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

Step-by-Step Solution

Answer

(a) \(v(t) = 3t^2 - 18t + 24\), \(a(t) = 6t - 18\). (b) Right: \(t < 2\) or \(t > 4\). (c) Left: \(2 < t < 4\). (d) Negative: \(t < 3\). (e) Diagram shows motion over time.

Step 1: Determine Velocity Function

The velocity function is the first derivative of the position function \(s(t)\). Given \(s(t) = t^3 - 9t^2 + 24t\), find the derivative \(v(t) = \frac{ds}{dt}\). Use the power rule for differentiation: \(\frac{d}{dt}[t^n] = nt^{n-1}\).\[v(t) = \frac{d}{dt}(t^3) - \frac{d}{dt}(9t^2) + \frac{d}{dt}(24t) = 3t^2 - 18t + 24\]

Step 2: Determine Acceleration Function

The acceleration function is the derivative of the velocity function \(v(t)\). Again, use the power rule to differentiate \(v(t) = 3t^2 - 18t + 24\).\[a(t) = \frac{dv}{dt} = 6t - 18\]

Step 3: Determine when Object is Moving Right

The object moves to the right when the velocity \(v(t) > 0\). Solve the inequality \(3t^2 - 18t + 24 > 0\).Factor the quadratic expression: \[3(t^2 - 6t + 8) = 3(t-2)(t-4)\]The solutions of \(3(t-2)(t-4) = 0\) are \(t = 2\) and \(t = 4\). Using a sign test or test values in the intervals, determine that the object moves to the right when \(t < 2\) or \(t > 4\).

Step 4: Determine when Object is Moving Left

The object moves to the left when the velocity \(v(t) < 0\). Using the factored expression from Step 3:The intervals where \(v(t) < 0\) are determined to be \(2 < t < 4\) through sign testing.

Step 5: Determine when Acceleration is Negative

The acceleration is negative when \(a(t) = 6t - 18 < 0\). Solve the inequality:\[6t - 18 < 0\]Add 18 to both sides:\[6t < 18\]Divide by 6:\[t < 3\]Thus, the acceleration is negative for \(t < 3\).

Step 6: Schematic Diagram of Motion

Plot the position function \(s(t) = t^3 - 9t^2 + 24t\), the velocity function \(v(t) = 3t^2 - 18t + 24\), and the acceleration function \(a(t) = 6t - 18\) on a graph. Highlight intervals to show when \(v(t) > 0\) (object moves right) and when \(v(t) < 0\) (object moves left). Mark \(t = 3\) as the point where acceleration changes from negative to positive.

Key Concepts

Velocity

Velocity is a key concept in understanding motion, as it describes how fast an object is moving and in which direction along a line. In calculus, we define velocity as the derivative of the position function with respect to time. This means, if you have a position function given as \(s(t) = t^3 - 9t^2 + 24t\), the velocity function \(v(t)\) is the first derivative of \(s(t)\). Here, we use the power rule in differentiation, which states that the derivative of \(t^n\) is \(nt^{n-1}\). Applying this rule, \(v(t)\) becomes:

• \( \frac{d}{dt}(t^3) = 3t^2 \)
• \( \frac{d}{dt}(-9t^2) = -18t \)
• \( \frac{d}{dt}(24t) = 24 \)

So, the velocity function is \(v(t) = 3t^2 - 18t + 24\). Velocity helps us answer when an object moves forward or backwards, depending on whether \(v(t) > 0\) or \(v(t) < 0\).

Acceleration

Acceleration is all about how the velocity of an object changes over time. It is the derivative of the velocity function. This means it describes the rate of change of the velocity function. For the velocity function identified earlier as \(v(t) = 3t^2 - 18t + 24\), we take another derivative to find acceleration \(a(t)\). Again, use the power rule:

• \( \frac{d}{dt}(3t^2) = 6t \)
• \( \frac{d}{dt}(-18t) = -18 \)
• \( \frac{d}{dt}(24) = 0 \)

Thus, the acceleration function is \(a(t) = 6t - 18\). This function tells us when an object is speeding up or slowing down. Negative acceleration, when \(a(t) < 0\), indicates the object is decelerating or slowing down. So for \(a(t) = 6t - 18\), the object decelerates when \(t < 3\).

Motion Along a Line

When talking about motion along a line, we track the movement of an object along a horizontal path. The position function \(s(t) = t^3 - 9t^2 + 24t\) represents the object's path over time. To fully describe the motion, we look at both velocity and acceleration:

• First, determine where the velocity is positive, \(v(t) > 0\), indicating the object is moving right. For \(v(t) = 3t^2 - 18t + 24\), the object moves right when \(t < 2\) or \(t > 4\).
• Then, identify where the velocity is negative, \(v(t) < 0\), which shows it moves left. This occurs between \(2 < t < 4\).
• Acceleration also impacts motion; if it's negative, the object's speed decreases. Here \(a(t) < 0\) for \(t < 3\).

Analyzing these intervals offers a complete picture of the object's directional motion over time.

Differentiation

Differentiation is a fundamental calculus operation used to find rates of change. It is crucial in analyzing motion problems because it allows us to derive velocity and acceleration from a position function. The power rule is a common method used in differentiation:

• You apply it to each term of a polynomial function, adjusting the exponent while multiplying by the original exponent.
• For instance, the derivative of \(t^3\) is \(3t^2\), \(-9t^2\) becomes \(-18t\), and \(24t\) is just \(24\).

By performing differentiation on the position function \(s(t)\), you obtain the velocity function \(v(t)\); performing it again on \(v(t)\) gives you the acceleration function \(a(t)\). Thus, differentiation is the key process behind determining how an object's position, velocity, and acceleration change over time.