1. SECTION 3.2 MOTION WITH CONSTANT ACCELERATION
 Objectives
Interpret position-time graphs for motion with constant acceleration.
Determine mathematical relationships among position, velocity, acceleration, and time.
Apply graphical and mathematical relationships to solve problems related to constant acceleration.

2. INTRO/VELOCITY WITH AVERAGE ACCELERATION
The definition of average acceleration can be manipulated similar to how average velocity can be manipulated to show the new velocity after a period of time given the initial velocity and the average acceleration.
a = vf – vi = v Average Acceleration Equation
tf – ti t
So Δv = aΔt and thus vf – vi = aΔt
Final Velocity – is equal to the initial velocity plus the product of the acceleration and time. It can be found using the equation
vf = vi + aΔt or vf = vi + at
Do Practice Problems p. 65 # 18-21

3. POSITION WITH CONSTANT ACCELERATION
You have learned that an object experiencing constant acceleration changes its velocity at a constant rate.
The area under the curve of a Velocity Time graph is the Displacement.
Example 3 p. 67
v = Δd / Δt => 75 = Δd / 1 => 75 m = Δd
v = Δd / Δt => 75 = Δd / 2 => 150 m = Δd
Do Practice Problems p. 67 # 22-25

4. POSITION WITH CONSTANT ACCELERATION
Δd = Δdrectangle + Δdtriangle = viΔt + ½ aΔt2
df – di = viΔt + ½ aΔt or df = di + viΔt + ½ aΔt2
Position with Average Acceleration – an object’s position at a time after the initial time is equal to the sum of its initial position, the product of the initial velocity and the time, and half the product of the acceleration and the square of the time.
df = di + viΔt + ½ aΔt2
Or from the old book the equation is
*** d = vit + ½ at2 ***

5. SOME OTHER FORMULAS v = d / t or d = vt v = ½ (vf + vi)
 d = ½ (vf + vi)t

6. AN ALTERNATIVE EXPRESSION
t = ∆v / a and d = vit + ½ at2
And if we combine them we get
d = vi(vf – vi) / a + ½ a[(vf – vi) / a ]2
and this will rearrange to the Velocity with Constant Acceleration
Velocity with Constant Acceleration – the square of the final velocity equals the sum of the square of the initial velocity and twice the product of the acceleration and the displacement since the initial time.
vf2 = vi2 + 2a(df – di) or *** vf2 = vi2 + 2ad ***

7. AN ALTERNATIVE EXPRESSION
Table 3.3 on p. 68 shows the 3 equations also the 3 from old book.
Do Example 4 p. 69
vf2 = vi2 + 2ad => 252 = (3.5)d => 625 = 7d => m = d
Do Practice Problems p. 69 # 26-29

8. AN ALTERNATIVE EXPRESSION
Do Example 5 p. 70
v = d / t And vf2 = vi2 + 2ad So total distance
25 = d / = (-8.5)d
11.25 m = d = -17d m
m = d
Do Practice Problems p. 71 # 30-33
Do 3.2 Section Review p. 71 # 34-41