New notations for AP Physics

Remember that these formulas are ONLY GOOD for objects under UNIFORM (Constant) ACCELERATION.

221 attvd i atvv if

advv if 222 These are still good equations, but remember now we may have problems with non-uniform acceleration so they are not our only option.

We are going to change the notation to more closely reflect a University Physics course for engineering students, but the equations are the same!

Many textbooks use the letter “s” to represent displacement instead of “d” as we did in Physics 1.

Others prefer to use letters that more accurately represent the position along an axis displacement such as “x”, “y” or “z”. We will be using these now…at least most of the time.

Remember that displacement is a change in position.

So now… 0

0

0

zzd

yyd

xxd

z

y

x

Where x, y, z are the final positions along those axes, and x0, y0, z0 are the initial positions along those axes.

For final velocity, we will just use “v”. Remember this is instantaneous velocity at any given time “t”, so…

For initial velocity, we will use “v0”. The “0” subscript refers to time “t=0” which would be the initial value.

vv f

Oi vv

Acceleration must be constant for these equations to valid.

Good news!! We will not change the notation for acceleration…we get to keep “a”!

aa

Time is time! We will not change time. We get to keep the good ole “t” for time.

Time is the interval over which the motion of interest takes place.

tt

horizontal or “x” – direction

)(2 02

02

0

221

00

xxavv

tavv

tatvxx

xxx

xxx

xx

vertical or “y” – direction

)(2 02

02

0

221

00

yygvv

gtvv

gttvyy

yy

yy

y

Remember that for projectiles, the horizontal and vertical motions must be separated and analyzed independently. Remember that “ax” is zero and “ay” is acceleration due to gravity “g”. CAUTION: The equations in this form assume “up” is “positive” and “down” is “negative”.

voθ

Almost every projectile problem can be solved by starting with the displacement equation to solve for time. In this case…

0sin 221

00 gttvyy

Since “y” and “y0” are both the same, vertical displacement y-y0 = 0.Now solve for time, this will be the time for the entire flight. Remember , if you want to find maximum height you will only use half the time. If you want to find range, use the total time.

ygttvy 221

0 sin0

Vertical displacement y-y0 = y.Now solve for time, this will be the time for the entire flight. Then you can use that time to find the horizontal distance at which it was at that height.

xtvx cos0 0

voθ

y

x

voθ

-y

x

ygttvy 221

0 sin0

xtvx cos0 0

Vertical displacement is “-y”Now solve for time, this will be the time for the entire flight. Then you can use that time to find the horizontal distance at which it was at that height.