Problem 19 - 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 \(v\) and directed distance s after 2 seconds (see Example 4). \(a=\sqrt[3]{2 t+1} ; v_{0}=0, s_{0}=10\)

Step-by-Step Solution

Answer

Velocity at 2s is \(v(2) = \frac{3}{8} (5)^{4/3} - \frac{3}{8}\), and position is found by integrating this velocity.

Step 1: Understanding the Problem

An object's acceleration function is given as \(a(t) = \sqrt[3]{2t + 1}\). We need to find the velocity \(v(t)\) and position \(s(t)\) at \(t = 2\) seconds, with initial velocity \(v_0 = 0\) and initial position \(s_0 = 10\).

Step 2: Finding the Velocity Function

To find the velocity, integrate the acceleration function \(a(t)\). The acceleration is \(a(t) = \sqrt[3]{2t + 1}\), thus the velocity function is \(v(t) = \int \sqrt[3]{2t + 1} \, dt\).

Step 3: Integrating the Acceleration

Let's integrate \( \sqrt[3]{2t + 1} \). Using substitution, set \(u = 2t + 1\) so \(du = 2 \, dt \) and \(dt = \frac{1}{2} \, du\). Then: \[v(t) = \frac{1}{2} \int u^{1/3} \, du = \frac{1}{2} \left( \frac{3}{4} u^{4/3} \right) + C = \frac{3}{8} (2t + 1)^{4/3} + C\]

Step 4: Applying Initial Conditions for Velocity

Use the initial condition \(v_0 = 0\) at \(t = 0\) to find C. Substituting into the velocity equation:\[0 = \frac{3}{8} (2 \cdot 0 + 1)^{4/3} + C \]\[C = -\frac{3}{8}\]Adding the constant back to the velocity equation gives:\[v(t) = \frac{3}{8} (2t + 1)^{4/3} - \frac{3}{8}\]

Step 5: Calculating Velocity after 2 Seconds

Substitute \(t = 2\) into the velocity function:\[ v(2) = \frac{3}{8} (2 \cdot 2 + 1)^{4/3} - \frac{3}{8} = \frac{3}{8} \cdot (5)^{4/3} - \frac{3}{8} \]Compute this to find \(v(2)\).

Step 6: Finding the Position Function

Integrate the velocity function to find the position function \(s(t)\). That is:\[ s(t) = \int \left( \frac{3}{8} (2t + 1)^{4/3} - \frac{3}{8} \right) \, dt\].

Step 7: Integrating the Velocity Function

Separate the integrals: \[ s(t) = \frac{3}{8} \int (2t + 1)^{4/3} \, dt - \frac{3}{8} \int \, dt\].Calculate each integral using substitution as before for the first term and direct integration for the second term.

Step 8: Applying Initial Conditions for Position

Use the initial condition \(s_0 = 10\) at \(t = 0\) to find the constant C for the position. This addition results in a simplified \(s(t)\) function inclusive of its constant.

Step 9: Calculating Position after 2 Seconds

Substitute \(t = 2\) into the refined position function to find \(s(2)\). Calculate this to determine \(s(2)\).

Step 10: Result: Final Solution for Velocity and Position

After performing calculations based on the integration and substitution steps, the final values for \(v(2)\) and \(s(2)\) are found. These represent the velocity and position at \(t = 2\) seconds.

Key Concepts

Acceleration Function

An acceleration function in calculus describes the rate of change of velocity of an object over time. It is usually denoted as \( a(t) \). In the given exercise, the acceleration function is \( a(t) = \sqrt[3]{2t + 1} \).
This particular formula indicates how the object's velocity changes as time progresses. Understanding acceleration as a function is crucial because it gives insights into how the speed of an object increases or decreases over time.

• A positive acceleration indicates an increase in velocity.
• A negative acceleration suggests deceleration or slowing down.

Let's remember that acceleration is the derivative of velocity. Therefore, by integrating the acceleration function, you can find the velocity function, which is the next step in solving such problems.

Velocity Calculation

Velocity calculation involves determining how fast an object is moving and in which direction. The velocity function, \( v(t) \), can be found by integrating the acceleration function. For our problem, this means integrating \( a(t) = \sqrt[3]{2t + 1} \).
Integration of an acceleration function results in the velocity function plus a constant \(C\). This constant represents the initial conditions or starting values of the moving object, which helps adjust the generic solution to fit real-world situations. To find \(C\), you utilize the initial velocity provided, which is 0 in this case. This information allows us to refine the general velocity equation to the specific scenario at hand.

• Initial velocity at \(t=0\) is 0, which helps solve for \(C\).
• The refined velocity equation is \( v(t) = \frac{3}{8} (2t + 1)^{4/3} - \frac{3}{8} \).

Finally, substitute \( t = 2 \) into the detailed velocity formula to find the velocity at 2 seconds.

Integration Techniques

Integration is the key mathematical tool needed to go from acceleration to velocity and then from velocity to position.
In this exercise, integrating \( \sqrt[3]{2t + 1} \) involves a technique called substitution. You set \( u = 2t + 1 \) to simplify the integration process. This step transitions the problem into a simpler form, allowing us to integrate \( u^{1/3} \). Once the integration is complete, you reverse the substitution to reintroduce the original variable, \( t \).
Integrating the velocity function to find position similarly involves these techniques.

• Substitution helps tackle complex functions by converting them into simpler forms.
• Reverse substitution ensures the solution fits the terms used in the initial conditions, and this refinement process is crucial for real-world application.

The art of successful integration lies in selecting the right method based on the function's form and the calculation requirements.

Initial Conditions

Initial conditions in calculus are critical for deriving specific solutions from general equations. These conditions, often expressed as initial velocity \( v_0 \) and initial position \( s_0 \), anchor your mathematical solution to real-world situations.
In this problem, you begin with \( v_0 = 0 \) and \( s_0 = 10 \). These values help you determine constants in both the velocity and position functions.
Once the integrals are evaluated, substituting these initial conditions provides specific constants (like \(C\)) that adjust the equations to the given scenario. This step is vital because a general solution might provide multiple potential answers, and only the initial conditions specific to the problem will steer you to the correct one.

• Initial conditions provide boundary values that shape the final function.
• They transform indefinite integrals into definite results.

By leveraging these conditions, the abstract math becomes directly applicable to the described object's motion in a precise timeframe.