Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

An object is moving along a coordinate line subject to the indicated acceleration a (in centimeters per second per second) with the initial velocity \(v_{0}\) (in centimeters per second) and directed distance \(s_{0}\) (in centimeters). Find both the velocity \(\underline{v}\) and directed distance \(s\) after 2 seconds (see Example 4). $$ a=(3 t+1)^{-3}, v_{0}=4, s_{0}=0 $$

Step-by-Step Solution

Answer

Velocity is approximately 4.163 cm/s; distance is approximately 8.190 cm.

Step 1: Integrate the Acceleration to Find Velocity Function

To find the velocity function, integrate the given acceleration function with respect to time. The acceleration \( a = (3t + 1)^{-3} \) needs to be integrated to find the velocity function \( v(t) \). So, we compute: \[ v(t) = \int (3t + 1)^{-3} \, dt \]. Apply the substitution method by letting \( u = 3t + 1 \), then \( du = 3 \, dt \) or \( dt = \frac{du}{3} \). This gives:\[ v(t) = \int u^{-3} \cdot \frac{1}{3} \, du = \frac{1}{3} \int u^{-3} \, du = \frac{1}{3} \cdot \left(-\frac{1}{2}u^{-2}\right) + C = -\frac{1}{6}(3t + 1)^{-2} + C \].

Step 2: Determine Constant of Integration for Velocity

Apply the initial condition \( v(0) = v_0 = 4 \) to find the constant \( C \). Substitute and solve: \[-\frac{1}{6}(3(0) + 1)^{-2} + C = 4\]\[-\frac{1}{6}(1)^{-2} + C = 4\]\[-\frac{1}{6} + C = 4\]\[C = 4 + \frac{1}{6} = \frac{24}{6} + \frac{1}{6} = \frac{25}{6} \]. Thus, the velocity function becomes \( v(t) = -\frac{1}{6}(3t + 1)^{-2} + \frac{25}{6} \).

Step 3: Evaluate the Velocity at 2 Seconds

Substitute \( t = 2 \) into the velocity function to find \( v(2) \):\[ v(2) = -\frac{1}{6}(3(2) + 1)^{-2} + \frac{25}{6} \]\[ = -\frac{1}{6}(7)^{-2} + \frac{25}{6} \]\[ = -\frac{1}{6} \times \frac{1}{49} + \frac{25}{6} \]\[ = -\frac{1}{294} + \frac{25}{6} \]Convert \( \frac{25}{6} \) to have a common denominator:\[ = -\frac{1}{294} + \frac{1225}{294} \]\[ = \frac{1224}{294} \]Simplify:\[ = \frac{204}{49} \] or approximately \(4.163\) cm/s.

Step 4: Integrate Velocity to Find Distance Function

Integrate the velocity function to find the distance function:\[ s(t) = \int \left(-\frac{1}{6}(3t + 1)^{-2} + \frac{25}{6}\right) \, dt \]\[ = -\frac{1}{6} \int (3t + 1)^{-2} \, dt + \frac{25}{6} \int 1 \, dt \]. Using the substitution from earlier \( u = 3t + 1 \), the integral becomes:\[ s(t) = -\frac{1}{6} \times (-\frac{1}{1})u^{-1} + \frac{25}{6}t + C_1 \]\[ = \frac{1}{6(3t+1)} + \frac{25}{6}t + C_1 \].

Step 5: Determine Constant of Integration for Distance

Apply the initial condition \( s(0) = s_0 = 0 \) to find \( C_1 \). Substitute and solve:\[ \frac{1}{6(3(0)+1)} + \frac{25}{6} \times 0 + C_1 = 0 \]\[ \frac{1}{6} + C_1 = 0 \]\[ C_1 = -\frac{1}{6} \]. Thus, the distance function is \( s(t) = \frac{1}{6(3t+1)} + \frac{25}{6}t - \frac{1}{6} \).

Step 6: Evaluate the Distance at 2 Seconds

Substitute \( t = 2 \) into the distance function to find \( s(2) \):\[ s(2) = \frac{1}{6(3(2) + 1)} + \frac{25}{6}(2) - \frac{1}{6} \]\[ = \frac{1}{6(7)} + \frac{50}{6} - \frac{1}{6} \]\[ = \frac{1}{42} + \frac{50}{6} - \frac{1}{6} \].Calculate using a common denominator of 42:\[ = \frac{1}{42} + \frac{350}{42} - \frac{7}{42} \]\[ = \frac{344}{42} \]Simplify:\[ = \frac{172}{21} \] or approximately \(8.190\) cm.

Key Concepts

Acceleration

Acceleration is a measure of how quickly the velocity of an object changes over time. It is often represented using the symbol \( a \). When an object speeds up or slows down, it experiences acceleration. In our exercise, the acceleration is given by the function \( (3t + 1)^{-3} \). This formula tells us how the acceleration changes as time \( t \) increases.

• Positive acceleration indicates speeding up.
• Negative acceleration indicates slowing down.

Knowing the acceleration is crucial because it helps in determining how the velocity of an object changes over time. It acts as the "link" connecting force and motion in physics.

Velocity

Velocity is a vector quantity that describes the speed of an object in a particular direction. It is denoted by \( v(t) \), indicating it is a function of time \( t \). In our problem, the initial velocity \( v_0 = 4 \) cm/s indicates that, initially, the object is moving at 4 cm/s. By integrating the acceleration, we get the velocity function.
The process involves finding \( v(t) \), which is derived to represent how quickly the position changes over time. Integrating acceleration gives us the velocity:\[ v(t) = -\frac{1}{6}(3t + 1)^{-2} + \frac{25}{6} \] This function considers both the initial conditions and the continuous effect of acceleration.

Integration

Integration is a fundamental concept in calculus used to determine things like the area under a curve or, in physical terms, to find quantities like displacement, velocity, or acceleration from their respective rate functions.

• Definite Integrals: These give a precise value over an interval.
• Indefinite Integrals: These include a constant of integration \( C \) and are used until limits are applied.

Here, integration changes the acceleration function to a velocity function and then from velocity to distance function. The process uses integration techniques like substitution to handle complicated functions, such as \( (3t + 1)^{-3} \). Understanding integration helps solve dynamic physical problems in which quantities change over time.

Coordinate System

A coordinate system is a way of defining the position of points. It provides a framework to describe motion, like a road map for where things are and where they move.
In this exercise, the motion of the object is along a single line, simplifying the scenario to a one-dimensional coordinate line. The coordinate system allows us to use scalar quantities, like distance (\( s \)) and velocity, easily. Usually, we use:

• The x-axis for one-dimensional motion horizontally.
• The y-axis for vertical motion, if needed.

Understanding how these coordinate systems work is crucial for contextualizing mathematical solutions within real-world problems.