Question

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=2+3 \sin t ; v(0)=1, s(0)=10$$

Short answer

Based on the given acceleration function \(a(t)=2+3 \sin t\), initial velocity \(v(0)=1\), and initial position \(s(0)=10\), the position function of the object is \(s(t)=t^2 + 3 \sin t + 4t + 10\).

Step-by-step solution

Integrate the acceleration function to find the velocity function

Integrate the given acceleration function, \(a(t)=2+3 \sin t\), to obtain the velocity function, \(v(t)\). Using the integral of \(2\) and \(\sin t\), we have: $$v(t)=\int(2+3 \sin t) dt = 2t - 3 \cos t + C_1$$

Apply the initial condition for velocity

Apply the initial condition \(v(0)=1\) to the velocity function we found in Step 1: $$1 = 2(0) - 3 \cos(0) + C_1$$ Solve for the constant \(C_1\): $$ C_1=1+3=4$$ The velocity function is, therefore, \(v(t)=2t-3 \cos t + 4\).

Integrate the velocity function to find the position function

Integrate the velocity function found in Step 2, \(v(t)=2t-3 \cos t + 4\), to obtain the position function, \(s(t)\). Using the integral of \(2t\), \(\cos t\), and \(4\), we have, $$s(t)=\int(2t-3 \cos t + 4) dt = t^2 + 3 \sin t + 4t + C_2$$

Apply the initial condition for position

Apply the initial condition \(s(0)=10\) to the position function we found in Step 3: $$10= (0)^2 + 3 \sin(0) + 4(0) + C_2$$ Solve for the constant \(C_2\): $$C_2=10$$ The position function is, therefore, \(s(t)=t^2 + 3 \sin t + 4t + 10\).

Key concepts

Acceleration Function

The acceleration function is a mathematical representation of how an object's velocity changes with time. In this context, acceleration is the rate of change of velocity. The given acceleration function is

• \( a(t) = 2 + 3 \sin t \)

This function indicates that the object's acceleration is composed of a constant component, \(2\), and a variable component that fluctuates with time, \(3 \sin t\). Consequently, the object's acceleration not only maintains a steady increase but also oscillates. Understanding the acceleration function is crucial because it allows us to determine how the velocity and position change over time. By integrating this function twice, we can find the velocity and position of the object.

Velocity Function

To obtain the velocity function from the acceleration function, we need to perform integration. The velocity function represents how fast an object is moving and is crucial for understanding motion.
Starting with the acceleration function \( a(t) = 2 + 3 \sin t \), we integrate it:

• \( v(t) = \int (2 + 3 \sin t) dt = 2t - 3 \cos t + C_1 \)

Here, \( C_1 \) is the integration constant. Integration is the reverse of differentiation and allows us to work backward from acceleration to velocity. Initially, we determine \( C_1 \) by using the initial condition \( v(0) = 1 \), which gives us \( 1 = 2 \times 0 - 3 \times \cos 0 + C_1 \) leading to \( C_1 = 4 \). Thus, the velocity function is:

• \( v(t) = 2t - 3 \cos t + 4 \)

This function gives us the velocity of the object at any time \( t \), combining both constant and oscillating changes.

Position Function

The position function gives a complete description of where an object is located along a line at any time \( t \). To find this, we integrate the velocity function. Using the velocity function we derived, \( v(t) = 2t - 3 \cos t + 4 \), we integrate to find:

• \( s(t) = \int (2t - 3 \cos t + 4) dt = t^2 + 3 \sin t + 4t + C_2 \)

Here, \( C_2 \) is another integration constant. The position function provides insights into the total path traveled by the object. By applying the initial condition for position \( s(0) = 10 \), we solve for \( C_2 \):

• \( 10 = 0^2 + 3 \sin 0 + 4 \times 0 + C_2 \)
• \( C_2 = 10 \)

Therefore, the position function is:

• \( s(t) = t^2 + 3 \sin t + 4t + 10 \)

This function fully describes the location of the object over time considering the initial conditions.

Integration

Integration is a fundamental concept in calculus that allows us to reverse the process of differentiation. In essence, when you integrate a function, you are finding its antiderivative.
In the context of motion, integration enables us to determine both the velocity and position when we begin with the acceleration function. For example, integrating the acceleration function \( a(t) = 2 + 3 \sin t \) leads to the velocity function:

• \( v(t) = \int (2 + 3 \sin t) dt = 2t - 3 \cos t + C_1 \)

Likewise, integrating the velocity function helps us find the position function:

• \( s(t) = \int (2t - 3 \cos t + 4) dt = t^2 + 3 \sin t + 4t + C_2 \)

Integration is an essential tool for understanding continuous change and is widely used in physics and engineering to solve real-world problems involving motion and forces.

Initial Conditions

Initial conditions are specific values given at the starting point of a problem that allow us to find particular solutions rather than general ones.
For problems in calculus involving motion, initial conditions are often given for both velocity and position. These conditions help us solve for the constants of integration that arise when we integrate functions. In this problem, we have:

• The initial velocity: \( v(0) = 1 \)
• The initial position: \( s(0) = 10 \)

These initial conditions are crucial because they let us determine unique solutions for both the velocity and position functions:

• The velocity function becomes \( v(t) = 2t - 3 \cos t + 4 \)
• The position function becomes \( s(t) = t^2 + 3 \sin t + 4t + 10 \)

Initial conditions ensure that the derived functions correctly describe the object's motion from the start of observation.