Problem 75 - Calculus and its applications

Question

Find \(v(t)\). $$ a(t)=4 t, \quad v(0)=20 $$

Step-by-Step Solution

Answer

The velocity function is \( v(t) = 2t^2 + 20 \).

Step 1: Understand the Problem

You are given the acceleration function \( a(t) = 4t \) and an initial condition \( v(0) = 20 \). You need to find the velocity function \( v(t) \).

Step 2: Integrate the Acceleration Function

To find the velocity function \( v(t) \), integrate the given acceleration function \( a(t) = 4t \). The integral of \( a(t) \) is \( v(t) = \int 4t \, dt \).

Step 3: Perform the Integration

Evaluate the integral \( \int 4t \, dt \). This gives \( v(t) = 2t^2 + C \), where \( C \) is the constant of integration.

Step 4: Use the Initial Condition

Use the initial condition \( v(0) = 20 \) to find \( C \). Substitute \( t = 0 \) and \( v(0) = 20 \) into \( v(t) = 2t^2 + C \). This gives \( 20 = 2(0)^2 + C \), so \( C = 20 \).

Step 5: Write the Velocity Function

Substitute \( C = 20 \) back into the expression for \( v(t) \). The velocity function is \( v(t) = 2t^2 + 20 \).

Key Concepts

Velocity Functions

In calculus, a velocity function is instrumental for understanding how the speed of an object changes over time. The velocity function, often represented as \( v(t) \), gives the rate of change of displacement concerning time.
It tells us both the speed and the direction of an object's motion at any given instant. To find a velocity function from an acceleration function, one typically needs to integrate the acceleration function. For the problem at hand, we're given that the acceleration function is \( a(t) = 4t \). Integrating this function gives us the velocity function, reflecting how velocity accumulates over time starting from the acceleration provided.

It's key to remember that

• Velocity is the first derivative of the displacement with respect to time.
• It is also the integral of acceleration with respect to time.

These relationships allow us to transition between different descriptions of motion easily.

Initial Conditions

Initial conditions play a crucial role in solving differential equations by helping to determine the constant of integration after performing an integration. In this exercise, we started with the acceleration function \( a(t) = 4t \) and integrated it to find the velocity function as \( v(t) = 2t^2 + C \).
The constant \( C \) is crucial because it accounts for whatever specific characteristics pertain to the initial state of motion.The given initial condition, \( v(0) = 20 \), specifies that at time \( t = 0 \), the velocity is 20. This information is used to solve for the constant \( C \) by substituting these values into the velocity function:
\( v(0) = 20 = 2(0)^2 + C \), which directly gives \( C = 20 \). Once we find \( C \), we substitute back to get a complete expression for the velocity function, tailored to the specific scenario defined by the problem's initial conditions. Initial conditions ensure that our mathematical model accurately reflects realistic and specific situations.

Acceleration Functions

Acceleration functions like \( a(t) = 4t \) are a key part of calculus and physics. They express how velocity changes over time. The function \( a(t) = 4t \) suggests that the acceleration is not constant. Instead, it increases linearly over time, meaning the rate of change of velocity increases as time progresses.
Acceleration is the derivative of velocity with respect to time. Therefore, to reverse-engineer the velocity from acceleration, we integrate the acceleration function.
This approach gives us insight into how an object's speed evolves and provides a deeper understanding of the motion itself.

• Acceleration represents how quickly an object "speeds up" or "slows down."
• It combines both direction and magnitude, similar to velocity.

Understanding acceleration functions is critical for analyzing motion in calculus, as these functions directly influence both the velocity and the eventual displacement of objects.