# How do we calculate instantaneous velocity in 2D?

Suppose a body is moving with a constant speed of $10~\mathrm{ms^{-1}}$ in negative $x$ direction in $x$-$y$ plane. Let $\vec r$ be the position vector.

Then what will be the instantaneous velocity vector?

I know it will be $-10\hat{i}$, but how do I calculate that using this:

$$\vec{v} = \dfrac{d\vec{r}}{dt}.$$

You first write your position vector as $\vec r = (x_0 -t\cdot10~\mathrm{m/s},y_0)$ and then take the derivative of that.

This produces $$\vec v = \frac{d\vec r}{dt}=(-10,0)~\mathrm{ms^{-1}}=-10\hat i\,~\mathrm{ms^{-1}}$$ which is what you want.

• why $x_0-10t$? how did you know that we had to attach that variable t? – Max Payne Jun 18 '15 at 11:05
• is it speed * time? – Max Payne Jun 18 '15 at 11:06
• @TimKrul: In a sense I’m working my way backwards. I'm really reconstructing the most generic position vector that will give the right velocity vector.And yes, for constant velocity then distance = speed * time – Mikael Fremling Jun 18 '15 at 11:58

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$Elaborating on Mikael's answer, note that equations like $\vec v = \dv{\vec r}{t}$ are sort of shorthand notations to make the life of a physicist easier. Note that there are two (three in 3d) equations in this condensed notation. What we mean by such equations is simply the following:

$$v_x = \dv{r_x}{t}\\ v_y = \dv{r_y}{t}\\ v_z = \dv{r_z}{t}\\$$

I used $r_i$ to be consistent with the notation, usually one would simply write $x,y$ or $z$. Why we do that is rather clear: to save time and effort. Instead of writing 3 equations you just write them in a vector equation. Since Mikael already answered the most part of your question I don't want to go into that.

• I know $$v_y = \dv{r_y}{t} \hat{j}$$ is 0 in this case. You missed out unit vectors. i wanted to ask what will be walue of $r_x$ so that we differentiate to get the answer. Thanks! – Max Payne Jun 18 '15 at 11:13
• I didn't miss the unit vectors because this is a component notation. The necessary basis vectors assumed to be understood. The value of $r_x$ is exactly what Mikael has written $r_x= x_0 - 10*t$ – Gonenc Jun 18 '15 at 11:38

If you have the position vector along a path $\vec{r}(q)$ parametrized by $q$, where $q$ can be time, angle, distance, or whatever then the derivatives are:

$$\vec{v}(q,\dot{q}) = \frac{\partial \vec{r}(q)}{\partial q} \dot{q}$$ $$v = \| \vec{v} \|$$ $$\vec{e} = \frac{ \vec{v}}{v}$$ $$\vec{a}(q,\dot{q},\ddot{q}) = \frac{\partial \vec{v}(q,\dot{q})}{\partial q} \dot{q} + \frac{\partial \vec{v}(q,\dot{q})}{\partial \dot{q}} \ddot{q}$$ $$\vec{a} = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n}$$

## General Example

An object follows an ellipse. Its location is defined by a parameter $\theta$:

1. Position vector $\vec{r}(\theta) = (10 \cos\theta, 3 \sin \theta)$
2. Velocity vector $\vec{v}(\theta,\dot{\theta}) = \frac{\partial (10 \cos\theta, 3 \sin \theta)}{\partial \theta} \dot{\theta} = (-10 \dot{\theta} \sin\theta, 3 \dot\theta \cos\theta)$
3. Speed $v=\|\vec{v}\| = \dot{\theta} \sqrt{10^2+(3^2-10^2)\cos^2\theta}$
4. Tangent vector $\vec{e}(\theta) = \frac{\vec{v}}{\|\vec{v}\|} = \frac{(-10 \sin\theta,3 \cos\theta)}{\sqrt{10^2+(3^2-10^2)\cos^2\theta}}$
5. If I know the speed to be $v=\sqrt{309}$ at $\theta=\frac{\pi}{3}$ then
• $\dot{\theta} = \frac{\sqrt{309}}{\sqrt{10^2+(3^2-10^2)\cos^2\frac{\pi}{3}}}=2$
• $\vec{v} =v \vec{e} = \frac{(-10 \sqrt{309} \sin\frac{\pi}{3},3 \sqrt{309} \cos\frac{\pi}{3})}{\sqrt{10^2+(3^2-10^2)\cos^2\frac{\pi}{3}}} = (-10\sqrt{3},3)$

## Simple Example

If your position vector is $\vec{r}(t) = (5-10 t,0)$ then the velocity vector is

$$\vec{v} = \frac{\partial \vec{r}}{\partial t} \dot{t} = (-10,0)$$

since the time rate of time is one $\dot{t}=1$