Problem 158 - Calculus Volume 1

Question

The position function \(s(t)=t^{3}-8 t\) gives the position in miles of a freight train where east is the positive direction and t is measured in hours. a. Determine the direction the train is traveling when s(t) = 0. b. Determine the direction the train is traveling when a(t) = 0. c. Determine the time intervals when the train is slowing down or speeding up.

Step-by-Step Solution

Answer

a. West; b. West; c. Speeding up: \((\infty, -\sqrt{8/3})\) and \((\sqrt{8/3}, \infty)\), slowing down: \((-\sqrt{8/3}, \sqrt{8/3})\).

Step 1: Find when the train is at position zero

To find when the position of the train is zero, we set the position function equal to zero and solve for \(t\):\[ s(t) = t^3 - 8t = 0 \]Factor the equation:\[ t(t^2 - 8) = 0 \]The solutions are: \( t = 0 \) or \( t^2 = 8 \).Solving for \( t^2 = 8 \):\[ t = \sqrt{8} \text{ or } t = -\sqrt{8} \]Thus, \( t = 0, \frac{2\sqrt{2}}{1} \), and \(-\frac{2\sqrt{2}}{1} \).

Step 2: Determine velocity for t values when s(t) = 0

Calculate the derivative of the position function to find the velocity function:\[ v(t) = s'(t) = 3t^2 - 8 \]Evaluate \( v(t) \) at the times when \( s(t) = 0 \):- At \( t = 0 \): \( v(0) = 3(0)^2 - 8 = -8 \), direction is west.- At \( t = \sqrt{8} \): \( v(\sqrt{8}) = 3(\sqrt{8})^2 - 8 = 16 \), direction is east.- At \( t = -\sqrt{8} \): \( v(-\sqrt{8}) = 3(-\sqrt{8})^2 - 8 = 16 \), direction is east.

Step 3: Determine when the acceleration is zero

Find the derivative of the velocity function to determine the acceleration:\[ a(t) = v'(t) = 6t \]Set acceleration to zero and solve for \(t\):\[ 6t = 0 \]Thus, \( t = 0 \).

Step 4: Determine velocity at t when a(t) = 0

Evaluate the velocity at \( t = 0 \) to determine direction:\[ v(0) = 3(0)^2 - 8 = -8 \]Thus, the train is traveling west when \( a(t) = 0 \).

Step 5: Determine intervals when the train is speeding up or slowing down

Identify when velocity and acceleration have the same sign (speeding up) and opposite signs (slowing down):- From the velocity function \( v(t) = 3t^2 - 8 \), find the critical points by setting \( 3t^2 - 8 = 0 \) to get \( t = \pm \sqrt{\frac{8}{3}} \).- Check the signs in the intervals \( (-\infty, -\sqrt{\frac{8}{3}}) \), \( (-\sqrt{\frac{8}{3}}, \sqrt{\frac{8}{3}}) \), and \( (\sqrt{\frac{8}{3}}, \infty) \) using a sign chart.- The train is speeding up when \( v(t) \) and \( a(t) \) have the same sign.

Key Concepts

Velocity Function

The velocity function is crucial for determining the direction and speed of an object over time. It is the derivative of the position function, indicating how fast the object's position is changing at any given moment. In simple terms, with the train's position function given by \( s(t) = t^3 - 8t \), the velocity function is its first derivative:

• \( v(t) = s'(t) = 3t^2 - 8 \)

To find out the train's motion when it reaches a certain position, we evaluate this function at different times. For instance, to check when the train is at position zero, we look at times \( t = 0, \sqrt{8}, \) and \(-\sqrt{8} \). Thus, calculating the velocity at these points helps us determine the direction:

• At \( t = 0 \), \( v(0) = -8 \), moving west.
• At \( t = \sqrt{8} \), \( v(\sqrt{8}) = 16 \), moving east.
• At \( t = -\sqrt{8} \), \( v(-\sqrt{8}) = 16 \), also moving east.

Understanding the velocity function is essential in physics to predict how fast an object moves in different directions over time.

Acceleration Function

Acceleration tells us how quickly the velocity of an object is changing. It's the derivative of the velocity function. When acceleration is zero, the velocity is constant at that moment, indicating that there is no change in speed. For our train problem, the velocity function is \( v(t) = 3t^2 - 8 \), so the acceleration function becomes:

• \( a(t) = v'(t) = 6t \)

By setting the acceleration function to zero, we can determine key points where velocity isn't increasing or decreasing:

• \( 6t = 0 \) gives us \( t = 0 \).

Evaluating the velocity at this time helps confirm the direction. When \( t = 0 \), the train's velocity \( v(0) = -8 \), indicating the train is moving west. Acceleration gives crucial insights into changes in an object's movement characteristics.

Solving Equations

Solving equations is a foundational skill in mathematics for finding unknown variables that make an equation true. The process involves manipulating the equation to isolate the variable. In the context of the train's movement, solving the position function equation, \( s(t) = t^3 - 8t = 0 \), is essential to find when the train's position is zero. This involves:

• Factoring the equation as \( t(t^2 - 8) = 0 \).
• Solving for \( t \) yields \( t = 0, \; \pm\sqrt{8} \).

Understanding how to solve these kinds of equations helps in determining specific time intervals important in analyzing any motion or trend. It's a necessary mathematical skill for applying calculus concepts to real-world problems.

Direction of Motion

Understanding the direction of motion is critical in analyzing an object's trajectory. With respect to our train problem, the direction is indicated by the sign of the velocity function. A positive velocity implies motion in the positive direction, which is east in this scenario, while a negative velocity implies motion west. We determined:

• At \( t = 0 \), \( v(t) = -8 \): The train moves west.
• At \( t = \sqrt{8} \), \( v(t) = 16 \): The train moves east.
• At \( t = -\sqrt{8} \), \( v(t) = 16 \): The train still moves east.

Gaining insight into an object's direction through velocity is a common application in physics and calculus, useful in numerous practical situations such as planning routes or analyzing movement.

Sign Charts

Sign charts are a visual way to analyze functions and determine the intervals where they are positive, negative, or zero. This is particularly useful when determining where a train, or any other object, is speeding up or slowing down. For the train problem, we use sign charts to examine the signs of the velocity \( v(t) \) and acceleration \( a(t) \):

• If both are positive or negative, the train speeds up.
• If one is positive and the other negative, the train slows down.

Creating sign charts involves checking these signs across different intervals formed by critical points. Therefore, understanding sign charts helps visualize behavior trends in functions, providing an intuitive grasp of acceleration and velocity interplay.