Question

Given the following velocity functions of an object moving along a line, find the position function with the given initial position. Then graph both the velocity and position functions. $$v(t)=2 t+4 ; s(0)=0$$

Short answer

Answer: The position function, s(t), is s(t) = t^2 + 4t.

Step-by-step solution

Integrate the velocity function with respect to time

To find the position function from the velocity function, we need to integrate the velocity function with respect to time \(t\). So, calculate the integral: $$s(t) = \int (2t+4) \ dt$$

Solve the integral

Now, we need to solve the integral: $$s(t) = \int (2t+4) \ dt = t^2+4t+C$$ Here, \(C\) is the integration constant.

Use the initial condition to find the integration constant

We are given that the initial position is \(s(0)=0\). Plug the initial condition into the position function to find the integration constant: $$0 = (0)^2 + 4(0) + C$$ $$C = 0$$

Write the position function

Now that we have found the integration constant, we can write the final position function: $$s(t) = t^2 + 4t$$

Graph the velocity and position functions

Now that we have both the velocity function, \(v(t)=2t+4\), and the position function, \(s(t)=t^2+4t\), we can graph them together. Make sure to label your axes and clearly differentiate the velocity and position functions on the graph. - The velocity function, \(v(t)=2t+4\), will be a straight line with slope 2 and intercept 4. - The position function, \(s(t)=t^2+4t\), will be a parabolic curve opening upwards. You can plot the functions using graphing software such as Desmos, GeoGebra, or a graphing calculator, or create a graph by hand.

Key concepts

Velocity Function

The velocity function is a way to express how quickly an object is moving along a straight line over time. In our case, we're examining the velocity function given by \(v(t) = 2t + 4\). This equation tells us two main things about the object's motion:

• **Slope (2)**: It represents the constant rate of increase in velocity, which means the object's speed increases by 2 units for every unit of time.
• **Intercept (4)**: This is the starting velocity when time \(t = 0\), showing the object is already moving with some initial speed rather than starting from rest.

In simpler words, every time the clock ticks, the object is not only moving faster but starting from a speed of 4 units. If you were to draw this on a graph, it would be a straight line sloping upwards.

Calculus

Calculus is the branch of mathematics that deals with rates of change and accumulation. It is split into two main parts, differential calculus and integral calculus. Here, we focus on how.

• **Differential Calculus**: It helps us find the velocity function from the position function by determining the rate at which position changes. Imagine if you wanted to know how fast you are moving at any given moment—that's what differential calculus tells you.
• **Integral Calculus**: Conversely, here we're interested in finding the position function from the velocity function, focusing on the essential concept of integration.

In this exercise, we use integral calculus to accumulate the total change in velocity over time, essentially reconstructing the path, or the position, from what's known about the speed.

Integration

Integration is the mathematical process used to find a function (position) given its rate of change (velocity). When we integrate the velocity function \(v(t) = 2t + 4\), we accumulate its effects over time to reconstruct the position function \(s(t)\).Here's a step-by-step of what happens during integration:- **Finding the Antiderivative**: We perform the integration \(s(t) = \int (2t + 4)\, dt\), which involves finding an antiderivative of the function, resulting in \(t^2 + 4t + C\).- **Integration Constant \(C\)**: This arises because integration can determine general forms of a function, leaving out the specifics unless given conditions (like initial positions).- **Applying Initial Conditions**: We utilize the condition \(s(0) = 0\) to solve for \(C\), ensuring the position function matches the known starting point. Here, we found \(C = 0\), simplifying the position function to \(s(t) = t^2 + 4t\).This process showcases how integration helps us not just understand an object's past path but predict where it will be going forward, given its speed changes over time.