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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.

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### Chapter 2:Motion in One DimensionEXAMPLES

Example 2.1Displacement

• x1 = 30 m x2 = 10 m

• Displacement is a VECTOR

• Suppose the person walks during 50 seconds.

• Displacement

Distance (d) = 100m

• Average velocity:

• Average Speed:

Xi

Xf

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

Instantaneous = average

• 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 2.5 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!!!

• Example 2.5 (Text book Page 31)

• Example 2.8 (Text book Page 37)

• 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)

• (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

• (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

• (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

• 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 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

• 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