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. $$v(t)=2 \sqrt{t} ; s(0)=1$$

Short answer

Question: Given the velocity function, $$v(t) = 2\sqrt{t}$$, find the position function considering an initial position. Then, describe the graphs of both the velocity and position functions. Answer: First, integrate the given velocity function with respect to time (t) to find the position function: $$s(t) = \int 2\sqrt{t} dt$$. The position function will be in the format $$s(t) = at^b + C$$, where $$C$$ is the initial position. The graphs of the velocity function and the position function will show how the object moves over time with different slopes and curvatures.

Step-by-step solution

Integrate the velocity function

To find the position function, integrate the given velocity function with respect to time (t): $$s(t) = \int v(t) dt= \int 2\sqrt{t} dt$$

Key concepts

Position Function

In the world of calculus, the position function provides crucial insights into the motion of an object. It tells us where an object is at any given time.
To find this function, you typically start with a velocity function, which describes how fast and in what direction the object is moving. By integrating the velocity function, we can derive the position function.
A key point to remember is the initial position, often denoted as \(s(0)\). This value is essential for determining the constant of integration when calculating the position function.
In our exercise, the integration of the velocity function \(v(t) = 2\sqrt{t}\) helps us find the position function \(s(t)\), showing us how the object moves over time.

Velocity Function

The velocity function \(v(t)\) describes the speed and direction of a moving object. It's like a snapshot of how fast something is going and which way.
This function is fundamental in physics and calculus as it relates directly to how things move. True to its name, velocity is about quickness and path.
In this exercise, given \(v(t)=2\sqrt{t}\), we can determine how the object’s velocity changes as time progresses. Here, as time increases, the velocity decreases, as the square root function grows slower compared to linear functions.
From this, we can infer that the movement is not constant but rather changes predictably with time.

Integration Techniques

When dealing with calculus problems, integrating functions is a powerful technique. Integration allows us to accumulate quantities, like finding the total distance traveled (position function) from a rate of speed (velocity function).
To integrate the velocity function \(v(t) = 2\sqrt{t}\), we leverage the power of substitution or transformation techniques. Specifically, we treat \(\sqrt{t}\) as \(t^{1/2}\), making it easier to integrate.
Following the rules of integration, increase the power by one (from \(1/2\) to \(3/2\)) and divide by this new power, resulting in \(\frac{2}{3} t^{3/2}\) plus the constant of integration \(C\). This process turns the velocity function into our desired position function, providing a complete picture of an object's movement.
Remember, finding \(C\) involves applying initial conditions, like \(s(0) = 1\), to our solution.