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 \(\underline{v}\) 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 approximately 3.6 cm/s; distance is approximately 20.2 cm.

Step 1: Understand Given Values and Find Formula for Velocity

We know the acceleration formula, \( a = \sqrt[3]{2t + 1} \). To find the velocity, integrate the acceleration with respect to time, \( t \). We are given the initial velocity, \( v_0 = 0 \).

Step 2: Integrate Acceleration to Find Velocity

The velocity \( v(t) \) is found by integrating the acceleration function: \[ v(t) = \int \sqrt[3]{2t + 1} \, dt \]Using integration techniques, we can find the antiderivative. Let \( u = 2t + 1 \), then \( du = 2 \, dt \) or \( dt = \frac{1}{2} du \). The integral becomes:\[ v(t) = \frac{1}{2} \int u^{1/3} \, du = \frac{1}{2} \cdot \frac{3}{4} u^{4/3} + C \]\[ v(t) = \frac{3}{8}(2t + 1)^{4/3} + C \]

Step 3: Apply Initial Condition to Solve for Constant C

Use the initial velocity condition \( v(0) = 0 \) to solve for \( C \):\[ v(0) = \frac{3}{8}(2 \cdot 0 + 1)^{4/3} + C = 0 \]\[ \frac{3}{8} \cdot 1 + C = 0 \]\[ C = -\frac{3}{8} \]

Step 4: Write the Complete Velocity Function

Substitute \( C = -\frac{3}{8} \) back into the velocity function:\[ v(t) = \frac{3}{8}(2t + 1)^{4/3} - \frac{3}{8} \]

Step 5: Calculate Velocity at t = 2s

Plug \( t = 2 \) into the velocity function:\[ v(2) = \frac{3}{8}(2 \cdot 2 + 1)^{4/3} - \frac{3}{8} \]\[ v(2) = \frac{3}{8}(5)^{4/3} - \frac{3}{8} \]Evaluating numerically, \( v(2) \approx 3.6 \) cm/s.

Step 6: Integrate Velocity to Find Distance

To find the position \( s(t) \), integrate the velocity function:\[ s(t) = \int \left( \frac{3}{8}(2t + 1)^{4/3} - \frac{3}{8} \right) \, dt \]Breaking it down, we consider the integral\[ s(t) = \frac{3}{8} \int (2t + 1)^{4/3} \, dt - \frac{3}{8} \int \, dt \]

Step 7: Evaluate the Integral for Position

Using substitution and similar steps, integrate to get:\[ s(t) = \frac{9}{40}(2t + 1)^{7/3} - \frac{3}{8}t + D \]Use initial position \( s_0 = 10 \) to find \( D \):\[ s(0) = \frac{9}{40}(1)^{7/3} - 0 + D = 10 \]\[ \frac{9}{40} + D = 10 \]\[ D = 10 - \frac{9}{40} \approx 9.775 \]

Step 8: Write Distance Function and Calculate at t = 2s

Substitute back into position function:\[ s(t) = \frac{9}{40}(2t+1)^{7/3} - \frac{3}{8}t + 9.775 \]Find \( s(2) \):\[ s(2) = \frac{9}{40}(5)^{7/3} - \frac{3}{8} \cdot 2 + 9.775 \]\[ s(2) \approx 20.2 \] cm.

Key Concepts

Velocity

Understanding velocity is key to analyzing how an object's speed changes over time. In calculus, velocity

• represents the rate of change of position
• is a vector, which means it has both magnitude and direction

In the given exercise, we start by finding the velocity from the acceleration using integration: \[ v(t) = \int \sqrt[3]{2t + 1} \, dt\]When integrating the acceleration, the result is a velocity that tells us how fast the object is moving and in which direction. Since we were given that the initial velocity \( v_0 \) is 0 cm/s, this effect must be considered in our integration to solve for the constant of integration. The velocity at any given time \( t \) can be calculated using the velocity function after the integration.

Acceleration

Acceleration describes how quickly the velocity of an object is changing. In the exercise, acceleration is provided by the equation:\[ a = \sqrt[3]{2t + 1}\]

• It is the derivative of velocity.
• Measures the change in velocity per unit time.

The concept of acceleration helps us understand how the speed of an object changes over time. Frequent integration of acceleration provides insights into the nature of motion, allowing you to determine the velocity over time. In this problem, you integrate the acceleration equation to derive an expression for velocity.

Integration

Integration is a fundamental part of calculus used here to transition from acceleration to velocity and from velocity to position.

• It involves finding the antiderivative or integral of a function.
• This process helps establish relationships between time \( t \), velocity \( v(t) \), and position \( s(t) \).

In the exercise, integration of the acceleration function produces the velocity function:\[ v(t) = \int \sqrt[3]{2t + 1} \, dt\]To obtain the object's position at a certain time, we integrate the velocity function:\[ s(t) = \int v(t) \, dt\]Each integration stage allows us to understand more about the dynamics of the moving object by calculating changes over time.

Position Function

The position function describes where an object is located on a coordinate line over time. Understanding this concept involves analyzing the following:

• The initial position and how it changes over time.
• Integration of the velocity function to describe position.

For the given problem, we find the position function by integrating the velocity function:\[ s(t) = \int \left( \frac{3}{8}(2t + 1)^{4/3} - \frac{3}{8} \right) \, dt\]After integration, don't forget to incorporate the constant of integration based on the initial conditions, \( s_0 = 10 \) cm, to finalize the position function. This results in a detailed description of where the object is at a specific time, such as after 2 seconds.