Section 2.13 Using Integrals in Physics
A strategy you will use to solve a variety of physics problems can be thought of as a sequence of steps called Chop-Multiply-Add. You can see these steps outlined in the video above for one example. Below is another example: if you know an object’s acceleration as a function of time and you want to find the change in velocity between some initial and final time.
Chop: You can start by chopping up a time interval into small pieces. Each piece of time is represented by the expression \(dt\text{,}\) where the \(d\) indicates that the actual amount of time is infinitesimally small.
Multiply: You want to know about velocity, but you know about acceleration. Although you want \(\Delta \vec{v}\text{,}\) the change in velocity over a whole time interval, you can start by trying to find an infinitesimal change in velocity \(d\vec{v}\text{.}\) By rewriting the definition of instantaneous acceleration as \(d \vec{v} = \vec{a}(t) dt\text{,}\) you can see that a way to get \(d\vec{v}\) is to multiply \(\vec{a}(t)\) and \(dt\text{.}\)
Add: Last, once you have the infinitesimal \(d\vec{v}\text{,}\) you need to add together every \(d\vec{v}\) over the full times of interest, which can be done with an integral:
\begin{equation*}
\int_{v_i}^{v_f}d \vec{v} = \int_{t_i}^{t_f} \vec{a}(t) dt
\end{equation*}
The left hand side of this equation can be easily evaluated to give
\begin{equation*}
\Delta \vec{v} = \int_{t_i}^{t_f} \vec{a}(t) dt
\end{equation*}
Exercises Activities
1.
The following two equations are both reasonable to write down for a displacement vector:
\begin{equation*}
1: \Delta \vec{r} = \int_{t_i}^{t_f}\vec{v}(t)dt
\end{equation*}
\begin{equation*}
2: \Delta \vec{r}= \vec{v}\Delta t
\end{equation*}
What is the difference between them? In what situations would you use the first equation? In what situations would you use the second equation?
