Question

Given the following velocity functions of an object moving along a line, find the position function with the given initial position. $$v(t)=2\sqrt{t};s(0)=1$$

Short answer

Answer: The position function s(t) is given by s(t) = (4/3)t^(3/2) + 1.

Step-by-step solution

Integrate the velocity function

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

Use substitution to simplify the integral

Let u = t, then du = dt. Replace t with u and dt with du in the integral: $$s(t)=\int 2\sqrt{u}du$$

Integrate

Now, integrate the function with respect to u: $$s(t)=\int 2\sqrt{u}du=2\int u^{1/2}du=2\frac{u^{3/2}}{\frac{3}{2}}+C=\frac{4}{3}{u}^{3/2}+C$$

Replace u with t

Replace u back with t to write the position function s(t) in terms of t: $$s(t)=\frac{4}{3}{t}^{3/2}+C$$

Use the initial condition to find the constant of integration

The initial condition given is s(0) = 1. Plug in t = 0 into the position function s(t) and set s(t) equal to 1: $$1=\frac{4}{3}(0{)}^{3/2}+C$$ Solve for C: $$C=1$$

Write the final position function

The position function s(t) with the constant of integration is: $$s(t)=\frac{4}{3}{t}^{3/2}+1$$

Key concepts

Velocity Function

A velocity function, in mathematics, describes the rate of change of an object's position with respect to time. This function gives you an idea of how fast an object is moving at any given time. When you're given a velocity function like $v(t)=2\sqrt{t}$, it tells us how the velocity of the object is related to time $t$.
For this specific function, as $t$ increases, the velocity also increases because $\sqrt{t}$ grows larger. Understanding the velocity function is a crucial first step when trying to determine the object's position. The velocity function is the derivative of the position function, meaning it shows how position changes over time.

Position Function

The position function provides the position of an object at any given time. It's derived from the velocity function by integrating it with respect to time. Integration is essentially about adding up all the tiny changes in velocity to find out the total change in position over a time period.
For instance, by integrating the velocity function $v(t)=2\sqrt{t}$, one finds the position function $s(t)=\frac{4}{3}{t}^{3/2}+C$. This expression tells you the object's position for any time $t$, but it includes an unknown constant, $C$. This form shows the cumulative effect of the velocity over time, but to determine the exact position, the constant must be solved for using other information, like the initial condition.

Constant of Integration

When you integrate a function, you often end up with a constant of integration—an unknown factor represented as $C$. This constant exists because integration can only determine changes in position, not the starting point.
This constant needs additional information, or an initial condition, to be determined. In the derived position function $s(t)=\frac{4}{3}{t}^{3/2}+C$, $C$ represents this undetermined starting position before you apply the initial condition. Determining the correct $C$ is crucial as it anchors the position function to a specific starting point, allowing you to predict the exact position at any time.

Initial Condition

An initial condition tells you some specific information about the system at a particular time, usually at $t=0$. It is an essential tool for finding out the constant of integration. For example, if $s(0)=1$, it tells us that when time $t$ is zero, the position of the object is already at 1.
By plugging this information into the position function $s(t)=\frac{4}{3}{t}^{3/2}+C$, you can solve for the constant $C$. Using the initial condition here, since $s(0)=1$ and $\frac{4}{3}(0{)}^{3/2}=0$, we find that $C=1$. This allows us to finalize the position function, incorporating both the variable part and the initial position.