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

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

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

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.

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

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 !!!

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)

d = vit +½at2 d = ½at2 vf = vi + at vf - vi = a v = a t t 2) 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!!!

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)

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

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?