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PPT – Chapter 2: Motion in One Dimension EXAMPLES PowerPoint presentation | free to download - id: 137dd5-N2VlM

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Chapter 2Motion in One DimensionEXAMPLES

Example 2.1 Displacement

- x1 30 m x2 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

Xf

Xi

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

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).

Example 2.4, cont

2.4 Acceleration

Example 2.5 Average Acceleration

- ax (), vx() Speeding Up!!
- ax (?), vx(?) Speeding Up!!

Example 2.6 Average Acceleration

- ax (), vx(?) Slowing Down!!
- ax (?), vx() Slowing Down!!

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!!

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

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!!!

Material for the Midterm

- Examples to Read!!!
- Example 2.5 (Text book Page 29)
- Example 2.8 (Text book Page 35)

2.6 Constant Acceleration

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)

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

Example 2.10, cont

- (2) From (B) ? (C) y(C) 0
- y(C) y(A) vyi(A) t ½ayt2
- 0 0 20.0 t 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

Example 2.10, cont

- (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
- 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

Example 2.11 (Problem 60 page 51)

- 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 -

?

?

?

.

Example 2.12 Question 11 (Homework)

- 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

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