1 / 21

Chapter 2: Motion in One Dimension EXAMPLES

Chapter 2: Motion in One Dimension EXAMPLES. Example 2.1 Displacement. x 1 = 30 m x 2 = 10 m Displacement is a VECTOR. Example 2.2 Average Velocity & Speed. Suppose the person walks during 50 seconds. Displacement Distance (d) = 100m Average velocity: Average Speed: . X i.

quinto
Download Presentation

Chapter 2: Motion in One Dimension EXAMPLES

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. Chapter 2:Motion in One DimensionEXAMPLES

  2. Example 2.1Displacement • x1 = 30 m x2 = 10 m • Displacement is a VECTOR

  3. Example 2.2 Average Velocity & Speed • Suppose the person walks during 50 seconds. • Displacement Distance (d) = 100m • Average velocity: • Average Speed: Xi Xf

  4. Example 2.3 Instantaneous & average velocities • If an objects moves at uniform velocity (constant), then: Instantaneous velocity and average velocity at any instant (t) are the same. Instantaneous = average

  5. Example 2.4 Instantaneous & average velocities • Are Instantaneous velocity and Average velocity at any instant t the same? NOT ALWAYS!!!! • Example: A car starts from rest, speed up to 50km/h, remains at that speed for a time. Slow down to 20 km/hr in a traffic jam, the finally stops. Traveling a total of 15 km in 30 min (0.5 hr).

  6. Example 2.4, cont

  7. 2.4 Acceleration

  8. Example 2.5 Average Acceleration • ax (+), vx(+) Speeding Up!! • ax (), vx() Speeding Up!!

  9. Example 2.6 Average Acceleration • ax (+), vx() Slowing Down!! • ax (), vx(+) Slowing Down!!

  10. Example 2.7 Conceptual Question • Velocity and acceleration are both vectors (they have magnitude & direction). • Are the velocity and the acceleration always in the same direction? NO WAY!!

  11. Example 2.8 Conceptual Question • Velocity and acceleration are both vectors (they have magnitude & direction). • Is it possible for an object to have a zero acceleration and a non-zero velocity? YES!!! Drive 65 miles/h on the Freeway

  12. Example 2.0 Conceptual Question • Velocity and acceleration are both vectors (they have magnitude & direction). • Is it possible for an object to have a zero velocity and a non-zero acceleration? YES!!! Start your car!!!

  13. Material for the Midterm • Examples to Read!!! • Example 2.5 (Text book Page 31) • Example 2.8 (Text book Page 37)

  14. 2.6 Constant Acceleration

  15. Example 2.10 Free Fall Example • Initial velocity at A is upward (+) and acceleration is g (– 9.8 m/s2) • At B, the velocity is 0 and the acceleration is g (– 9.8 m/s2) • At C, the velocity has the same magnitude as at A, but is in the opposite direction • The displacement is – 50.0 m (it ends up 50.0 m below its starting point)

  16. Example 2.10, cont • (1) From (A) → (B) Vyf(B) = vyi(A) + ayt(B)  0 = 20m/s + (–9.8m/s2)t(B) t = t (B) = 20/9.8 s = 2.04 s ymax = y(B) = y(A) +vyi(A)t + ½ayt2 y(B) = 0 + (20m/s)(2.04s) + ½(–9.8m/s2)(2.04s)2 y(B) = 20.4 m

  17. Example 2.10, cont • (2)From (B) → (C): y(C) = 0 y(C) = y(A)+ vyi(A) t – ½ayt2 • 0 = 0 + 20.0t – 4.90t2 (Solving for t): t(20 – 4.9t) = 0 • t = 0 or t(C) = t = 4.08 s vyf(C) = vyi(A) + ayt (C) • vyf(C) = 20m/s + (– 9.8m/s2)(4.08 s) • vyf(C) = –20.0 m/s

  18. Example 2.10, cont • (3) From (C) → (D) Using position (C) as the reference point the tat (D) position is not 5.00s. It will be: t (D) = 5.00 s – 4.08 s = 0.96 • vyf(D) = vyi(C) + ayt (D) • vyf(D) = -20m/s + (– 9.8m/s2)(0.96 s) vyf(D) = – 29.0 m/s  y(D) = y(C) +vyi(C)t + ½ayt2 • y(D) = 0 – (29.0m/s)(0.96s) – (4.90m/s2)(0.96s)2 = – 22.5 m • y(D) = –22.5 m

  19. Example 2.11 (Problem #66 page 54) • From the free fall of the rock the distance will be: • From the sound de same distance will be: • But: t1 + t2 = 2.40s  t1 = 2.40 – t2 • Replacing (t1 ) into the first equation and equating to the second:    .

  20. Example 2.12 Objective Question #13 • A student at top of the building of height h throws one ball upward with speed vi and then throws a second ball downward with the same initial speed, vi . How do the final velocities of the balls compare when they reach the ground? After Ball 1 reaches maximum height it falls back downward passing the student with velocity –vi . This velocity is the same as Ball 2 initial velocity, so after they fall through equal height h, their impact speeds will also be the same!!! BALL 2 BALL 1 +vi - vi - vi h h

  21. Material for the Midterm • Material from the book to Study!!! • Objective Questions: 2-13-16 • Conceptual Questions: 6-7-9 • Problems: 3-6-11-16-17-20-29-42-44

More Related