Just as distance and displacement have different definitions despite their similarities, so do speed and velocity. Speed is a scalar quantity that indicates “how fast an object moves,” while velocity is a vector quantity that refers to “the rate at which an object changes its position.” This notion may seem perplexing, but remember that vectors are always related to their orientation. Furthermore, in Physics, we are more concerned with velocity rather than speed since we also want to quantify the direction. However, before we can master this concept, we must first learn how to solve some problems without worrying its direction.
Velocity is usually represented by the symbol $\vv{v}$. Velocity, especially in real-world applications, may be more challenging to quantify than it seems, especially when referring to the instantaneous velocity.
Let us have a simple example to illustrate this notion. For instance, a man traveled between cities. His velocity would undoubtedly vary due to traffic, curves, and other unanticipated situations. Yes, the man would be able to see his instantaneous velocity on his odometer, but calculating it would be daunting since it will require some Calculus. Even so, not all parameters can be accounted for.
However, in this lesson, we are more concerned with what we call the average velocity. To determine the average velocity between two points, we can divide the total displacement by the total time to finish the course. Average velocity is denoted as,
$$\begin{align} \text{average velocity} &= \overline{\vv{v}}=\frac{\Delta \vv{s}}{\Delta t} \label{eq:average velocity}\\ \text{average speed} &= \overline{v}=\frac{ \text{total distance} }{\Delta t} \end{align}$$
$\example{1}$ If a car travelled 550 km in 10 hours. Determine (a) the car’s average velocity, (b) how far will it go in 11 hours at the same velocity, and (c) how long will it take to reach 350 km at the same velocity.
$\solution$
$\For{a}$
To determine the average velocity use $\eref{eq:average velocity}$,
$\For{b}$
To determine the displacement at the the same velocity, rearrange $\eref{eq:average velocity}$.
Since $\overline{\vv{v}} = 55 \un{km/hr}$,
\[\begin{align*} &= (55 \un{km/\cancel{\mathrm{hr}}})(11\,\cancel{\text{hrs}}) \\ &= 605 \un{km} \tagans \end{align*}\]$\For{c}$
To determine the time, rearrange $\eref{eq:average velocity}$,
$\example{2}$ A class have their field trip, it took them 2 hours to reach the first stop which is 100 km from its starting point, then another 3 hours to reach the second stop which is 165 km from the first stop. Determine the average velocity of the trip.
$\solution$
The average velocity is defined as the total total displacement divided by the total time it took to reach its final destination. Thus,
\[\begin{align*} \overline{\vv{v}} &= \frac{\Delta \vv{s}}{\Delta t} \\ &= \frac{ 100\un{km}+165\un{km} }{ 2\un{hrs}+3\un{hrs} } \\ &= \frac{ 265\un{km} }{ 5\un{hrs} } \\ &= 53\un{km/hr} \tagans \end{align*}\]Instantaneous Velocity
Suppose a stray dog chased a student, his initial instinct tells him to run as fast as possible, say he is running at a rate of 10 m/s, but as time passed by, he starts slowing down to a rate of 4 m/s, then somehow realized that the dog already stopped chasing him, so he then stopped running.
Instantaneous velocity, is the velocity at a certain instant, so at the time he hit his velocity at 10 m/s that was his instantaneous velocity at that certain moment. And at the time he stopped, he has an instantaneous velocity of 0 m/s. Hence, instantaneous velocity is denoted as,
$$\begin{align} \vv{v} = \frac{ds}{dt} = \lim_{\Delta t \to 0} \frac{\Delta\vv{s}}{\Delta{t}} \end{align}$$
But we will not dwell on this for now, since we’ll need calculus to evaluate this type of problem.