Velocity with respect to position

Question

We are given an expression for an object's position with respect to time.

To find its velocity at $t=0,$ can we put $t=0$ into the position then divide by zero?

If not, why?

Answer 1

No. It doesn't work that way. Instead, the velocity is the derivative of the position function with respect to time $t$. For example a position function of a particle that travels in a straight line. Say, $f(t) = t^2 + 4t + 4$ in feet, then the velocity at $t = 0$ second is the derivative of $f$ at $t = 0$. In other words, $v(0) = f'(0) = 2t+4|_{t=0} = 4$ ft/sec.

Answer 2

I assume you are thinking of velocity as the ratio $$\frac{\text{change in distance}}{\text{change in time}} = \frac{d(t_1) - d(t_0)}{t_1 - t_0},$$ and then you want to plug in $t_1 = t_0$ for 'instantenous' change, but unfortunately that's not how it works. In fact, we have the limit: $$d'(t_1) = \lim_{t_0 \to t_1} \frac{d(t_1) - d(t_0)}{t_1 - t_0},$$ which we call the velocity at time $t_1$.

Answer 3

Suppose distance as a function of time $$f(t)=at^2+bt+c$$ $($This is just an example, the function can be of any degree$)$

Now we know that $$\overrightarrow{v}=\frac{d\overrightarrow{s}}{d\overrightarrow{t}}$$ Differentiating this we get, $$f'(t)=2at+b$$ Now this is the velocity of the particle as a function of time. Put $t=n$ to find out the velocity at that instant. In your case $n=0$ Therefore velocity at $t=0$ is $$2a×0+b=b$$ This is how it works

Answer 4

You correctly know that $$\text{average velocity over }[t_1,t_2]=\frac{\text{position}_{t_2}-\text{position}_{t_1}}{t_2-t_1},\tag1$$ and you invalidly conclude that $$\text{velocity at }t_a=\frac{\text{position at }t_a}{t_a}.\tag{2}$$

But formula $(2)$ must be wrong, because:

1. it says that at $t=7,$ a car parked at position $x=7$ has velocity $1\;$!
2. it claims that position at $t$ is sufficient information to determine at velocity at $t;$ that is, an object in a given spot has the same velocity regardless of its actual motion!

Furthermore, formula $(2)$ is never able to give velocity at $t=0.$

The correct formula for (instantaneous) velocity is $$\text{velocity at }t_a=\frac{\mathrm d}{\mathrm dt}(\text{position at }t)\Bigg|_{t_a},\tag{✔2}$$ which considers how the position of the object is changing across the instant $t_a.$