1. parabolic
y = kx2
exponential proportion
distance is proportional to time squared
d = kt2
cm/s2= k
Unit analysis
constant acceleration

2. 2*distance is directly proportional to time squared
linear
direct proportion
2*distance is directly proportional to time squared
2d = kt2
y = kx
Unit analysis
cm/s2= k
constant acceleration

3. velocity is directly proportional to time
linear
direct proportion
velocity is directly proportional to time
vf = kt
y = kx
Unit analysis
cm/s2= k
constant acceleration

4. The work of Galileo
Galileo set up a plane with a small angle of inclination.
He rolled a ball down the plane and measured the times for different distances. (water clock)
He then extrapolated his results for acceleration down the plane to the vertical.

5. Why use an incline to study the effects of gravity?
What would have been a better way to measure acceleration due to gravity?
drop something !!!!!
Why didn’t Galileo do this?
He couldn’t measure the time

6. All objects fall at the same rate REGARDLESS of their mass !!!
Earth’s Gravity (g)
9.8 m/s2 = 980 cm/s2 = 32 ft/s2
Acceleration due to gravity on the Earth’s surface is uniform
Galileo showed -
All objects fall at the same rate REGARDLESS of their mass !!!

7. 2d t2
1.7
1.300
6.8
3.881
15.3
9.242
27.2
17.48
42.5
25.91 1)
It shows that the ball
rolled with a constant
acceleration
b) Slope ≈ 1.6 m/s2
g = 1.6 ÷ sin 10°
g ≈ 9.21 m/s2
6.0% 2d
(m) d)
t2 (s2)

8. d = vit +½at2 d = ½at2 vf = vi + at vf - vi = a v = a t t2)
First of all, where do these equations come from?
d = vit +½at2
d = ½at2
vi = 0 (starts from rest)
vf = vi + at
vf - vi = a
v = a
rearranging t t
Galileo couldn’t measure velocity!!!

9. First of all, where do these equations come from?3)
First of all, where do these equations come from?
= vavg
d = vavgt d t
Can’t plug a final
velocity in for
an average velocity
vf = vi + at
vf = at
vi = 0 (starts from rest)

10. vf = vi + at 55 = 22 + (a)(11) 2d = (vf + vi)t 2d = (55+22)(11)
4) Given:
vf = vi + at
vi = 22 m/s
55 = 22 + (a)(11)
a = 3.0 m/s2
vf = 55 m/s
t = 11 s
a = ? m/s2
2d = (vf + vi)t
d = ? m
2d = (55+22)(11)
d = 424 m
5) Given:
2d = (vf + vi)t
vi = 0 m/s
d = 190 m
vf = 40 m/s
2d = (40+0)(9.5)
t = 9.5 s
d = ? m

11. What’s the deal with that?
g = a ÷ sinθ
9.4°
6) Given:
vi = 12 m/s
d = vit +½at2
a = -1.6 m/s2
t = 6 s; 9 s
d = 12(6) + ½(-1.6)(6)2
d = 43.2 m
d = ? m
d = 12(9) + ½(-1.6)(9)2
d = 43.2 m
What’s the deal with that?
It’s at the same spot two times
(once on the way up the ramp, and once on the way down)
Who can tell me the angle of the incline?