Question

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

Short answer

Answer: The position function for the object is \(s(t) = -\frac{3}{4} \sin 2t + \frac{5}{2}t + 10\).

Step-by-step solution

Integrate the acceleration function to find the velocity function

To find the velocity function, we need to integrate the given acceleration function with respect to time: $$v(t) = \int a(t) dt = \int 3 \sin 2 t dt$$ To integrate, we will perform a substitution: let \(u = 2t\), then \(du = 2 dt\). Now we can rewrite our integral as: $$v(t) = \frac{3}{2} \int \sin u du$$ The integral of \(\sin u\) is \(-\cos u\). Therefore, we have: $$v(t) = -\frac{3}{2} \cos u + C_1$$ Now we need to replace \(u\) with our original variable \(t\): $$v(t) = -\frac{3}{2} \cos 2t + C_1$$ Now we will find the constant \(C_1\) using the initial velocity condition \(v(0) = 1\): $$1 = -\frac{3}{2} \cos(0) + C_1$$ Solving for \(C_1\), we get: $$C_1 = 1 + \frac{3}{2}$$ So, the velocity function is: $$v(t) = -\frac{3}{2} \cos 2t + \frac{5}{2}$$

Integrate the velocity function to find the position function

Now we need to integrate the velocity function with respect to time to obtain the position function: $$s(t) = \int v(t) dt = \int \left(-\frac{3}{2} \cos 2t + \frac{5}{2}\right) dt$$ We will integrate term by term: $$s(t) = -\frac{3}{2} \int \cos 2t dt + \frac{5}{2} \int dt$$ To integrate \(\cos 2t\), we will perform another substitution: let \(u = 2t\), then \(du = 2 dt\). Now we can rewrite our integral as: $$s(t) = -\frac{3}{4} \int \cos u du + \frac{5}{2} \int dt$$ The integral of \(\cos u\) is \(\sin u\). Therefore, we have: $$s(t) = -\frac{3}{4} \sin u + \frac{5}{2}t + C_2$$ Now we need to replace \(u\) with our original variable \(t\): $$s(t) = -\frac{3}{4} \sin 2t + \frac{5}{2}t + C_2$$ Now we will find the constant \(C_2\) using the initial position condition \(s(0) = 10\): $$10 = -\frac{3}{4} \sin(0) + \frac{5}{2}(0) + C_2$$ Solving for \(C_2\), we get: $$C_2 = 10$$ So, the position function is: $$s(t) = -\frac{3}{4} \sin 2t + \frac{5}{2}t + 10$$

Key concepts

Integration

Integration is a fundamental concept in calculus. It is the process of finding the integral, which can be thought of as the opposite of differentiation. In simpler terms, while differentiation helps us find the rate of change, integration helps us accumulate values.
When dealing with physical movements, integration allows us to go from acceleration to velocity and then from velocity to position.

• The indefinite integral represents a family of functions with an integration constant, often denoted as \( C \). This constant comes from the fact that the derivative of any constant is zero, so it cannot be recovered by integration alone.
• Definite integrals, on the other hand, compute a number and do not include \( C \).

In our given problem, we used integration twice. First, we integrated the acceleration function to find velocity, and then we again integrated the velocity function to find the position. The integration steps involved substitutions, such as \( u = 2t \), to simplify the integral expressions. This technique makes finding the antiderivative more manageable.

Velocity function

The velocity function represents the rate of change of position with respect to time. It tells us how fast an object is moving and in which direction. If you know the acceleration, you can find the velocity by integrating the acceleration function.
In our example:

• The acceleration function \( a(t) = 3 \sin 2t \) was integrated giving us the velocity function \( v(t) = -\frac{3}{2} \cos 2t + \frac{5}{2} \).
• The integration took into account the initial velocity \( v(0) = 1 \) to solve for the constant \( C_1 \).

This step is crucial because velocity at any time \( t \) offers insights into how fast the object is traveling and influences how the position is calculated in subsequent steps.

Position function

The position function details an object's location along a line at any given time \( t \). If velocity shows how fast the object moves, position tells you where the object is exactly located.
By integrating the velocity function, we acquire the position function. In this exercise, we went from velocity \(-\frac{3}{2} \cos 2t + \frac{5}{2}\) to the position function:

• Integration performed term by term resulted in \( s(t) = -\frac{3}{4} \sin 2t + \frac{5}{2}t + 10 \).
• The initial position \( s(0) = 10 \) helped us identify the unknown constant \( C_2 \).

The position function is pivotal as it grants a complete picture of motion beyond mere speed, incorporating where the object is over time, dependent on the initial conditions.