PPP Chapter 5-3 Acceleration

Uniform Acceleration: Acceleration is constant V vs t plot is a straight line Acceleration is the slope of that line What is the acceleration of Airplane A? Airplane B?

Instantaneous Acceleration: --the acceleration of an object at an instant of time --the slope of the tangent to the curve on a velocity – time graph

Simply put, if the object is Speeding up in the positive direction or positive Slowing down in the negative direction acceleration Slowing down in the positive direction negative Speeding up in the negative direction acceleration In uniform motion (remember, that means constant velocity— speed and direction do not change) the acceleration is zero

Hey, do you think this might be important?

Acceleration is still of the same magnitude and in the same direction when v is positive, zero or negative In the instant of time (duration = zero sec.) the ball is reversing direction and v = 0

The speed is steadily changing (getting less and less in the positive Direction) before reaching the top or increasing (speed getting more and more in the negative direction) after reaching the top--- So, there must be a constant acceleration involved Think about it! To satisfy yourself that this is true, Take small increments of time before or after the top: a =  v  t Big Note: After all, acceleration only has meaning when talking about change in velocity, sot there must be a finite (non-zero) time increment involved

Maybe it makes more sense if we look at it up and down like this:

That is, your final velocity, v, equals whatever velocity you started With (v 0 ) plus whatever velocity you gain or lose from the accel. This makes total sense!

Let’s develop the equations of motion (these equations will be our Bible in our study of kinematics): I will write down equations of motion, explain each term and explain some pitfalls. We learned that d = vt This equation is true but we must understand under what conditions we can apply it. the object starts from zero (that is, d0 = 0.00) the velocity is constant (no acceleration) v represents the average velocity, or v bar

If the object has a non zero starting point, then we must add in d 0 as a term, thus: d = d 0 + vt [Eq 1] Note that the above equation will work in a situation with acceleration if we know or can calculate the average velocity. Remember that the average of any two things is their sum divided by 2. So, the average velocity is: Average velocity = v bar = ½ ( v 0 + v 1 ) [Eq 2]

So, we now have an equation that we can use to calculate displacement if we have initial position, final velocity and initial velocity: d = d 0 + ½ ( v 0 + v 1 ) t [Eq 3] Big Note: Eq 1 will NOT work for accelerated motion (although students frequently try to apply it in that situation) but Eq 3 WILL work under certain conditions (namely, that you have initial and final velocities).

But what if you DON’T have initial and final velocities? What if they give you the acceleration instead? Then, our displacement equation must have 3 terms (this equation is derived on p 99 and I have posted a derivation on yellow paper on the classroom bulletin board). I will not derive it here but, rather, I will explain the meaning of each term:

d = d 0 + v 0 t + ½at 2 [Eq 4]

TA DA!! Finally ready to list the equations of motion, now that we know how they are derived:

v = at d = ½ vt d = ½ at 2 v 2 = 2ad These equations simplify greatly when the conditions: d 0 = 0 and v 0 = 0

Fin 5- 3