**pre reqs:** *none*

Often I will do some type of analysis that I think is quite cool. But there is a problem. I keep having to make a choice. Either go into all the little details, or skip over them. My goal for this blog is to make each post such that someone could learn some physics, but I also don't want it to go too long. So, instead of continually describing different aspects of basic physics - I will just do it once. Then, when there is a future post using those ideas, I can just refer to this post. Get it?

Fine. On with the first idea - kinematics. Kinematics typically means a description of motion (not what causes that motion). In particular, kinematics looks at position, velocity, and acceleration. In this post, I will try to stay in one dimension. This will make things look simpler without really losing too much. Later, when I talk about vectors, I will make it all better.

**Position**

In one dimension, position is a location on the axis (x or y or whatever you want to call it). Note that this really doesn't tell you much because the origin of this axis (I will call it x) is completely arbitrary. Where is x = 0 meters in the room you are in now? Anything you say could be correct. The coordinate system is just make believe, you know, like Peter Pan.

So, position isn't that useful. **Change** in position IS important. Also, change in position is independent of your coordinate system.

**Velocity**

How fast the position changes is a measure of velocity. In the x-direction only, this can be written as:

![v x](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/v-x.jpg)

A couple of points:

- Since this is the velocity in the x-direction, it can be positive or negative.

- Don't confuse this with the velocity vector, which can only have a positive magnitude. Technically, this v-x only has a positive magnitude only.

- What about speed vs. average velocity? Usually, the definition is that speed is how far you have gone divided by the time. Average velocity is your change in position over time. Here is an example of how these two would be different.

![speed velocity](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/speed-vel…)

Suppose in this diagram, something moved from one red dot to the other following the purple path. The average velocity would simply be:

![avg velocity](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/avg-veloc…)

While the average speed would be:

![avg speed](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/avg-speed…)

Ok. Enough about that. There are much more things to talk about.

**Average Velocity and Position**

It may seem obvious, but I will say it anyway. Using average velocity, I can write an expression that determines the final position given the initial position, the change in time and the average velocity:

![x2](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/x2.jpg)

-Note that this is true. Most people will say that this only works if the velocity is constant, but since I put AVERAGE velocity, it always works. I will come back to this shortly.

-What about instantaneous velocity? This is the velocity when the time interval gets really small.

**Acceleration**

Just as velocity is the change in position, acceleration is the change in velocity. This would make a great ACT analogy question (for any ACT writers that read this).

Average acceleration is:

![accel](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/accel.jpg)

I could also talk about average and instantaneous accelerations, but truthfully many useful situations can be approximated as constant acceleration. In such a case, the average and instantaneous accelerations are the same.

**Kinematic Equations**

You will often see much stuff in textbooks about kinematic equations. The kinematic equations are some equations that relate position, velocity, acceleration and time. Most people will claim that these are derived from calculus (which they can be), but that isn't necessary. Let me start with average velocity:

![kinematics](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/kinematic…)

If the velocity is changing, then I can write:

![avg velocity](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/avg-veloc…)

and substitute this in for average velocity:

![kinematics](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/kinematic…)

Now, I want to get rid of the v_{2}. I can use the definition of acceleration:

![v2](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/v2.jpg)

and substituting this in:

![kinematics](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/kinematic…)

And that is the common kinematics equation everyone thinks of (not using calculus). The next step would be to algebraically manipulate the above equations to get one that is independent of time - I will not go through that exercise, but I will list it:

![kinematics](http://scienceblogs.com/dotphysics/wp-content/uploads/2008/09/kinematic…)

So that is it. The "kinematic equations". Use them wisely.

**Update** I just want to make one thing clear. The above kinematics equations are only valid when the assumption that the average velocity is v_{1} plus v_{2} divided by two. This is obviously true when there is a constant acceleration. This is also mostly true when the time interval is very short. Ok, I feel better now.

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You can continue this game. After acceleration comes jerk, followed by snap, crackle and pop. Once you use numerical integrations these 'taylor coefficients' come in handy sometimes.

i need help i a question about kinematics

here it is:

a motorist is traveling at 10m/s when he sees a deer in the road 41m ahead.

if the maximum negative accelaration of the vehicle is -7m/s^-2, what is the maximum time needed by the driver ( delta t) that will avoid him to hit the deer

we don't understand ur question . sorry

BY ; MUHAMMAD AWAIS

`CHEF AUGSITATIV`

OF PYSICS

Hi shaima,

the min. time needed would be 1,43 seconds if negative acceleration is constant and instantly available @ -7m/s^2

answer this :

cyclist A starts with initial displacement 0 and moves with velocity 3 m/s. at the same time cyclist B starts from a point with displacement 200m and moves with velocity -2m/s.

when does A meet B and where are they when this happen ?

It will happen 120m form the original position and the time will be 40s because you sum 3+2 (because is the sum of each absolute valor) and you divide it with 200m, I mean 200/5 and you know this is correct because when you multiply each velocity with the time and then you sum them the result will be 200. It is something like this (3*40)+(2*40)= 120+80=200 @KAS