1. Chapter 2: Motion in One Dimension EXAMPLES

2. Example 2.1 Displacement x 1 = 30 m x 2 = 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: XfXf XiXi

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

5. 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). Example 2.4 Instantaneous & average velocities

6. Example 2.4, cont

7. 2.4 Acceleration

8. Example 2.5 Average Acceleration a x (+), v x (+) Speeding Up!! a x (  ), v x (  ) Speeding Up!!

9. Example 2.6 Average Acceleration a x (+), v x (  ) Slowing Down!! a x (  ), v x (+) 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. Examples to Read!!! Example 2.5(Text book Page 31) Example 2.8(Text book Page 37) Material for the Midterm

14. 2.6 Constant Acceleration

15. Initial velocity at A is upward (+) and acceleration is g (– 9.8 m/s 2 ) At B, the velocity is 0 and the acceleration is g (– 9.8 m/s 2 ) 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) Example 2.10 Free Fall Example

16. (1) From (A) → (B) V yf(B) = v yi(A) + a y t (B)  0 = 20m/s + (–9.8m/s 2 )t (B) t = t (B) = 20/9.8 s = 2.04 s y max = y (B) = y (A) + v yi(A) t + ½a y t 2 y (B) = 0 + (20m/s)(2.04s) + ½(–9.8m/s 2 )(2.04s) 2 y (B) = 20.4 m Example 2.10, cont

17. (2) From (B) → (C): y (C) = 0 y (C) = y (A) + v yi(A) t – ½a y t 2  0 = 0 + 20.0 t – 4.90t 2 (Solving for t): t(20 – 4.9t) = 0  t = 0 or t (C) = t = 4.08 s v yf(C) = v yi(A) + a y t (C)  v yf(C) = 20m/s + (– 9.8m/s 2 )(4.08 s)  v yf(C) = –20.0 m/s

18. (3) From (C) → (D) Using position (C) as the reference point the t at (D) position is not 5.00s. It will be: t (D) = 5.00 s – 4.08 s = 0.96  v yf(D) = v yi(C) + a y t (D)  v yf(D) = -20m/s + (– 9.8m/s 2 )(0.96 s) v yf(D) = – 29.0 m/s  y (D) = y (C) + v yi(C) t + ½a y t 2  y (D) = 0 – (29.0m/s)(0.96s) – (4.90m/s 2 )(0.96s) 2 = – 22.5 m  y (D) = –22.5 m Example 2.10, cont

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: t 1 + t 2 = 2.40s  t 1 = 2.40 – t 2 Replacing (t 1 ) 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 v i and then throws a second ball downward with the same initial speed, v i. 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 –v i. 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!!! +v i - v i BALL 1 h h BALL 2 - v i

21. 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 Material for the Midterm