Problem 25 - 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)=\left\\{\begin{array}{ll} t / 2 & \text { if } 0 \leq t \leq 2 \\ 1 & \text { if } 2

Step-by-Step Solution

Answer

The position of the object at time \( t = 4 \) is 3 units.

Step 1: Understand the Problem

We are given a piecewise velocity function and need to find the position of an object at time \( t = 4 \), given that it starts at the origin \((s(0) = 0)\). To do this, we will integrate the velocity function over the given intervals to find the displacement.

Step 2: Integrate over the First Interval (0 to 2)

For \( 0 \leq t \leq 2 \), the velocity function is \( v(t) = \frac{t}{2} \). The position function is the integral of the velocity function, so we calculate:\[s(t) = \int v(t) \, dt = \int \frac{t}{2} \, dt = \frac{1}{2} \left( \frac{t^2}{2} \right) = \frac{t^2}{4}\]Evaluating from \( t = 0 \) to \( t = 2 \):\[s(2) = \frac{2^2}{4} = 1\]This gives the position at \( t = 2 \) as 1 unit.

Step 3: Integrate over the Second Interval (2 to 4)

For \( 2 < t \leq 4 \), the velocity function is \( v(t) = 1 \). We integrate to find the additional displacement:\[s(t) = \int v(t) \, dt = \int 1 \, dt = t\]Evaluate from \( t = 2 \) to \( t = 4 \):\[s(4) - s(2) = (4 - 2) = 2\]The additional displacement from time \( t = 2 \) to \( t = 4 \) is 2 units.

Step 4: Add Displacements to Find Total Position at t = 4

The total displacement is the sum of the displacements from both intervals. From Step 2, the object moved 1 unit during \( 0 \leq t \leq 2 \). From Step 3, it moved an additional 2 units during \( 2 < t \leq 4 \). Thus, the total position at \( t = 4 \) is:\[s(4) = 1 + 2 = 3\]

Key Concepts

Understanding Integration

In calculus, integration is a fundamental concept used to find quantities such as areas under curves and total accumulations. When we integrate a velocity function with respect to time, we find the displacement, which helps us determine the object's position along its path. In this exercise, the given velocity function tells us how fast an object is moving at any given time.

• The integral of a function calculates the area under the curve of that function within specified limits.
• For velocity, integration gives us the total distance or displacement over an interval of time.
• Each segment of a piecewise function needs to be integrated separately over its respective interval to get a complete understanding of the object's motion.

These steps help us translate the abstract movement described by a velocity function into a tangible position value, starting from an initial point such as the origin in this problem.

Velocity Function in Piecewise Form

A velocity function expresses the rate of change of an object's position with respect to time, often as a function of time itself. The function provided in the problem is a piecewise velocity function, which means it is defined by different expressions over different intervals.

• For the interval from 0 to 2, the velocity function is given by \( v(t) = \frac{t}{2} \), indicating that velocity increases linearly with time.
• For the interval from 2 to 4, the velocity function is constant at 1, representing uniform motion.

Understanding how to interpret and apply each piece helps in accurately calculating the object's displacement for each time segment. By dealing with each interval separately, we can precisely track how the object's velocity changes over time.

Calculating the Position Function

Once the velocity function is integrated over its respective intervals, we obtain what is known as the position function. The position function tells us the exact position of an object at any given time. Starting from the origin at the time \( t = 0 \), the displacement found through integration gives the total position.In this problem:

• Integrating \( v(t) = \frac{t}{2} \) from \( t = 0 \) to \( t = 2 \) results in a displacement of 1 unit by the time the object reaches \( t = 2 \).
• Integrating \( v(t) = 1 \) from \( t = 2 \) to \( t = 4 \) yields an additional displacement of 2 units.
• Adding these two displacements gives us the total position at \( t = 4 \), which is 3 units from the origin.

Each segment of the piecewise velocity function contributes to the total displacement, providing a complete position trajectory for the object.