Problem 30 - Calculus Early Transcendentals

Question

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. $$a(t)=\frac{20}{(t+2)^{2}}, v(0)=20, s(0)=10$$

Step-by-Step Solution

Answer

Answer: The position function of the object is: $$s(t) = -20\ln(t+2) + 30t + 10 + 20\ln2$$ The velocity function of the object is: $$v(t) = -\frac{20}{t+2} + 30$$

Step 1: Integrate acceleration function to find velocity function

To find the velocity function, we need to integrate the acceleration function, \(a(t)\), with respect to time. So we have: $$v(t) = \int a(t) dt = \int \frac{20}{(t+2)^{2}} dt$$ Now, let's perform the integration: $$v(t) = -\frac{20}{t+2} + C$$ We are given the initial velocity, \(v(0) = 20\). We can use this information to find the value of the integration constant, \(C\): $$20 = -\frac{20}{0+2} + C \implies C = 30$$ So the formula for the velocity function is: $$v(t) = -\frac{20}{t+2} + 30$$

Step 2: Integrate velocity function to find position function

Now, we have the velocity function, and we need to integrate it to find the position function, \(s(t)\). So we have: $$s(t) = \int v(t) dt = \int \left(-\frac{20}{t+2} + 30\right) dt$$ Let's perform the integration: $$s(t) = -20\ln(t+2) + 30t + D$$ We are given the initial position, \(s(0) = 10\). We can use this information to find the value of the integration constant, \(D\): $$10 = -20\ln(0+2) + 30(0) + D \implies D = 10 + 20\ln 2$$ So the formula for the position function is: $$s(t) = -20\ln(t+2) + 30t + 10 + 20\ln2$$ The position and velocity functions for the object moving along a straight line are: Position function: $$s(t) = -20\ln(t+2) + 30t + 10 + 20\ln2$$ Velocity function: $$v(t) = -\frac{20}{t+2} + 30$$

Key Concepts

Integration

Integration is a fundamental concept in calculus that involves finding an antiderivative or the original function from its derivative. In practical terms, when provided with acceleration, integrating allows you to find the velocity function, and further integration of the velocity will yield the position function. This process is essential for understanding the motion of objects, especially when velocity or acceleration is not constant.

For example, in our exercise, we are given an acceleration function:

• \(a(t) = \frac{20}{(t+2)^{2}}\)

To find the velocity, we integrate this function with respect to time \(t\). The same logic is applied to find the position from velocity.

When integrating, don't forget to include the constant of integration \(C\) (or \(D\) when integrating twice), because any constant will vanish once you differentiate the function. Utilize known initial conditions, such as initial velocity or position, to solve for these constants and find the particular solution for your problem.

Velocity

Velocity is the rate of change of an object's position with respect to time. It has both a magnitude (speed) and a direction. For different situations, velocity may change in complex ways, such as when an object accelerates or slows down.

In our original exercise, to find the velocity function, we do this by integrating the acceleration function \(a(t)\), which gives us:

• \(v(t) = -\frac{20}{t+2} + C\)

Using the initial velocity condition \(v(0) = 20\), we solve for \(C\) to get a particular solution:

• \(v(t) = -\frac{20}{t+2} + 30\)

This equation now describes how velocity changes over time. Understanding this can give insights into how fast an object is moving and in what direction at any given time.

Acceleration

Acceleration refers to the change in velocity of an object over time. It is a vector quantity, which means it has magnitude and direction. Positive acceleration represents an increase in velocity, while negative acceleration, also known as deceleration, represents a decrease.

The original problem gives an acceleration function:

• \(a(t) = \frac{20}{(t+2)^{2}}\)

This function shows how acceleration decreases as time \(t\) increases. The task is to integrate this function to obtain the velocity.

Understanding acceleration is crucial, as it's not only necessary to determine how the velocity changes but also helps interpret real-world motion problems, including how quickly a car might speed up or slow down on a road.

Position Function

The position function describes the location of an object at any given time based on its velocity and acceleration. This is typically represented as \(s(t)\) in calculus.

To find the position function, we integrate the velocity function:

• \(s(t) = \int v(t) \: dt\)
• \(s(t) = -20\ln(t+2) + 30t + D\)

Here, \(D\) is the constant of integration. Using the initial position, \(s(0) = 10\), allows us to solve for \(D\) and obtain:

• \(s(t) = -20\ln(t+2) + 30t + 10 + 20\ln2\)

With this function, one can predict the location of the object at any time \(t\). It combines the effects of initial position, velocity, and continuous acceleration, offering a complete picture of the object's movement along a path.