Problem 76 - Calculus and its applications

Question

Find \(v(t)\). $$ a(t)=6 t, \quad v(0)=30 $$

Step-by-Step Solution

Answer

The velocity function is \(v(t) = 3t^2 + 30\).

Step 1: Understand the Problem

We need to find the velocity function \(v(t)\). We are given the acceleration function \(a(t) = 6t\) and the initial velocity \(v(0) = 30\). Velocity is the integral of acceleration.

Step 2: Set Up the Integral

The velocity function \(v(t)\) is the integral of the acceleration function \(a(t)\). This is represented by the equation:\[ v(t) = \int a(t) \, dt \] Substituting the given acceleration, the equation becomes:\[ v(t) = \int 6t \, dt \]

Step 3: Integrate the Acceleration Function

Perform the integration on the right side:\[ v(t) = \int 6t \, dt = 6 \int t \, dt \]The integral \(\int t \, dt\) is \(\frac{t^2}{2}\). So:\[ v(t) = 6 \left( \frac{t^2}{2} \right) = 3t^2 \]

Step 4: Apply the Initial Condition

We know \(v(0) = 30\). Substitute \(t = 0\) into the velocity function:\[ v(0) = 3(0)^2 + C = 30 \]Since \(3(0)^2 = 0\), it simplifies to:\[ C = 30 \]

Step 5: Combine the Results

Now that we have determined \(C = 30\), substitute back into \(v(t)\):\[ v(t) = 3t^2 + 30 \]

Key Concepts

Integration

In calculus, integration is a fundamental concept used to find quantities based on a rate of change. This is especially useful when determining values such as area under a curve or finding the original function from its derivative. In this context, when we know the acceleration function, which is a derivative of velocity, integration allows us to find the velocity function itself.The integration of a function essentially reverses differentiation. For example, given an acceleration function, integrating it with respect to time ( t ) will yield the velocity function. To represent it mathematically:

• If a(t) is the acceleration function, then the velocity function v(t) can be found as \( v(t) = \int a(t) \, dt \) .

Integration constantly pops up in physics and engineering when dealing with motion and rates of change.

Velocity Function

The velocity function, denoted as v(t) , describes how fast an object is moving over time. It is the first derivative of the position function concerning time and the antiderivative or integral of the acceleration function.Given our exercise, the velocity function is derived from an acceleration formula a(t) = 6t . By integrating 6t with respect to time (t) , we find the equation:

• \( v(t) = \int 6t \, dt = 3t^2 + C \)

Where C is the integration constant that will later be determined using an initial condition. The velocity function thus provides a mathematical model for predicting future velocities based on a known rate of acceleration.

Initial Condition

Initial conditions are used to solve for any constants that appear after integration. When calculating integrals, an extra constant ( C ) is included because the integral of a function is not unique without additional information.In our scenario, the given initial condition is that at time t = 0 , the velocity ( v(0) ) is 30. By plugging this into the integrated formula, we solve for C:

• \( v(0) = 3(0)^2 + C = 30 \)
• This simplifies to \( C = 30 \)

This is crucial as it personalizes the velocity function to fit the specific motion conditions under consideration. In physical applications, such initial conditions are vital to create equations that accurately represent real-world phenomena.

Acceleration Function

The acceleration function tells us how quickly an object's velocity is changing over time. It is the derivative of the velocity function concerning time (t) .In this task, we start with the acceleration function a(t) = 6t. It signals that acceleration is directly proportional to time— as time progresses, acceleration increases.From this relationship, by integrating 6t, we derive the velocity function.

• Thus, \( v(t) = \int 6t \, dt = 3t^2 + C \)

Understanding the acceleration function is key in physics, especially to predict motion behaviors and to reverse-engineer velocity or position data. This helps engineers and physicists design and interpret systems related to dynamic movements.