Hello it's a me again drifter1! We have exams in a week and that's why I don't have so much time because I have to study, but here I am again to post about **Physics **again! Today we will get into **2-dimensional motion** or **plane motion** and more specifically get into** position, velocity and acceleration vectors **in such a motion. So, without further do, let's get started!

## Getting into 2 dimensions

In my previous posts we only talked about one dimension or one axis and movement that happens only on that specific axis. All the vectors where one-dimensional and that made them have only 2 possible directions that can be left/right or up/down depending on the problem, and they could also be zero.

But, using one axis and one dimension is not enough to describe the physics phenomena around us, and so we will have to get to 2 or even 3 dimensions, depending on the problem that we want to solve.

To get into 2 dimensions the **position now needs to be a position vector**, mostly called r, that has a value that represents the distance to the center O(0, 0, ..., 0).

So, the **position **now is:

**r = xi + yj + zk + ...** if we need even more dimensions

This can also be written as:

**r = (x, y, z)** in the classic format of a **Point**

## Velocity vector

Using this new vector representation of position, but the same exacty logic we used in one dimension, we can now describe the velocity in 2 or even more dimensions.

The **Change of position is Δr = r2 - r1**.

Using it we can now specifiy the** average velocity** as:

**u avg = Δr / Δt **= r2 - r1 / t2 - t1

Because r1 and r2 are now vectors we can clearly see that the result of that equation is an vector that describes the velocity better as an vector and doesn't give us a numeric value as in 1D.

By calculating the** limit when Δt -> 0** of Δr, **Δr becomes a tangent of the line** **that describes the position in time**, and so describes the **direction of the instantaneous velocity** and so equals the instantaneous velocity.

So, the **instantaneous velocity** is being calculated by:

**u = lim Δt -> 0 (Δr/Δt) = dr/dt**

To find the actual velocity value we can either calculate the meter of that vector u or can also find the velocity in each dimension separately.

The (instantaneous)** velocity in each dimension** is:

ux = dx/dt

uy = dy/dt

uz = dz/dt

etc...

Where** ux, uy, uz, ... are the vector components** of u

And this means that we can representate the **velocity **as:

**u = dr/dt = dx/dt i + dy/dt j + dz/dt k + ...**

using the unit vectors: i, j, k, ...

That way we can now also see that the** actual velocity value **and so the **velocity vector meter** is equal to:

**|u|= root(ux^2 + uy^2 + uz^2 + ...)**

In 2d this looks like this:

**|u| = root(ux^2 + uy^2)**

where uy/ux gives us the tangent of the **angle **of vector u to the x'x axis:

**tan a = uy/yx**

So, the only thing that changes now in tasks is that the position is also being given as an function or that the position is not only a specific value of one axis, but is a point in 2 or 3-dimensional space!

## Acceleration vector

Acceleration in 2 or even more dimensions is being described by the same exact way as before, but now clearly is an "vector".

The **average acceleration** is:

**a avg = Δu/Δt = u2 - u1 / t2 - t1**

that is the exact same equation we had in 1 dimension.

The **instantaneous acceleration** is:

**a = lim Δt -> 0 (Δu/Δt) = du/dt**

that is also the same as before

Well, then why bother?

The thing that now changes is that we can get into acceleration in each dimension separately, which means that the **acceleration **can be described by:

**a = dux/dt i + duy/dt j + duz/d k + ...**

or even:

**a = ax i + ay j + az k + ...**

where:

ax = d^2x/dt^2

ay = d^2y/dt^2

az = d^2z/dt^2

etc..

that is the **second derivative of position with time**.

The actual meter can again be found using the root-equation.

## Full-On Example

Suppose the position with time is given by the following equations:

x = 3 + 2t^2

y = 10t + 0.25t^3

a) Where is the object at t = 2seconds?

b) Calculate the change of position and average velocity at the range of t = 0s to t = 2s

c) Write an equation that describes the velocity with time and find the velocity at t = 2s

d) Calculate the average acceleration at the range of t = 0s to t = 2s

e) Write an equation that describes the acceleration with time and find the acceleration at t = 2s

a)

At t = 2s the object is at:

x = 3 + 2*2^2 = 10m

y = 10*2 + 0.25*2^3 = 22m

So, r = 10i + 22y and the meter is:

|r| = root(x^2 + y^2) = root(10^2 + 22^2) = 24.6m

b)

For t = 0 we have that:

r0 = xi + yj = (3+2*0^2)i + (10*0+0.25*0^3) = 3i + 0j

For t =2 we have that:

r2 = xi + yj = (3+2*2^2)i + (10*2+0.25*2^3) = 11i + 22j

So, Δr = r2 - r1 = (11-3)i + (22-0)j = 8i + 22j

The average velocity is:

u avg = Δr/Δt = 8u + 22j / 2 - 0 =>

u avg =4i + 11j

where ux avg = 4m/s and uy avg = 11 m/s

c)

The component vectors are:

ux = dx/dt = 2*2t = 4t

uy = dy/dt = 10 + 0.25*3*t^2 = 10 + 0.75t^2

So, the instantaneous velocity is:

u = 4ti + (10 + 0.75t^2)j

At t = 2s we have that:

u2 = 4*2i + (10 + 0.75*2^2)j =>

u2 = 8i + 13j

and the value is:

|u2| = root(8^2 + 13^2) = 15m/s

with tan a = ux/uy = 13/8 => a = ~58 degrees

d)

To calculate the average acceleration we have to find the value for each component vector and so for each dimension separately.

ax avg = Δux/Δt = 8 - 0 / 2 - 0 = 4 m/s^2

ay avg = Δuy/Δt = 13 - 10 / 2 - 0 = 1.5 m/s^2

e)

The acceleration equation can be found by finding each dimensions equation using differentiation.

So,

ax = dux/dt = 4 m/s^2

ay = duy/dt = 0.75*2t = 1.5t m/s^2

So, the equation for acceleration is:

a = 4i + 1.5tj

At t = 2s we have that:

ax2 = 4 m/s^2, cause it's static

ay2 = 1.5*2 = 3 m/s^2

So, a = 4i + 3j and the meter is |a| = root(ax^2 + ay^2) = root(4^2 + 3^2) = 5 m/s^2

We can also calculate the angle to x'x that is:

tan b = 3/4 => b = ~37 degrees

And this is actually it for today and I hope you enjoyed it!

Next time we will get into missile or bullet motion, that is one of the basic plane motions (the other one being circular motion) that you have to know to understand 2-dimensional physics!

See ya soon...Bye!