Problem 24 - Calculus

Question

The velocity function for an object is given. \(A s-\) suming that the object is at the origin at time \(t=0\), find the position at time \(t=4\). $$ v(t)=1+2 t $$

Step-by-Step Solution

Answer

The position of the object at time \( t = 4 \) is 20.

Step 1: Understand the Problem

We are given the velocity function \( v(t) = 1 + 2t \). We need to find the position of the object at \( t = 4 \). Since it's given that the object is at the origin at time \( t = 0 \), the initial position \( s(0) = 0 \).

Step 2: Find Position as a Function of Time

The position function \( s(t) \) is the integral of the velocity function \( v(t) \). We can write it as: \[ s(t) = \int v(t) \, dt = \int (1 + 2t) \, dt \].

Step 3: Integrate the Velocity Function

Calculate the integral: \[ \int (1 + 2t) \, dt = \int 1 \, dt + \int 2t \, dt \].The solutions to these are:\( \int 1 \, dt = t \) and \( \int 2t \, dt = t^2 \).So the position function is: \[ s(t) = t + t^2 + C \], where \( C \) is the constant of integration.

Step 4: Use Initial Condition to Find Constant

We were given that \( s(0) = 0 \). Plug \( t = 0 \) into the position equation: \[ s(0) = 0 + 0^2 + C = 0 \].Solving this gives \( C = 0 \). Hence, the position function is \[ s(t) = t + t^2 \].

Step 5: Calculate the Position at t = 4

Use the position function \( s(t) = t + t^2 \) to find the position at \( t = 4 \):\[ s(4) = 4 + 4^2 = 4 + 16 = 20 \].

Key Concepts

Velocity Function

In calculus, the velocity function represents how an object's speed and direction change over time. It's essentially the derivative of the position function. If you think of walking on a path, velocity tells you how fast and in which direction you're moving.
For example, when given a velocity function such as \( v(t) = 1 + 2t \), this means:

• The object starts at a speed of 1 unit per time interval when \( t = 0 \).
• With each increment in time \( t \), the speed increases by 2 units.

Understanding the velocity function helps you predict future positions or determine past positions based on changes in speed.

Position Function

The position function describes an object's location at any given point in time. To derive this from a velocity function, we calculate the integral. In the example, if \( v(t) = 1 + 2t \), integrating this gives the position function:\[s(t) = \int (1 + 2t) \, dt = t + t^2 + C\]Where \( C \) is a constant determined by initial conditions. Initially at the origin \( t=0 \), the position \( s(0) \) is zero, meaning:\[C = 0\]Thus, the position function becomes \( s(t) = t + t^2 \). This function tells you where the object will be at any future time \( t \). Knowing this, you can figure out where the object will be when \( t = 4 \), which in this exercise, resulted in a position of 20 units.

Integration

Integration is the mathematical process of finding a function given its derivative. It is essentially the reverse of differentiation. By integrating a velocity function, we obtain the position function. This process involves antidifferentiating components of the velocity function independently.
For our example of \( v(t) = 1 + 2t \), the integration steps were:

• Finding the integral of 1, which results in \( t \).
• Finding the integral of \( 2t \), resulting in \( t^2 \).

Combining these results, we formed \( s(t) = t + t^2 + C \). The constant \( C \) is solved using given conditions, ensuring the position function correctly reflects the object's path.
Integration helps connect instantaneous rates of change, like velocity, to cumulative quantities, like position. This concept is vital in physics and engineering, where predicting motion and position based on changing circumstances is crucial.