Velocidade: Aceleração e Aceleração instantanea

Remember the definition of speed: it's the rate of change in position over time. Knowing that, let's start with a basic example:

To answer this question, we use the famous equation $$\text{distance}=\text{speed}\times \text{time},$$ which gets us $$\text{speed}=\frac{120\text{ miles}}{3\text{ hours}}=40\text{ mph}.$$

Great! But there's one piece of information that this doesn't give us: the direction. Jim's average speed of $30$ miles per hour doesn't tell us whether he was driving north, south, east, west, or any other direction!

This is the main difference between speed and velocity: while speed tells us just the numerical value of the rate of change, the velocity also tells us the direction. (In general, quantities like speed are called scalars, while quantities like velocity are called vector quantities. So, we denote speed with a regular variable $s$, while for velocity we add a vector above the symbol: $\overrightarrow{v}$.)

[One other note before we begin: directions technically exist in any number of dimensions, but for the sake of single-variable calculus we will only introduce velocity in one dimension.]

Now let's look at the case when Jim drives in one dimension, or in other words, a line.

In this case, Jim can only travel left or right. So, conventionally we set the velocity to be positive when Jim travels to the right, and negative when Jim travels to the left.

Let's look back at the previous example:

To summarize: if Jim is driving $30$ miles per hour to the right, his velocity would be $30$mph, whereas if he drives to the left, his velocity would be $-30$mph.

Great! But so far, we've only been talking about average velocity. But what if Jim's speed wasn't constant? What if Jim drove $110$ miles in the first hour, and then the remaining $10$ miles in the next two hours? This is why we have to distinguish between average and instantaneous velocity.

To find out the exact difference, let's define these terms rigorously: suppose Jim's position is now given as a function of time- say $\overrightarrow{x}(t)$. As before, we can define average speed as the total displacement divided by the total time. So, the average velocity from time $t=t_i$ to $t=t_f$ is given by $$\overrightarrow{v}_{\text{av}}=\frac{\overrightarrow{x}(t_f)-\overrightarrow{x}(t_i)}{t_f-t_i}.$$ Observe that if Jim's position at $t=t_f$ is at the left of his initial position, then the numerator of the expression would be negative, thus accounting for the direction.

Now you try!

By the expression for average velocity above, we have $$\overrightarrow{v}(t)_{\text{av}}=\frac{\overrightarrow{x}(4)-\overrightarrow{x}(2)}{4-2}=\frac{-16-(-4)}{4-2}=-6.$$

Now, let's talk about instantaneous velocity. As its name suggests, instantaneous velocity tells us about Jim's speed and direction at a specific time. To start, recall the expression for average velocity: $$\overrightarrow{v}_{\text{av}}=\frac{\overrightarrow{x}(t_f)-\overrightarrow{x}(t_i)}{t_f-t_i}.$$ What happens to the average velocity as $t_f$ becomes very close to $t_i$ (i.e. the change in time becomes really small)? We can rigorously write this as $$\lim_{t_f\to t_i} \frac{\overrightarrow{x}(t_f)-\overrightarrow{x}(t_i)}{t_f-t_i}.$$ But wait a second! This is the definition of the derivative of $\overrightarrow{x}(t)$ at $t=t_i$, provided that the function is differentiable at that point. So, we define this quantity as the instantaneous velocity at $t=t_i$: $$\overrightarrow{v}_{\text{inst}}=\lim_{t_f\to t_i} \frac{\overrightarrow{x}(t_f)-\overrightarrow{x}(t_i)}{t_f-t_i}=\frac{d\overrightarrow{x}}{dt}\bigg|_{t=t_i}.$$ This definition makes sense intuitively, since an instantaneous change would have to be around a single point.

Let's try an example!

To recap: since $$\frac{d\overrightarrow{x}}{dt}=\frac{d}{dt} (-t^2)=-2t,$$ the instantaneous velocity at $t=4$ is equal to $$\overrightarrow{v}_{\text{inst}}=-2t\bigg|_{t=4}=-8.$$

Here's another problem:

Surprisingly, the third choice is right! To see why, let's graph $\overrightarrow{x}(t)$:

When $t=4$, we notice a ``corner'' on the position-time curve. This tells us that at that specific point, this function is not differentiable, meaning that $$\frac{d\overrightarrow{x}(t)}{dx}\bigg|_{t=4}$$ is not well-defined. Thus, we conclude that there is no instantaneous velocity at that point.

The above example shows us that an instantaneous velocity doesn't have to exist at every instance. So be careful of that differentiability criterion!

If $p(x)$ is the position function of an object and $v(x)$ is the velocity function of the object, then

$\frac{d}{dx}p(x) = v(x)$ or $p'(x) = v(x)$

This briefing will show relevant calculations in one dimension.

Do note that for a specific time $t$, the velocity of the particle at time $t$ is equivalent to the derivative of $p$ when $x = t$; the only extra step added here is that you plug in $x = t$ into the velocity function.