function. We differentiate \(v(t) = 3t^2 - 18t + 24\) with respect to \(t\): \[ a(t) = \frac{d}{dt}(3t^2 - 18t + 24) = 6t - 18. \]

Step 3: Determine When the Object is Moving to the Right

An object moves to the right when its velocity is positive, \(v(t) > 0\). Set the velocity function \(v(t) = 3t^2 - 18t + 24\) greater than zero and solve for \(t\): \[ 3t^2 - 18t + 24 > 0. \] Factor out the quadratic expression: \[ 3(t^2 - 6t + 8) > 0. \] Recognizing it as a quadratic we can factor, we find \(t^2 - 6t + 8 = (t-2)(t-4)\). Hence, \[ 3(t-2)(t-4) > 0. \] Through sign analysis, we find the object moves right for \(t < 2\) or \(t > 4\).

Step 4: Determine When the Object is Moving to the Left

An object moves to the left when its velocity is negative, \(v(t) < 0\). From the factored form, \[ 3(t-2)(t-4) < 0. \] The object moves left between \(t = 2\) and \(t = 4\), obtained through sign analysis.

Step 5: Determine When the Acceleration is Negative

Acceleration is negative when \(a(t) < 0\). Solve \(6t - 18 < 0\): \[ t < 3. \] Thus, the acceleration is negative for \(t < 3\).

Step 6: Draw the Schematic Diagram of Motion

We plot the critical points and signs of \(v(t)\) and \(a(t)\): 1. For \(0 < t < 2\), \(v(t) > 0\) (moving right), \(a(t) < 0\). 2. For \(2 < t < 4\), \(v(t) < 0\) (moving left). 3. For \(t > 4\), \(v(t) > 0\) (moving right). This can be represented on a number line with arrows indicating the direction of movement.

Key Concepts

Velocity Function

The velocity of an object in motion is crucial as it tells us how fast and in which direction the object is moving at any given time. In calculus, the velocity function, often denoted as \( v(t) \), is derived from the position function \( s(t) \). Simply put, the velocity function is the first derivative of the position function with respect to time \( t \).

• Take for example the function \( s(t) = t^3 - 9t^2 + 24t \).
• To find the velocity \( v(t) \), we differentiate it: \[ v(t) = \frac{d}{dt}(t^3 - 9t^2 + 24t) = 3t^2 - 18t + 24. \]

This resultant equation is our tool to determine crucial information about the object's motion.
For instance, by setting \( v(t) \) greater than zero \(( v(t) > 0)\), we can find out when the object is moving to the right, which happens when the velocity is positive. Conversely, setting \( v(t) \) less than zero \(( v(t) < 0)\) shows when the object moves to the left, indicating negative velocity.

Acceleration Function

Acceleration represents the rate of change of velocity with respect to time, and it helps us understand how quickly the velocity is changing. The acceleration function, \( a(t) \), is the derivative of the velocity function. This means it's the second derivative of the position function \( s(t) \).
Continuing with our example where \( v(t) = 3t^2 - 18t + 24 \), to find acceleration, we differentiate \( v(t) \) as follows:

• \[ a(t) = \frac{d}{dt}(3t^2 - 18t + 24) = 6t - 18. \]

The acceleration function helps identify the periods when the acceleration is negative or positive.
By setting the function \( a(t) \) to be less than zero \(( a(t) < 0)\), you can determine intervals of negative acceleration, meaning the object is slowing down during those times.

Quadratic Inequality

Quadratic inequalities are statements involving a quadratic expression that is either greater than or less than zero. Solving these inequalities requires a combination of factoring, testing intervals, or using methods like sign analysis. They are essential in determining specific intervals in motion problems.
In our exercise, the function \( v(t) = 3t^2 - 18t + 24 \) becomes an inequality to find when the object is moving. We expressed it as:

• \[ 3(t^2 - 6t + 8) > 0 \]
• Which factors to \((t-2)(t-4)\).

We then use these factors to determine where the function is positive or negative.
By testing intervals around the roots \( t = 2 \) and \( t = 4 \), we can establish when the object moves right, \( t < 2 \) or \( t > 4 \), and when it moves left, \( 2 < t < 4 \). These steps elucidate the dynamics of the quadratic inequality in analyzing motion.