**Where are we going with this?**The information on this page relates to the skills needed to investigate and evaluate the graphical and mathematical relationship (using either manual graphing or computers) of one-dimensional kinematic parameters (distance, displacement, speed, velocity, acceleration) with respect to an object's position, direction of motion, and time.

**Acceleration and Displacement**

*(Okay, NOW we're moving!)*

There it is… at rest.

Then, it begins to move! Bam! Now, it's somewhere else! Where? Where is it?

Let's start with a reference point and an object some distance from it.

An object is 3 meters from a point… (That is to say, an object has an initial distance of 3 meters from a point.)

So, if it doesn't move, then (duh) the final distance from the reference point is the same as the initial distance…

df = di

That's easy…

Now, suppose it moves. And we'll call that motion ∆d… then

df = di + ∆d

If we understand that ∆d is a function of the rate of motion, velocity, and how long it moves, then

∆d = vave • t

So, we can substitute…

df = di +vave • t

Okay, so far so good. Now, if an object is accelerating, then the velocity is changing. We can find the average velocity as…

vave = (vf + vi)/2

where is

*vf*final velocity and*vi*is initial velocity.But wait! There's more!

The final velocity can be found! Where

*vf*is final velocity and*vi*is initial velocity and*∆v*is the change in velocity thenvf = vi + ∆v

and since

∆v = at

then

vf = vi +at

We should plug that into… something…

df = di +vave • t

df = di +(vf + vi)/2• tdf = di +((vi +at) +vi)/2• t

And then, math-magic occurs and…

df = di + (vit) + (1/2 at^{2}) <--- Parenthesis added for clarity

The above equation will become the starting point for many motion problems. It is worth committing to memory or recording in a convenient place for reference.

Generally, it will be written without the parenthesis as…

df = di +vit + 1/2 at^{2}

______________________

Seriously! How about doing that again with colors?

df = di + ∆d

∆d = vave • t

vave = (vf + vi)/2

vf = vi + ∆v

∆v = at

vf = vi +at

df = di + ∆d

df = di +vave • t

df = di +(vf + vi)/2• tdf = di +((vi +at) +vi)/2• t

And then, math-magic occurs and…

df = di + (vit) + (1/2 at^{2}) <--- Parenthesis added for clarity

____________________________

Math magic, anyone?

**Physics of Motion: Deriving the Distance Equation**

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