motion in one dimension chapter 2. displacement and velocity ∙ ∙ a → b displacement is the...

Motion in One Dimension

Chapter 2

Displacement and velocity∙ ∙A → B

• Displacement is the change of position of an object.• Displacement in SI units is measured in meters.• Coordinate systems are used to describe motion.• x axis for horizontal displacement• y axis for vertical displacement • Displacement → ∆x = xf – xi

• Displacement = change in position = final position – initial position

Distance and displacement Distance and displacement are two quantities which may seem to mean the

same thing, yet have distinctly different definitions and meanings.

• Distance is a scalar quantity which refers to "how much ground an object has covered" during its motion.

• Displacement is a vector quantity which refers to "how far out of place an object is"; it is the object's change in position.

• To test your understanding of this distinction, consider the following motion depicted in the diagram below. A person walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.

• Even though this person has walked a total distance of 12 meters, his displacement is 0 meters. During the course of his motion, he has "covered 12 meters of ground" (distance = 12 m). Yet when he is finished walking, he is not "out of place" - i.e., there is no displacement for his motion (displacement = 0 m). Displacement, being a vector quantity, must give attention and regard to direction. The 4 meters east is canceled by the 4 meters west; and the 2 meters south is canceled by the 2 meters north.

4 m

2 m

4 m

2 m

http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1b.html

http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L1b.html

Displacement can be positive or negative

Horizontal displacement (∆x)

→ Positive displacement

← Negative displacement

Vertical displacement (∆y)

↑ Positive displacement

↓ Negative displacement

Velocity• Velocity is a vector quantity that measures

how fast something moves from one point to another.

• A vector quantity is a physical quantity that has both magnitude and direction.

• A scalar quantity is a physical quantity that has only magnitude and no direction.

• What is the difference between speed and velocity?

Average velocity is displacement in a given time

• To calculate the average velocity of an object, you must know the object’s displacement.

• You must also know the time the object left its initial position and the time it arrived at its final position.

• The average velocity is equal to the displacement divided by the time during which the displacement occurred.

• The SI unit of velocity is meters per second (m/s)

Average velocity

v avg = __Δx_

Δt

• Average velocity = change in position

• change in time

???????

• P/1 If an arc of 60º on Circle I has the same length as an arc of 45º on Circle II, the ratio of the area of circle I to that of Circle II is:

A) 19:9

B) 9:16

C) 4:3

D) 3:4

60º 45º

▲θ1 = ▲s1/ r1 ▲θ2 = ▲s2/ r2

S1 = S2

π/ 3 r1 = π/4 r2

r1 = ¾ r2

A1 = π (r1)²A2 = π (r2)²

Average velocityExample # 1: During a race on level ground, Kelly covers 950 m in 2 min. while running due east. Find Kelly’s average velocity

• Given: Δx = 950 m to the east• Δt = 2 min.• Unknown: v avg = ?• Solution: Use the average velocity equation:

v avg = Δx / Δt

v avg = 950 m / 2 min.

v avg = 950 m / 120 s

v avg = 7.92 m/s to the east

Physics Set # 3 August ____, 2008Student Name: _____________________________________________ Class Period: ______P/1 Ramiro takes 34 min to walk eastward along a straight road to a store 2 km away.

What is his average velocity in m/s?

Answer: _____________

P/2 Victoria is 140 km due south of Houston. If Ramses rides from Houston to Victoria on his bike in 6 h, what is his average velocity?

Answer: ____________

P/3 If the bus stop is 0.68 km down the street from the museum and it takes you 9.5 min to walk north from the bus stop to the museum entrance, what is your average velocity?

Answer: ___________

P/4 Miguel drives his car with an average velocity of 24 m/s toward the east.

How long will it take him to drive 560 km on a perfectly straight highway?

Answer: ___________

P/5 How much time will Miguel save by increasing his average velocity to 26 m/s?

Answer: __________

P/6 A bus traveled south along a straight path for 3.2 h with an average velocity of 88 km/h, stopped for 20 min, then traveled south for 2.8 h with an average velocity of 75 km/h

a) What is the average velocity for the total trip? Answer: __________

b) What is the displacement for the total trip? Answer: __________

P/7 Ramiro takes 44 min to walk eastward along a straight road to a store 3 km away.


Answer: _____________

Summer School Physics Set # 3 June 4, 2008Student Name: _____________________________________________ Class Period: ______P/1 Ramiro takes 34 min to walk eastward along a straight road to a store 2 km away.


Given: ∆x = 2 km → 2,000 m

∆t = 34 min → 2,040 s

Unknown: v avg = ?

Solution: v avg = ∆x / ∆t

v avg = (2,000 m) /(2,040 s) = 0.98 m/s

Answer: 0.98 m/s due east

P/2 Victoria is 140 km due south of Houston. If Ramses rides from Houston to Victoria on his bike in 6 h, what is his average velocity?

Given: ∆x = 140 km → 140,000 m

∆t = 6 h → 21,600 s

Unknown: v avg = ?


v avg = (140,000 m) /(21,600 s) = 6.48 m/s

Answer: 6.48 m/s due south

P/3 If the bus stop is 0.68 km down the street from the museum and it takes you 9.5 min to walk north from the bus stop to the museum entrance, what is your average velocity?

Given: ∆x = 0.68 km → 680 m

∆t = 9.5 min → 570 s

Unknown: v avg = ?


v avg = (680 m) /(570 s) = 1.19 m/s

Answer: 1.19 m/s due north

P/7 Ramiro takes 44 min to walk eastward along a straight road to a store 3 km away.


Answer: _____________

P/4 Miguel drives his car with an average velocity of 24 m/s toward the east.


Given: v avg = 24 m/s due east

∆x = 560 km →560,000 m

Unknown: ∆t = ?


∆t = ∆x / v avg = 560,000 m/ 24 m/s

∆t = 23,333.33 s

Answer: 23,333.33 s

P/5 How much time will Miguel save by increasing his average velocity to 26 m/s?

Given: v avg = 26 m/s due east

∆x = 560 km →560,000 m

Unknown: ∆t = ?


∆t = ∆x / v avg = 560,000 m/ 26 m/s

∆t = 21,538.46 s

Answer: 21,538.46 s


a) What is the average velocity for the total trip? Answer: __________

b) What is the displacement for the total trip? Answer: __________


a) What is the average velocity of the bus?B) What is the total displacement?

V1 = 88 km/h∆t 1 = 3.2 h

V2 = 75 km/h∆t2 = 2.8 h

20 min

Acceleration

• Acceleration measures changes in velocity• Acceleration describes the rate of change of

velocity in a given time interval.• The SI units for acceleration is meters per square

second (m/s²)

Average acceleration = change of velocity/change in time

a avg = ∆v / ∆t = (vf – vi) / (tf – ti)

Average accelerationExample # 1The Rockwell Bus from the Brownsville Urban System comes to a normal stop from 12 m/s to 0 m/s in 5 s. Find the average acceleration of the Bus.

• Given: vi = 12 m/s

vf = 0 m/s

∆t = 5 s• Unknown: a avg = ?• Solution: use the equation for average acceleration

a avg = ∆v /∆t

∆v = vf – vi = 0 m/s – 12 m/s = - 12 m/s

a avg = - 12 m/s / 5 s

a avg = - 2.4 m/s²

Acceleration

• Acceleration has direction and magnitude

• Acceleration is a vector quantity.

vi a Motion

+ + Speeding up

- - Speeding up

+ - Slowing down

- + Slowing down

- or + 0 Constant velocity

0 - or + Speeding up from rest

0 0 Remaining at rest

Velocity and acceleration

Summer School Physics Set # 4 June 4, 2008Student Name: ___________________________________ Class Period: ______

P/1 A bus comes to a sudden stop to avoid hitting a dog.

Its initial velocity was 9 m/s, and it came to a stop in 1.5 s.

Find the average acceleration of the bus. Answer: __________

P/2 A car traveling initially at 7 m/s accelerates to a velocity of 12 m/s in 2 s.

What is the average acceleration of the car? Answer: __________

P/3 Turner’s treadmill starts with a velocity of – 1.2 m/s and speeds up at regular intervals during half-hour intervals. After 25 min, the treadmill has a velocity of – 6.5 m/s.

What is the average acceleration during this time period? Answer: __________

P/4 If a treadmill starts at a velocity of – 2.7 m/s and has a velocity of – 1.3 m/s after 5 min.

What is the average acceleration of the treadmill?

Answer: __________

P/5 With an average acceleration of – 0.50 m/s², how long will it take a cyclist to bring a bicycle with an initial velocity of + 13.5 m/s to a complete stop? Answer: __________

P/6 If a treadmill starts at a velocity of – 2.2 m/s and has a velocity of – 1.4 m/s after 3 min.

What is the average acceleration of the treadmill? Answer: __________

P/7 Miguel drives his car with an average velocity of 27 m/s toward the west.


Answer: ___________

Displacement depends on acceleration, initial velocity and time

v avg = ∆ x / ∆tFor an object moving with constant acceleration, the average velocity is

equal to the average of the initial velocity and the final velocityv avg = (vi+ vf) / 2

Displacement with constant acceleration:∆ x = ½ (vi + vf) ∆t

Velocity(m/s)

Time (s)

v avg

v final

Final velocity depends on initial velocity, acceleration and time

a avg. = (vf – vi) / ∆t

(a avg)∙ ∆t = (vf – vi)

Velocity with constant acceleration:

vf = vi + (a avg)∙ ∆t

Final velocity = initial velocity + (acceleration X time interval)

Final velocity depends on initial velocity, acceleration and time

Displacement with constant acceleration:∆x = ½ (vi + vf) ∆t

Velocity with constant acceleration:vf = vi + (a avg)∙ ∆t

Combining the two equations:∆x = ½ [vi + (vi + (a avg)∙ ∆t ) ] ∆t

Displacement with constant acceleration ∆x = vi ∆t + ½ (a avg) (∆t )²

Extra PointsP/1Albert is a keen dog admirer and over the years has had a number of dogs. He has had an Alsatian, a Dalmatian, a Poodle and a Great Dane, but not necessarily in that order. Albert had Jamie first. The Dalmatian was an adored pet before the Great Dane. Sammy, the Alsatian, was the second dog Albert loved. Whitney was housed before the Poodle and Jimmy was not a Great Dane.

Can you tell each of the dogs' name and the order in which Albert had them?

Answer: Albert had Jamie the Dalmatian first, then Sammy the Alsatian, Whitney the Great Dane and finally, Jimmy the Poodle.

P/2 A car traveled from London at a speed of 40mph. Its fuel consumption was 30 mpg. It had a 5 gallon tank which was full when it started, but at that very moment began to leak fuel. After 60 miles the car stopped with a completely empty tank.

How many gallons per hour was it losing?

Answer: 2 gallons per hour. The car traveled 60 miles before stopping, this means it used 2 gallons. The car took 1.5 hours to travel 60 miles. Therefore the car lost 3 gallons in 1.5 hours = 2 gallons per hour.

Set # 5 AnswersP/1 A car with an initial speed of 23.7 km/h accelerates at a uniform

rate of 0.92 m/s² for 3.6 s.

Find the final speed and the displacement of the car during this time.

Given: vi = 23.7 km/h → vi = 6.58 m/s a = 0.92 m/s² ∆t = 3.6 s Unknown: vf = ? , ∆x = ? Solution: vf = a∙∆t + vi → vf = (0.92 m/s²) (3.6 s) + 6.58 m/s

vf = 3.31 m/s + 6.58 m/s

vf = 9.89 m/s ∆x = ½ (vi+vf)∙∆t

∆x = ½ (6.58 m/s + 9.89 m/s) (3.6 s)

∆x = 29.65 m

Set # 5 AnswersP/2 A car with an initial speed of 4.30 m/s accelerates at the rate of 3 m/s².Find the final speed and the displacement after 5 s.

• Given: vi = 4.30 m/s, a = 3 m/s² ∆t = 5 s• Unknown: vf = ?, ∆x = ?

Set # 5 AnswersP/3 A car can finish a 400 m race in 6 s. What is the minimum constant acceleration necessary?

• Given: ∆x = 400 m• ∆t = 6 s• Unknown: a = ?

Set # 5 AnswersP/4 A car starts from rest and travels 5 s with a uniform acceleration of – 1.5 m/s².What is the final velocity of the car? How far does the car travel in this time interval?

• Given: vi = 0 m/s, ∆t = 5s, a = - 1.5 m/s²• Unknown: vf = ?, ∆x = ?

Set # 5 AnswersP/5 A driver of a car traveling at – 15 m/s applies the brakes, causing a uniform acceleration of 2 m/s²If the brakes are applied for 2.5 s. What is the velocity of the car at the end of the braking period?. How far has the car moved during the braking period?

• Given: vi = - 15 m/s, a = 2 m/s, ∆t = 2.5 s• Unknowns: vf = ? and ∆x = ?


P/1 A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s² for 3.6 s.


Answer: vf = _______

∆x = __________

P/2 A car with an initial speed of 4.30 m/s accelerates at the rate of 3 m/s².

Find the final speed and the displacement after 5 s.

Answer: vf = __________

∆x = __________

P/3 A car can finish a 400 m race in 6 s.

What is the minimum constant acceleration necessary?

Answer: a = __________

P/4 A car starts from rest and travels 5 s with a uniform acceleration of – 1.5 m/s².

What is the final velocity of the car? How far does the car travel in this time interval?

Answer: vf = __________

∆x = __________

P/5 A driver of a car traveling at – 15 m/s applies the brakes, causing a uniform acceleration of 2 m/s²

If the brakes are applied for 2.5 s. What is the velocity of the car at the end of the braking period?.

How far has the car moved during the braking period?

Answer: vf = __________

∆x = __________


P/1 A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92 m/s² for 3.6 s.


Given: vi = 23.7 km/h → vi = 6.58 m/s

a = 0.92 m/s²

∆t = 3.6 s

Unknowns: vf = ? , ∆x = ?

Solution: vf = a∙∆t + vi → vf = (0.92 m/s²) (3.6 s) + 6.58 m/s

vf = 3.31 m/s + 6.58 m/s

vf = 9.89 m/s

∆x = ½ (vi+vf)∙∆t

∆x = ½ (6.58 m/s + 9.89 m/s) (3.6 s)

∆x = 29.65 m

P/2 A car with an initial speed of 4.30 m/s accelerates at the rate of 3 m/s².

Find the final speed and the displacement after 5 s.

Given: vi = 4.30 m/s, a = 3 m/s² ∆t = 5 s

Unknowns: vf = ?, ∆x = ?

Solution: vf = a∙∆t + vi → vf = (3 m/s²) (5 s) + 4.3 m/s

vf = 15 m/s + 4.30 m/s

vf = 19.30 m/s

∆x = ½ (vi+vf)∙∆t

∆x = ½ (4.30 m/s + 19.30 m/s) (5 s)

∆x = 59 m


If the brakes are applied for 2.5 s. What is the velocity of the car at the end of the braking period?.


Answer: vf = __________

∆x = __________

P/3 A car can finish a 400 m race in 6 s.

What is the minimum constant acceleration necessary?

Given: ∆x = 400 m

∆t = 6 s

Unknown: a = ?

Solution:

1. First find the average velocity: v avg = ∆x / ∆t

v avg = 400 m/ 6 s = 66.66 m/s

2. Using the formula for average velocity, find the final velocity:

v avg = (vi+vf)/2 vi = 0 m/s

vf = (2) (66.66 m/s) = 133.32 m/s

3. Use the acceleration formula:

a = (vf- vi) / ∆t

a = (133.32 m/s – 0 m/s) / 6s = 22.22 m/s²

Answer: a = 22.22 m/s²

P/4 A car starts from rest and travels 5 s with a uniform acceleration of – 1.5 m/s².

What is the final velocity of the car? How far does the car travel in this time interval?

Given: vi = 0 m/s ∆t = 5s a = - 1.5 m/s²

Unknowns: vf =?, ∆x = ?

Solution: a = (vf-vi)/ ∆t → vf = a· ∆t + vi

vf = (- 1.5 m/s²) (5s) + 0 m/s

vf = - 7.5 m/s

∆x = (vi+vf)·∆t = (0 m/s – 7.5 m/s) (5s)

2 2

∆x = - 18.75 m


If the brakes are applied for 2.5 s.

What is the velocity of the car at the end of the braking period?.


Given: vi = - 15 m/s a = 2 m/s ∆t = 2.5 s

Unknowns: vf = ?

∆x = ?

Solution:

1. From the formula for acceleration, we can find the final velocity:

a = (vf – vi) / ∆t → vf = a ∆t + vi

vf = (2 m/s) (2.5 s) + (- 15 m/s)

vf = 5 ms – 15 m/s = - 10 m/s

2. Find the displacement:

∆x = (vi+vf)·∆t = (-15 m/s – 10 m/s) (2.5s)

2 2

∆x = - 31.25 m


P/1 A car on a wet road can achieve an acceleration of only – 1 m/s² without sliding.

Find the required stopping distance for a speed of 48 km/h

Answer: ______________

P/2 A plane starting at rest at one end of a runway undergoes a constant acceleration of 4.8 m/s² for 20 s.

What is its speed at takeoff? Answer: ______________

How long must the runway be for the plane to be able to take off? Answer: ______________

P/3 Nathan accelerates his skateboard along a straight path from rest to 15 m/s in 3.4 s

a) What is Nathan’s acceleration? Answer: ______________

b) What is Nathan’s displacement during this time interval? Answer: ______________

c) What is Nathan’s average velocity during this time interval? Answer: ______________

P/4 Maria Teresa’s car accelerates at a rate of + 2.60 m/s². How long does it take for her car to accelerate from a speed of 88.5 km/h to a speed of 102 km/h?

Answer: _____________

P/5 If you live 10 km from St. Joseph’s Academy and your car takes 0.53 h to reach your school, what is the average velocity (in m/s) of your car?

Answer: _____________

Falling objects• Freely falling bodies undergo constant acceleration• In the absence of air resistance, all objects dropped near

the surface of a planet fall with the same constant acceleration. Such motion is referred as free fall.

• The free fall acceleration is denoted with the symbol g• At the surface of Earth the magnitude of g is approximately

9.81 m/s².• This acceleration is directly downward, toward the center

of the Earth.• The downward direction is negative.• The acceleration of objects in free fall is g = - 9.81 m/s²

Introduction to Free Fall• A free-falling object is an object which is falling under the sole

influence of gravity. Thus, any object which is moving and being acted upon only by the force of gravity is said to be "in a state of free fall."

• This definition of free fall leads to two important characteristics about a free-falling object:

1.Free-falling objects do not encounter air resistance.

2. All free-falling objects on Earth accelerate downwards at a rate of approximately -9.81 m/s²

Free fall acceleration

• This numerical value for the acceleration of a free-falling object is such an important value that it has been given a special name.

• It is known as the acceleration of gravity – the acceleration for any object moving under the sole influence of gravity. As a matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it – the symbol g. The value for g = - 9.81 m/s²

• There are slight variations in this numerical value (to the second decimal place) which are dependent primarily upon altitude.

• Acceleration is the rate at which an object changes its velocity. Between any two points in an object's path, acceleration is the ratio of velocity change to the time taken to make that change.

• To accelerate at 9.81 m/s² means to change your velocity by 9.81 m/s

each second.

If gravity had a value of -10 m/s²

-

-

-

-

-

-

-

-

Gravity is = – 9.81 m/s²• Free-falling objects are in a state of acceleration• Specifically, they are accelerating at a rate of -9.81 m/s². • This is to say that the velocity of a free-falling object is

changing by 9.81 m/s every second. • If dropped from a position of rest, the object will be traveling:

From: a = Δv / Δt solving for Δv Δv = a·Δt

when t = 0 s → velocity = 0 m/s acceleration = - 9.81 m/s²

t = 1 s v = (a·Δt ) = (-9.81 m/s²) (1 s) = - 9.81 m/s a = - 9.81 m/s²

t = 2 s v = (a·Δt ) = (-9.81 m/s²) (2 s) = - 19.62 m/s a = - 9.81 m/s² t = 3 s v = (a·Δt ) = (-9.81 m/s²) (3 s) = - 29.43 m/s a = - 9.81 m/s² t = 4 s v = (a·Δt ) = (-9.81 m/s²) (4 s) = - 39.24 m/s a = - 9.81 m/s²

Free falling objectsStudent Name: _________________________ Sep/13/07

Δt

(s)

a

(m/s²)

vi

(m/s)

vf

(m/s)

Δy

(m)

0 - 9.81 0 0 0

1 - 9.81 0

2 - 9.81 0

3 - 9.81 0

4 - 9.81 0

5 - 9.81 0

6 - 9.81 0

7 - 9.81 0

8 - 9.81 0

9 - 9.81 0

10 - 9.81 0

What goes up, must come down• When we throw an object up in the air, it will continue to move upward

for a short time, stop momentarily at the peak, and then change direction and begin to fall.

• Because the object changes direction, it may seem like the velocity and the acceleration are both changing.

• Objects thrown into the air have a downward acceleration as soon as they are released.

• When an object is thrown up in the air, there is an instant when the velocity is equal to 0 m/s.

• Although the velocity is zero at the instant that the object reaches the peak, the acceleration is equal to – 9.81 m/s².

• This acceleration is equal to – 9.81 m/s² at every instant regardless of the magnitude or direction of the velocity,

• When an object is thrown up in the air it has a positive velocity and a negative acceleration, that means that the object is slowing down.

• At the top of its path, the object’s velocity has decreased until it is zero. Although it is impossible to see this because it happens so quickly, the object is actually at rest for an instant, and even tough the velocity is zero at this point, the acceleration is still – 9.81 m/s².

• The object then begins moving down, and when is moving down it has a negative velocity, and since the acceleration is still negative, the object will be speeding up.

Example # 1 A ball is thrown up in the air with an initial velocity of 44 m/s. Find the velocity and displacement of the ball in the first 10 s of this motion.

Vi = 44 m/s

Time Velocity

vf = vi + a •Δt

Acceleration

Gravity

Displacement

Δy = ½ (vi+vf) •Δt

0 s 0 m/s - 9.81 m/s² 0 m

1 s = 44 m/s + (- 9.81 m/s²) (1 s)

= 34.19 m/s

- 9.81 m/s² = ½ (44 m/s + 34.19 m/s) (1s)

= 39.095 m

2 s =44 m/s + (- 9.81 m/s²) ( 2s)

= 24.38 m/s

- 9.81 m/s² = ½ (44 m/s + 24.38 m/s) (2s)

= 68.38 m

3 s = 44 m/s + (- 9.81 m/s²) (3s)

= 14.57 m/s

- 9.81 m/s² = ½ (44 m/s + 14.57 m/s) (3 s)

= 87.855 m

Time Velocity Acceleration Displacement

0 s 0.00 m/s - 9.81 m/s² 0.000 m 1 s 34.19 m/s - 9.81 m/s² 39.095 m 2 s 24.38 m/s - 9.81 m/s² 68.380 m 3 s 14.57 m/s - 9.81 m/s² 87.855 m 4 s 4.76 m/s - 9.81 m/s² 97.520 m 5 s - 5.05 m/s - 9.81 m/s² 97.375 m 6 s -14.86 m/s - 9.81 m/s² 87.420 m 7 s -24.67 m/s - 9.81 m/s² 67.655 m 8 s -34.48 m/s - 9.81 m/s² 38.080 m 9 s -44.29 m/s - 9.81 m/s² - 1.305 m10s -54.10 m/s - 9.81 m/s² - 50.500 m

What goes up, must come down

Vi = Vf =

V = _____


P/1 A ball is thrown up in the air with an initial velocity of 64 m/s. Find the velocity and displacement of the ball in the first 10 s of this motion:

Answer:

t = 0 s v = _________ ∆y = __________

t = 1 s v = _________ ∆y = __________

t = 2 s v = _________ ∆y = __________

t = 3 s v = _________ ∆y = __________

t = 4 s v = _________ ∆y = __________

t = 5 s v = _________ ∆y = __________

t = 6 s v = _________ ∆y = __________

t = 7 s v = _________ ∆y = __________

t = 8 s v = _________ ∆y = __________

t = 9 s v = _________ ∆y = __________

t = 10 s v = _________ ∆y = __________

P/2 A ball is thrown up in the air with an initial velocity of 34 m/s. Find the velocity and displacement of the ball in the first 6 s of this motion:

Answer:

t = 0 s v = _________ ∆y = __________

t = 1 s v = _________ ∆y = __________

t = 2 s v = _________ ∆y = __________

t = 3 s v = _________ ∆y = __________

t = 4 s v = _________ ∆y = __________

t = 5 s v = _________ ∆y = __________

t = 6 s v = _________ ∆y = __________

Free FallExample # 1: Sofia hits a volleyball so that it moves with an initial velocity of 8 m/s straight upward. If the volleyball starts from 2 m above the floor, how long will it be in the air before it strikes the floor?

2 m

Vi = 8 m/s

X axis

Y axis

Example # 1: Sofia hits a volleyball so that it moves with an initial velocity of 8 m/s straight upward. If the volleyball starts from 2 m above the floor, how long will it be in the air before it strikes the floor?

• Given: vi = 8 m/s a = - 9.81 m/s²

∆y = - 2 m• Unknowns: vf = ? ∆t = ?• Solution: vf² = vi² + 2a∙∆y

vf² = (8 m/s)² + 2 (- 9.81 m/s²) (- 2 m)

vf² = 64 m²/s² + 39.24 m²/s²

vf² = 103.24 m²/s²→ vf = √103.24 m²/s²

vf = 10.16 m/s

P/1 Martin drops a wrench from the top of a tower 75 m tall.With what velocity does the wrench strike the ground?

Given: vi = 0 m/s

∆y = - 75 m

a = - 9.81 m/s²

Unknown: vf (at the time that it strikes the ground)

vf² = vi² + 2a∆y

vf² = (0 m/s)² + (2)(- 9.81 m/s²)(- 75 m)

vf² = 1,471.50 m²/s²

vf = √ 1,471.50 m²/s²

vf = - 38.36 m/s

P/2Ambar throws a softball straight up into the air. The ball was in the air for a total of 5.5 s before it was caught at its original position.A) What was the initial velocity of the ball?B) How high did it rise?

P/9 A bus traveled south for 1.4 h with an average velocity of 78 km/h. Stopped for 20 min, then traveled south for 1.8 h with an average velocity of 65 km/h. It stopped again for 15 min, and then traveled west for 1.4 h with an average velocity of 50 km/h

∆t1 =1.4 h v1 =78 km/h

20 min

∆t2 = 1.8 h v2 = 65 km/h

15 min

∆t3= 1.4 h v3 = 50 km/h

Physics Set # 8 September 15, 2007Student Name: ___________________________________ Class Period: ______

P/1 A bus traveled south along a straight path for 1.7 h with an average velocity of 82 km/h, stopped for 25 min, then traveled west for 2.5 h with an average velocity of 75 km/h. It stopped for 42 min and then traveled south for 1.2 h with an average velocity of 60 km/h.

a) What is the average velocity for the total trip? Answer: _________

b) What is the displacement for the total trip? Answer: _________

P/2 A red car traveled north along a straight path for 2.2 h with an average velocity of 72 km/h, stopped for 45 min, then traveled west for 1.5 h with an average velocity of 65 km/h. It stopped for 20 min and then traveled north for 1.1 h with an average velocity of 70 km/h.



P/3 A blue car traveled north along a straight path for 2.4 h with an average velocity of 72 km/h, stopped for 35 min, then traveled north for 1.5 h with an average velocity of 65 km/h. It stopped for 40 min and then traveled north for 1.1 h with an average velocity of 80 km/h.






P/5 A car on a wet road can achieve an acceleration of only – 2.5 m/s² without sliding.

Find the required stopping distance for a speed of 82 km/h

Answer: __________

P/1 Marifer Rivera went to the top of the Boca Chica Tower Building and dropped a baseball. How long will it take to the baseball to hit the

ground?. The Boca Chica Tower Building is 75 m high

A) - 4 s

B) - 3 s

C) 4 s

D) 3 s

Sofia hits a volleyball so that it moves with an initial velocity of 12 m/s straight upward. If the volleyball starts from 4 m above the floor.How long will it be in the air before it strikes the floor?

A) 1 s

B) - 3 s

C) 3 s

D) 4 s


What is the displacement for the total trip?

A) 366.80 km

B) 291.06 km

C) 150.01 km

D) 185.02 km

P/4 A rock is thrown straight upward with an initial velocity of 24.5 m/s.What is the rock displacement after 1 s?

• A) 9.81 m

• B) 19.60 m

• C) 24.50 m

• D) 29.40 m

P/5 A baseball is thrown straight upward with an initial velocity of 19.6 m/s What time interval elapses between the rock’s being thrown and its return to its original lunch point?

• A) 4 s

• B) 5 s

• C) - 4 s

• D) 10 s

P/6 Maria Elena released a coin from rest at the top of a tower and the coin hits the ground 1.5 s later. What is the velocity of the coin as it hits the ground?

A) -15 m/s

B) -21 m/s

C) -31 m/s

D) -39 m/s

P/7 When there is no air resistance, objects of

different masses

A) Fall at equal acceleration with similar displacements

B) Fall at different acceleration with different displacements

C) Fall at equal acceleration with different displacements

D) Fall at different acceleration with similar displacements

P/ 14 A rock is thrown from the top of a cliff with an initial speed of 18 m/s.If the rock hits the ground after 9 s, what is the height of the cliff?

• Given: Vi = 18 m/s, a= -9.81 m/s², ∆t = 9 s• Unknown: ∆y = ?• Solution: First find vf, then find displacement

vf = vi + a∙∆t

Vf = 18 m/s + (-9.81 m/s²) (9 s)

Vf = - 70.29 m/s

Find displacement:

∆y = ½ (vi+vf) ∆t

∆y = ½ (18 m/s -70.29 m/s) (9 s)

∆y =- 235.05 m

• Lesson 1: Vectors - Fundamentals and Operations•

Vectors and Direction• A study of motion will involve the introduction of a variety of quantities which are used to describe the physical

world. Examples of such quantities include distance, displacement, speed, velocity, acceleration, force, mass, momentum, energy, work, power, etc. All these quantities can by divided into two categories - vectors and scalars. A vector quantity is a quantity which is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity which is fully described by its magnitude. The emphasis of this unit is to understand some fundamentals about vectors and to apply the fundamentals in order to understand motion and forces which occur in two dimensions.

• Examples of vector quantities which have been previously discussed include displacement, velocity, acceleration, and force. Each of these quantities are unique in that a full description of the quantity demands that both a magnitude and a direction are listed. For example, suppose that your teacher tells you that "a bag of gold is located outside the classroom; to find it, displace yourself 20 meters." This statement may provide yourself enough information to pique your interest; yet, not enough information to find the bag of gold. The displacement required to find the bag of gold has not been fully described. On the other hand, suppose your teacher tells you that "a bag of gold is located outside the classroom; to find it, displace yourself from the center of the classroom door 20 meters in a direction 30 degrees to the west of north." This statement provides a complete description of the displacement vector - it lists both magnitude (20 meters) and direction (30 degrees to the west of north) relative to a reference or starting position (the center of the classroom door). Vector quantities are not fully described unless both magnitude and direction are listed.

• Vector quantities are often represented by scaled vector diagrams. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Vector diagrams were introduced and used in earlier units to depict the forces acting upon an object; such diagrams are

http://www.physicsclassroom.com/Class/1DKin/U1L1b.html

http://www.physicsclassroom.com/Class/1DKin/U1L1b.html

http://www.physicsclassroom.com/Class/1DKin/U1L1c.html

http://www.physicsclassroom.com/Class/1DKin/U1L1d.html

http://www.physicsclassroom.com/Class/1DKin/U1L1e.html

http://www.physicsclassroom.com/Class/newtlaws/u2l2a.html

http://www.physicsclassroom.com/Class/1DKin/U1L2c.html

motion in one dimension chapter 2. displacement and velocity ∙ ∙ a → b displacement is the...

Documents