Integrals in Kinematics

In this lesson we talk about where integrals appear in kinematics. We will see integrals in one simple example, but the same idea appears in many other contexts in kinematics.

Let's suppose we have a particle $A$ moving along the $x$ axis, and suppose that it's velocity is $v(t)=4$ and initial position is $x=0$.

Easy enough! This is just our simple $\text{distance} = \text{rate} \times \text{time}$.

But what if our velocity is more complex, like $v(t)=t^2$, then what do we do?

Now let's suppose we have a particle $B$ moving along the $x$ axis with $v(t)=t^2$ and initial position is $x=0$. Let's try to find how far the particle travels from $0$ to $3$.

...so let's estimate that the velocity of $B$ for $0\le t< 1$ is $0$. (Of course, this is an absurd assumption, but we will see soon this is not so absurd)

...so let's estimate that the velocity of $B$ for $1\le t< 2$ is $1$.

...so let's estimate that the velocity of $B$ for $2\le t< 3$ is $4$.

Correct! We can approximate that $B$ travels $0\times 1s = 0$ units from $t=0$ to $1$, $1\times 1s = 1$ units from $t=1$ to $2$, and $4\times 1s = 4$ units from $t=2$ to $3$. This gives 5 unites total.

Great! Let's suppose we use $n$ equally spaced intervals from $0$ to $3$.

The velocity on the $k$th time interval will be $(3k/n)^2$ and the size of each time interval is $3/n$, so the total distance traveled on the $k$th time interval is $(3/n)\cdot(3k/n)^2 = 27k^2/n^3$ $$\sum_{k=0}^n \frac{9k^2}{n^3}$$ And for $n=10$ this is about $7.70$, and for $n=100$, this is about $8.87$.

If we really want to make this completely accurate, we should take an infinite number of intervals, and this is where integrals comes in. Our $3/n$ interval sizes will become $dt$, and the velocity on that interval will be just $t^2$, so we get the exactly value of the distance traveled is $$\int_0^3 t^2dt = \frac{t^3}{3}\mid_0^3 = \frac{27}{3}-0=\boxed{9}$$