1. 8/25/2015PH 105 5 0.1 Problem on uniform acceleration (from Ch. 2): A bike moving at constant velocity 5 m/s passes a car stopped at a light, as the light turns green and the car begins to accelerate at 2 m/s 2. (a) When does the car pass the bike (in sec.)? PH 105-003/4 ---- Friday, Aug. 22, 2007 Equations v f = v i + a  t x f = x i + v i t + ½ a t 2

2. 8/25/2015PH 105 A.10 m from light B.15 m C.20 m D.25 m E.40 m F.50 m A bike moving at constant velocity 5 m/s passes a car stopped at a light, as the light turns green and the car begins to accelerate at 2 m/s 2. (a) When does the car pass the bike? Equations v f = v i + a  t x f = x i + v i t + ½ a t 2 (b) Where does this happen?

3. 8/25/2015PH 105 Problem on uniform acceleration: A bike moving at constant velocity 5 m/s passes a car stopped at a light, as the light turns green and the car begins to accelerate at 2 m/s 2. (a) When does the car pass the bike? (b) Where does this happen? A.Equations B.v f = v i + a  t C.x f = x i + v i t + ½ a t 2 (c) How fast is the car moving then (in m/s)? 10 0.1

4. 8/25/2015PH 105 Components of a Vector A component is a projection of a vector along an axis –Any vector can be completely described by its components It is useful to use rectangular components –These are the projections of the vector along the x- and y-axes PH 105-003/4 Friday, Aug. 29, 2007

5. 8/25/2015PH 105 Math Review Trigonometry  c a b sin  = a/c, cos  = b/c tan  = a/b c 2 =a 2 + b 2 (Pythagorean Theorem) Example:  V V = 785 km/h, find V x and V y vyvy VxVx V x = Vsin  = 489 km/h V y = Vcos  = 614 km/h CHECK: 785 2 = 489 2 + 614 2

6. Position and Displacement The position of an object is described by its position vector, The displacement of the object is defined as the change in its position PH 105-003/4 Wednesday, September 5, 2007

7. Instantaneous Velocity The instantaneous velocity is the limit of the average velocity as Δ t approaches zero –As the time interval becomes smaller, the direction of the displacement approaches that of the line tangent to the curve

8. Average Acceleration

9. Motion with Uniform Acceleration in 3D The velocity vector can be obtained from the definition of acceleration: The position vector can be expressed as a function of time:

10. Simplest case of uniformly- accelerated 2D motion: Projectile motion (free fall) a = constant = (0, -g, 0) = -g j a x = 0, a y = -g, a z = 0 PH 105-003/4 Friday, September 7, 2007 v x = v ix + a x t x = x i + v ix t + ½ a x t 2 v y = v iy + a y t y = y i + v iy t + ½ a y t 2 v x = v ix (constant) x = x i + v ix t v y = v iy – gt (NOT const.) y = y i + v iy t – ½ g t 2 active figure 4.07

11. Clicker Question The dome light breaks off in a car moving at a constant velocity 30 m/s. It will hit the floor of the car: A.Directly under its starting point B.Behind its starting point C.Ahead of its starting point v = 30 m/s A B C

12. Assumptions of Projectile Motion The free-fall acceleration is constant over the range of the motion –It is directed downward –This is the same as assuming a flat Earth over the range of the motion –It is reasonable as long as the range is small compared to the radius of the Earth The effect of air friction is negligible With these assumptions, an object in projectile motion will follow a parabolic path (y ~ x 2 )

13. Sample clicker question (test- taking skills):  40  20     ? A.0 B.1 C.2 D.3 E.4 F.5 G.6 H.7 PH 105-003/4 -- Monday, September 10, 2007

14. 8/25/2015PH 105 Sample clicker question (test-taking skills): A man walks 40 meters in 20 seconds. What is his average velocity, in m/s? A.0 B.1 C.2 D.3 E.4 F.5 G.6 H.7

15. Range of a Projectile v x = v ix (constant) x = x i + v ix t v y = v iy – gt (NOT const.) y = y i + v iy t – ½ g t 2 means where does it return to y=0? 0 = 0 + v iy t – ½ g t 2 v iy = ½ g t t = 2v iy /g is when, not where! x = 0 + v ix t = 2 v ix v iy /g = 2 v i cos  v i sin  /g

16. Range of a Projectile

17. Uniform Circular Motion Trajectory is a circle, so r i = r i = r Use definitions of a, v:  v = a  t,  r = v  t If  small,  v ~ v  r ~ r  so a  t  ~ v  r, v  t~r  Divide: a/v = v/r, or a = v 2 / r

18. 8/25/2015PH 105 Clicker question (vectors) A man walks 30 m north, then 40 m east. The magnitude of his total displacement is (in m) A.10 B.30 C.40 D.50 E.70

19. 8/25/2015PH 105 Clicker Question: relative motion A UA student is driving (horizontally) 40 m/s in a convertible when 1 cm diameter hail starts to fall, vertically at 30 m/s. The speed with which the hail strikes her is A.30 m/s B.40 m/s C.50 m/s D.60 m/s E.70 m/s