Problem 78 - Calculus and its applications

Question

Find \(s(t)\). $$ a(t)=-6 t+7, \text { with } v(0)=10 \text { and } s(0)=20 $$

Step-by-Step Solution

Answer

The position function is \(s(t) = -t^3 + \frac{7}{2}t^2 + 10t + 20\).

Step 1: Integrate acceleration to find velocity

To find the velocity function \(v(t)\), we need to integrate the given acceleration function \(a(t) = -6t + 7\). The general formula is \(v(t) = \int a(t) \, dt\). So, we integrate:\[v(t) = \int (-6t + 7) \, dt = -6\frac{t^2}{2} + 7t + C_1\]Simplified, the velocity function is:\[v(t) = -3t^2 + 7t + C_1\]

Step 2: Solve for the constant of integration in velocity

We know that \(v(0) = 10\). Using the velocity function from Step 1, substitute \(t = 0\):\[v(0) = -3(0)^2 + 7(0) + C_1 = 10\]Thus, \(C_1 = 10\). Therefore, the velocity function is\[v(t) = -3t^2 + 7t + 10\]

Step 3: Integrate velocity to find position

Now, we need to find the position function \(s(t)\) by integrating the velocity function \(v(t) = -3t^2 + 7t + 10\). So, we integrate:\[s(t) = \int (-3t^2 + 7t + 10) \, dt = -3\frac{t^3}{3} + 7\frac{t^2}{2} + 10t + C_2\]Simplified, the position function is:\[s(t) = -t^3 + \frac{7}{2}t^2 + 10t + C_2\]

Step 4: Solve for the constant of integration in position

We know that \(s(0) = 20\). Using the position function from Step 3, substitute \(t = 0\):\[s(0) = -0 + 0 + 0 + C_2 = 20\]Thus, \(C_2 = 20\). Therefore, the position function is\[s(t) = -t^3 + \frac{7}{2}t^2 + 10t + 20\]

Key Concepts

Integration

Integration is a fundamental concept in calculus that is often seen as the reverse operation of differentiation. When we have a function that describes a rate of change, like acceleration, integration helps us find a function that describes the accumulated quantity, such as velocity or position.

In this particular problem, we start with the acceleration function and find the velocity by integrating it. The formula to find the velocity function from acceleration is:

• \(v(t) = \int a(t) \, dt\)

This means we calculate the integral of the acceleration function to find the velocity. Similarly, to find the position function from velocity, we integrate the velocity function:

• \(s(t) = \int v(t) \, dt\)

Integration gives us not only the general form of these functions but also constants of integration that we determine using initial conditions.

Acceleration

Acceleration describes how velocity changes over time. It is a vector quantity, which means it has both magnitude and direction. In this problem, the acceleration function is given by \(a(t) = -6t + 7\).

The negative sign in front of \(-6t\) indicates that this part of the function is decelerating the motion. Meanwhile, the constant \(+7\) acts as a positive force, contributing to increase in velocity. To understand the motion of a particle, examining the acceleration gives insights into how quickly the velocity is changing and in which directions.

When working with problems involving acceleration, knowing the initial conditions such as initial velocity or position (as given here with \(v(0)=10\) and \(s(0)=20\)) is crucial. This helps in solving for the constants of integration after performing the required integrations.

Velocity

Velocity is a measure of how fast something is moving and in which direction. It’s derived from the integral of the acceleration function, as demonstrated in our exercise.

For this problem, we take the acceleration \(a(t) = -6t + 7\) and integrate it to obtain the velocity function. Specifically, integrating gives:

• \(v(t) = -3t^2 + 7t + 10\)

The constants of integration, like \(C_1\), are determined using known values, such as \(v(0)=10\). Velocity is essential because it serves as a bridge between acceleration and position. The information that velocity provides helps us understand motion patterns over time. Consider that positive and negative values indicate direction, speed, and shifts in motion.

Position Function

The position function describes the trajectory or path of an object over time, often denoted as \(s(t)\). It gives us the actual position of the particle at any given time \(t\).

In our task, finding the position function involved integrating the velocity function \(v(t) = -3t^2 + 7t + 10\). By doing so, we obtained:

• \(s(t) = -t^3 + \frac{7}{2}t^2 + 10t + 20\)

Each term in the position function provides insight about the motion of the object. The cubic and quadratic components signify changes in acceleration and velocity over time.

The initial position \(s(0)=20\) helps confirm the constant of integration \(C_2\), ensuring that the function starts at the correct location on a graph. Understanding the position function is crucial for predicting where an object will be at a certain time.