Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

An object moves along a horizontal coordinate line in such a way that its position at time \(t\) is specified by \(s=t^{3}-3 t^{2}-24 t-6 .\) Here \(s\) is measured in centimeters and in seconds. When is the object slowing down; that is, when is its speed decreasing?

Step-by-Step Solution

Answer

The object slows down when \(t \in (-\infty, -2) \cup (1, 4)\).

Step 1: Find the velocity function

The velocity of the object is the derivative of the position function. Given the position function \(s = t^3 - 3t^2 - 24t - 6\), find the derivative \(v(t) = \frac{ds}{dt}\).\[ v(t) = \frac{d}{dt}(t^3 - 3t^2 - 24t - 6) = 3t^2 - 6t - 24 \]

Step 2: Find the acceleration function

The acceleration of the object is the derivative of the velocity function. Compute the derivative of \(v(t)\) to find the acceleration function \(a(t)\).\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 6t - 24) = 6t - 6 \]

Step 3: Determine when the object is slowing down

An object is slowing down when the velocity and acceleration have opposite signs. This means finding when \(v(t)\) and \(a(t)\) have opposite signs.1. Solve \(v(t) = 3t^2 - 6t - 24 = 0\) for \(t\) to find critical points. \[ t^2 - 2t - 8 = 0 \] Solving this quadratic equation gives roots \(t = 4\) and \(t = -2\). 2. Solve \(a(t) = 6t - 6 = 0\) for \(t\). \[ t = 1 \] 3. Test intervals between critical points: \((-\infty, -2), (-2, 1), (1, 4), (4, \infty)\). - For each interval, choose a test point and determine the signs of \(v(t)\) and \(a(t)\). - For \((-\infty, -2)\), \(t = -3\): \(v(-3) > 0\), \(a(-3) < 0\) \(\Rightarrow\) slowing down - For \((-2, 1)\), \(t = 0\): \(v(0) < 0\), \(a(0) < 0\) \(\Rightarrow\) speeding up - For \((1, 4)\), \(t = 2\): \(v(2) < 0\), \(a(2) > 0\) \(\Rightarrow\) slowing down - For \((4, \infty)\), \(t = 5\): \(v(5) > 0\), \(a(5) > 0\) \(\Rightarrow\) speeding up

Step 4: Conclusion

The object is slowing down when \(t \in (-\infty, -2) \cup (1, 4)\). These time intervals indicate when the velocity and acceleration of the object are of opposite signs.

Key Concepts

Velocity and Acceleration

In calculus, velocity and acceleration are key concepts used to describe motion. When an object moves, its position changes over time, and these changes can be measured using derivatives. The velocity function is the first derivative of the position function. It tells us the speed and direction of the object at any given time.

Acceleration, on the other hand, is the rate of change of velocity. It is found by taking the derivative of the velocity function. Acceleration indicates how quickly the object is speeding up or slowing down.

• Velocity: First derivative of position function, denoted as \(v(t)\).
• Acceleration: Derivative of velocity function, denoted as \(a(t)\).

Understanding these concepts is crucial for analyzing motion. For an object to slow down, its velocity and acceleration must have opposite signs. This concept helps identify when the object is decelerating.

Derivative of Position Function

The derivative of a position function provides valuable insights into the motion of an object. Given a position function, such as \(s(t) = t^3 - 3t^2 - 24t - 6\), its derivative represents velocity.

To find this, we apply basic derivative rules. For our example function:

• The derivative of \(t^3\) is \(3t^2\).
• The derivative of \(-3t^2\) is \(-6t\).
• The derivative of \(-24t\) is \(-24\).
• The derivative of a constant \(-6\) is 0.

Therefore, the velocity function \(v(t)\) becomes \(3t^2 - 6t - 24\). This derived function allows us to determine how fast and in what direction the object is moving at any time \(t\). It forms the basis for further calculations related to acceleration and solving motion-based problems.

Solving Quadratic Equations

Quadratic equations often emerge in calculus and physics, especially when dealing with motion-related problems. They take the form \(ax^2 + bx + c = 0\) and can be solved using various methods including factoring, completing the square, or the quadratic formula.

In our example, the velocity function \(v(t) = 3t^2 - 6t - 24\) was set to zero to find critical points. By simplifying, we get \(t^2 - 2t - 8 = 0\):

• Factoring gives us \((t - 4)(t + 2) = 0\).
• This results in two solutions: \(t = 4\) and \(t = -2\).

These solutions help identify time values where the function's sign might change, which is crucial for pinpointing intervals of acceleration or deceleration. Understanding how to solve such equations is a fundamental skill in calculus applications.