1 / 19

5 0.1

PH 105-003/4 ---- Friday, Aug. 22, 2007. 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.)?. 5 0.1.

evania
Download Presentation

5 0.1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. PH 105-003/4 ---- Friday, Aug. 22, 2007 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/s2.(a) When does the car pass the bike (in sec.)? • 5 • 0.1 Equations vf = vi + a Dt xf = xi + vi t + ½ a t2 PH 105

  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/s2.(a) When does the car pass the bike? • 10 m from light • 15 m • 20 m • 25 m • 40 m • 50 m (b) Where does this happen? Equations vf = vi + a Dt xf = xi + vi t + ½ a t2 PH 105

  3. 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/s2.(a) When does the car pass the bike?(b) Where does this happen? • 10 • 0.1 (c) How fast is the car moving then (in m/s)? • Equations • vf = vi + a Dt • xf = xi + vi t + ½ a t2 PH 105

  4. PH 105-003/4Friday, Aug. 29, 2007 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

  5. Math Review Trigonometry c a sinQ = a/c, cosQ = b/c tanQ = a/b Q b c2 =a2 + b2 (Pythagorean Theorem) V vy Example: Q V = 785 km/h, find Vx and Vy Vx Vx = VsinQ = 489 km/h Vy = VcosQ = 614 km/h CHECK: 7852 = 4892 + 6142 PH 105

  6. PH 105-003/4Wednesday, September 5, 2007 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

  7. Instantaneous Velocity • The instantaneous velocity is the limit of the average velocity as Δt 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. PH 105-003/4Friday, September 7, 2007 Simplest case of uniformly-accelerated 2D motion:Projectile motion (free fall) • a = constant = (0, -g, 0) = -g j • ax = 0, ay = -g, az = 0 vx = vix + ax t x = xi + vix t + ½ ax t2 vy = viy + ay t y = yi + viy t + ½ ay t2 vx = vix (constant) x = xi + vix t vy = viy – gt (NOT const.) y = yi + viy t – ½ g t2 active figure 4.07

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

  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 ~ x2)

  13. PH 105-003/4 -- Monday, September 10, 2007 Sample clicker question (test-taking skills):A man walks 40 meters in 20 seconds. What is his average velocity, in m/s? • 0 • 1 • 2 • 3 • 4 • 5 • 6 • 7

  14. Sample clicker question (test-taking skills): A man walks 40 meters in 20 seconds. What is his average velocity, in m/s? • 0 • 1 • 2 • 3 • 4 • 5 • 6 • 7 PH 105

  15. Range of a Projectile vx = vix (constant) x = xi + vix t vy = viy – gt (NOT const.) y = yi + viy t – ½ g t2 means where does it return to y=0? 0 = 0 + viy t – ½ g t2 viy = ½ g t t= 2viy/g is when, not where! x= 0 + vix t = 2 vix viy /g = 2 vicosq vi sinq /g

  16. Range of a Projectile

  17. Uniform Circular Motion • Trajectory is a circle, so ri = ri = r • Use definitions of a, v: Dv = a Dt, Dr = v Dt • If Dq small, Dv ~ v Dq, Dr ~ r Dq, so aDt~ vDq, Dr, vDt~rDq Divide: a/v = v/r, or a = v2 / r

  18. Clicker question (vectors)A man walks 30 m north, then 40 m east. The magnitude of his total displacement is (in m) • 10 • 30 • 40 • 50 • 70 PH 105

  19. Clicker Question: relative motionA 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 • 30 m/s • 40 m/s • 50 m/s • 60 m/s • 70 m/s PH 105

More Related