Problem 77 - Calculus and its applications

Question

Find \(s(t)\). $$ a(t)=-2 t+6, \text { with } v(0)=6 \text { and } s(0)=10 $$

Step-by-Step Solution

Answer

The position function is \( s(t) = -\frac{t^3}{3} + 3t^2 + 6t + 10 \).

Step 1: Understanding Acceleration and Initial Conditions

We are given the acceleration function, which is \[ a(t) = -2t + 6. \]We need to find the position function \( s(t) \), and we are provided with initial conditions: the initial velocity \( v(0) = 6 \) and the initial position \( s(0) = 10 \). To find \( v(t) \), we integrate \( a(t) \). Then, we'll integrate \( v(t) \) to find \( s(t) \).

Step 2: Integrating the Acceleration Function

To find the velocity function \( v(t) \), we integrate the acceleration function:\[ v(t) = \int (-2t + 6) \, dt = -t^2 + 6t + C_v, \]where \( C_v \) is the constant of integration. To find \( C_v \), we use the initial condition \( v(0) = 6 \).

Step 3: Applying Initial Condition to Velocity

Substitute \( t = 0 \) into \( v(t) = -t^2 + 6t + C_v \):\[ v(0) = -0^2 + 6(0) + C_v = 6 \Rightarrow C_v = 6. \]So, the velocity function is:\[ v(t) = -t^2 + 6t + 6. \]

Step 4: Integrating the Velocity Function

Next, to find the position function \( s(t) \), we integrate the velocity function:\[ s(t) = \int (-t^2 + 6t + 6) \, dt = - \frac{t^3}{3} + 3t^2 + 6t + C_s, \]where \( C_s \) is the constant of integration. To find \( C_s \), use the initial condition \( s(0) = 10 \).

Step 5: Applying Initial Condition to Position

Substitute \( t = 0 \) into \( s(t) = -\frac{t^3}{3} + 3t^2 + 6t + C_s \):\[ s(0) = -\frac{0^3}{3} + 3(0)^2 + 6(0) + C_s = 10 \Rightarrow C_s = 10. \]So the position function is:\[ s(t) = -\frac{t^3}{3} + 3t^2 + 6t + 10. \]

Key Concepts

Acceleration Function Integration

In calculus, understanding the connection between acceleration, velocity, and position is crucial. The process begins with the acceleration function, which tells us how the velocity of an object changes over time. Given an acceleration function, such as \(-2t + 6\), the next step is to find the velocity function. This is done through integration. Integration is the reverse process of differentiation, and it combines all the small changes in acceleration to produce the overall effect on velocity. When integrating the acceleration function \(-2t + 6\), you need to add a constant of integration, usually denoted as \(C_v\), resulting in the velocity function:

• \( v(t) = \int (-2t + 6) \, dt = -t^2 + 6t + C_v \)

This constant \(C_v\) is essential for accurately representing the system, as without it, our velocity function wouldn't account for the initial condition given at a specific point in time.

Initial Conditions in Calculus

Initial conditions are vital in solving problems involving integration. They allow us to find the precise form of a function after integration by determining the constants of integration.For velocity

• When we integrate an acceleration function to find velocity, adding a constant \(C_v\) is fundamental. To find \(C_v\), we use the initial velocity condition, \(v(0) = 6\). By substituting this initial condition into the velocity equation, we solve for \(C_v\):
• \( v(0) = -0^2 + 6(0) + C_v = 6 \ \Rightarrow C_v = 6 \)

This confirms our velocity function as \(v(t) = -t^2 + 6t + 6\).For position

• Similarly, when finding the position function \(s(t)\), the initial position condition \(s(0) = 10\) helps us solve for another constant \(C_s\).
• This ensures the accurate depiction of the object's path over time, anchoring the function to the specificed initial point \((t = 0)\).

Initial conditions essentially "anchor" the solutions to real-world scenarios, providing meaningful, practical results.

Position Function Determination

After integrating the velocity function to find the position function, determining the trajectory of the object becomes clear. Let's integrate the velocity function \(-t^2 + 6t + 6\):

• \( s(t) = \int (-t^2 + 6t + 6) \, dt = -\frac{t^3}{3} + 3t^2 + 6t + C_s \)

The result \(-\frac{t^3}{3} + 3t^2 + 6t\) gives us the main part of the position function, but it's not complete until we account for \(C_s\). By applying the initial condition for position \(s(0) = 10\):

• \( s(0) = -\frac{0^3}{3} + 3(0)^2 + 6(0) + C_s = 10 \)
• Solving for \(C_s\) gives us \(C_s = 10\).

Ultimately, the position function becomes \(s(t) = -\frac{t^3}{3} + 3t^2 + 6t + 10\), effectively describing the object's position at any moment based on the initial conditions.