Problem 41 - Thomas Calculus

Question

Give the acceleration \(a=d^{2} s / d t^{2}\) , initial velocity, and initial position of a body moving on a coordinate line. Find the body's position at time \(t\). \(a=32, \quad v(0)=20, \quad s(0)=5\)

Step-by-Step Solution

Answer

The body's position at time \( t \) is \( s(t) = 16t^2 + 20t + 5 \).

Step 1: Understand the Given Information

We are provided with the acceleration of a body as a constant value \( a = 32 \), the initial velocity \( v(0) = 20 \), and the initial position \( s(0) = 5 \). We will use these to find the position of the body at any time \( t \).

Step 2: Integrate Acceleration to Find Velocity

Since acceleration \( a = 32 \) is constant, we integrate it with respect to time \( t \) to find the velocity function \( v(t) \). Thus, \( v(t) = \int 32 \, dt = 32t + C_1 \).

Step 3: Apply Initial Velocity Condition

We know \( v(0) = 20 \). Substitute into our velocity equation: \( 20 = 32(0) + C_1 \). Solving for \( C_1 \), we get \( C_1 = 20 \). Therefore, \( v(t) = 32t + 20 \).

Step 4: Integrate Velocity to Find Position

Next, we integrate the velocity function \( v(t) = 32t + 20 \) to find the position function \( s(t) \). Thus, \( s(t) = \int (32t + 20) \, dt = 16t^2 + 20t + C_2 \).

Step 5: Apply Initial Position Condition

Using \( s(0) = 5 \), substitute into the position equation: \( 5 = 16(0)^2 + 20(0) + C_2 \). Solving for \( C_2 \), we find \( C_2 = 5 \). Therefore, the position function is \( s(t) = 16t^2 + 20t + 5 \).

Step 6: Conclusion: Position at Time \( t \)

The position of the body at time \( t \) is given by the function \( s(t) = 16t^2 + 20t + 5 \).

Key Concepts

Acceleration

Acceleration is the rate at which an object's velocity changes over time. In most problems involving linear motion, this is measured in meters per second squared (m/s²). In this exercise, we're given a constant acceleration value of 32 m/s². This means that every second, the velocity of the object increases by 32 m/s.
It's important to remember:

• Acceleration is a vector quantity, meaning it has both magnitude and direction.
• A positive acceleration indicates an increase in velocity, while a negative acceleration (often called deceleration) shows a decrease.

Understanding acceleration is crucial for predicting how fast an object will move at future times.

Velocity function

The velocity function describes how an object's speed and direction change over time. To derive this function from the given acceleration, we integrate the acceleration with respect to time. In our example, the acceleration is constant at 32 m/s², leading to:
\[v(t) = \int 32 \, dt = 32t + C_1\]
This integration process essentially sums up the continuous nature of acceleration over time to find the velocity.

• The integration constant, \(C_1\), is found using initial conditions. In our scenario, we know the initial velocity \(v(0) = 20\) m/s, leading to \(C_1 = 20\).
• This results in the velocity function: \(v(t) = 32t + 20\).

The velocity function helps us understand not only how fast something will be going, but also how its speed evolves during its journey.

Position function

To determine where an object will be located at any time, we use the position function. This is found by integrating the velocity function. The given velocity function is \(v(t) = 32t + 20\), so we integrate it as follows: \[s(t) = \int (32t + 20) \, dt = 16t^2 + 20t + C_2\]This function gives a quadratic equation, indicating how the position evolves over time in response to velocity changes.

• The constant of integration, \(C_2\), is determined with the given initial position \(s(0) = 5\).
• Solving for \(C_2\) ensures that the initial conditions match the problem context.

Thus, the full position equation becomes \(s(t) = 16t^2 + 20t + 5\). This statement gives a snapshot of an object's position at any specified time, allowing predictions about its future location on the coordinate line.