Problem 18 - 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). \(a=(1+t)^{-4}, v_{0}=0, s_{0}=10\)

Step-by-Step Solution

Answer

Velocity after 2 seconds is \(\frac{26}{81}\) cm/s and distance is \(\frac{284}{27}\) cm.

Step 1: Understand the Problem

We need to find the velocity \(v\) and the directed distance \(s\) after 2 seconds for an object with initial velocity \(v_0 = 0\), initial distance \(s_0 = 10\), and acceleration \(a(t) = (1+t)^{-4}\).

Step 2: Integrate the Acceleration Function

The velocity function \(v(t)\) is the integral of the acceleration function with respect to time. We start by integrating \(a(t) = (1 + t)^{-4}\). \[ v(t) = \int (1 + t)^{-4} \, dt \] The antiderivative of \((1 + t)^{-4}\) is \(-\frac{1}{3}(1 + t)^{-3} + C\), where \(C\) is the constant of integration.

Step 3: Determine the Constant of Integration for Velocity

Use the initial condition \(v_0 = 0\) when \(t = 0\) to find \(C\). \[ v(0) = -\frac{1}{3}(1 + 0)^{-3} + C = 0 \] This simplifies to \(-\frac{1}{3} + C = 0\), so \(C = \frac{1}{3}\). Thus, the velocity function is \[ v(t) = -\frac{1}{3}(1+t)^{-3} + \frac{1}{3} \]

Step 4: Evaluate the Velocity at t = 2

Substitute \(t = 2\) into the velocity function to find \(v(2)\). \[ v(2) = -\frac{1}{3}(1+2)^{-3} + \frac{1}{3} \] \[ v(2) = -\frac{1}{3}(3)^{-3} + \frac{1}{3} \] \[ v(2) = -1/81 + 1/3 \] Find a common denominator: \[ v(2) = -1/81 + 27/81 = 26/81 \] Thus, \(v(2) = \frac{26}{81} \text{ cm/s}\).

Step 5: Integrate the Velocity Function for Distance

Now integrate the velocity function to find the distance \(s(t)\). \[ s(t) = \int \left(-\frac{1}{3}(1+t)^{-3} + \frac{1}{3} \right) \, dt \] Each term is integrated separately: - The antiderivative of \(-\frac{1}{3}(1+t)^{-3}\) is \(\frac{1}{6}(1+t)^{-2}\). - The antiderivative of \(\frac{1}{3}\) is \(\frac{1}{3}t\). So, \[ s(t) = \frac{1}{6}(1+t)^{-2} + \frac{1}{3}t + K \] where \(K\) is the constant of integration.

Step 6: Determine the Constant of Integration for Distance

Use the initial condition \(s_0 = 10\) when \(t = 0\) to find \(K\). \[ s(0) = \frac{1}{6}(1+0)^{-2} + 0 + K = 10 \] This simplifies to \(\frac{1}{6} + K = 10\). Solve for \(K\): \[ K = 10 - \frac{1}{6} = \frac{60}{6} - \frac{1}{6} = \frac{59}{6} \] Thus, the distance function is: \[ s(t) = \frac{1}{6}(1+t)^{-2} + \frac{1}{3}t + \frac{59}{6} \]

Step 7: Evaluate the Distance at t = 2

Substitute \(t = 2\) into the distance function to find \(s(2)\). \[ s(2) = \frac{1}{6}(1+2)^{-2} + \frac{1}{3}(2) + \frac{59}{6} \] \[ s(2) = \frac{1}{6}(3)^{-2} + \frac{2}{3} + \frac{59}{6} \] Convert fractions to a common denominator and compute: \[ s(2) = \frac{1}{54} + \frac{12}{18} + \frac{531}{54} \] \[ s(2) = \frac{1}{54} + \frac{36}{54} + \frac{531}{54} = \frac{568}{54} = \frac{284}{27} \approx 10.52 \text{ cm} \]

Key Concepts

Integration Techniques

Integration is a fundamental technique in calculus, used to find functions such as velocity and distance from acceleration. Let's unpack it further. The process of integration essentially reverses differentiation, allowing us to build functions from their rates of change. In the context of this exercise, we start with the acceleration function, and our task is to find the velocity by integrating acceleration with respect to time.

Here are some tips to keep in mind when dealing with integration:

• Always look for patterns in the function you are integrating. Recognizing these patterns will help in selecting the right integration technique, such as substitution or by-parts.
• Remember to add a constant of integration, since integration can only determine the antiderivative up to a constant.
• Use initial conditions to solve for this constant. These conditions are extra pieces of information that give meaning to the abstract integral.

In this exercise, we dealt with integrating a simple power function: \( \int (1 + t)^{-4} \, dt \), which resulted in an antiderivative of \(-\frac{1}{3}(1 + t)^{-3} + C\), where \(C\) is the constant of integration initially unknown without further information.

Initial Conditions in Calculus

Dealing with initial conditions is crucial in ensuring the accuracy of integration, especially in real-world problems. Initial conditions in calculus provide the specific values at a certain point to solve constants in the integrated functions, like velocity or distance.

In this problem, we have two initial conditions: initial velocity \(v_0 = 0\) and initial distance \(s_0 = 10\). These conditions help bind our abstract solutions to real values:

• For the velocity function, the initial condition helps in finding the constant \(C\) after integrating acceleration: \(v(0) = -\frac{1}{3}(1+0)^{-3} + C = 0\).
• For the distance function, the condition \(s(0) = 10\) enables calculating the constant \(K\) after integrating the velocity.

Initial conditions make sure that the solutions to the velocity and distance functions are specific to this particular scenario, not just any similar problem with different starting factors.

Distance and Velocity Functions

Understanding the relationship between distance, velocity, and acceleration is key in solving kinematics problems like the one in this exercise. Let's break down these concepts further:

• Velocity: It is the rate of change of distance with respect to time. Finding velocity involves integrating the acceleration function. In this exercise, the function \(v(t)\) tells us how the object's speed changes over time.
• Distance: It accumulates as a result of velocity over time. Hence, the distance function \(s(t)\) is found by integrating the velocity function. This gives you the positional change of the object from its start point.
• Time Specifics: After forming these functions, evaluating them at specific time points (like 2 seconds) helps find exact numeric outcomes, crucial for practical applications.

In our example, after calculating both velocity and distance equations, plugging \(t = 2\) provides numerical solutions: - Velocity \(v(2) = \frac{26}{81}\) cm/s, - Distance \(s(2) \approx 10.52\) cm. These results give insight into how fast and how far the object has moved after two seconds of motion described by the provided acceleration function.