One Dimensional Kinematics - Chapter Outline

Lesson 1 : Describing Motion with Words

Lesson 2 : Describing Motion with Diagrams

Lesson 3 : Describing Motion with Displacement vs. Time Graphs

Lesson 4 : Describing Motion with Velocity vs. Time Graphs

Lesson 5 : Free Fall and the Acceleration of Gravity

Lesson 6 : Describing Motion with Equations

1.Describing Motion with Words

1. Know the Language of Kinematics

2. Differentiate Scalars and Vectors

3. Understand Distance and Displacement

4. Be able to calculate Speed and Velocity

5. Be able to calculate Acceleration

Introduction to the Language of Kinematics

• Mechanics - the study of the motion of objects.

• Kinematics is the science of describing the motion of objects using words, diagrams, numbers, graphs, and equations. Kinematics is a branch of mechanics.

Scalars and Vectors

• Scalars are quantities that are fully described by a magnitude (or numerical value) alone.

• Vectors are quantities that are fully described by both a magnitude and a direction.

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

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

• Example: consider the motion depicted in the diagram below. A physics teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.

the physics teacher has walked a total distance of 12 meters, her displacement is 0 meters.

Distance and Displacement

example• Use the diagram to determine the resulting displacement and the

distance traveled by the skier during these three minutes.

The skier covers a distance of(180 m + 140 m + 100 m) = 420 m and has a displacement of 140 m, rightward.

example• What is the coach's resulting displacement and distance

of travel?

The coach covers a distance of(35 yds + 20 yds + 40 yds) = 95 yards

and has a displacement of 55 yards, left.

Check Your Understanding• 1. What is the displacement of the cross-country

team if they begin at the school, run 10 miles and finish back at the school?

• 2. What is the distance and the displacement of the race car drivers in the Indy 500?

The displacement of the runners is 0 miles.

The displacement of the cars is somewhere near 0 miles since they virtually finish where they started. Yet the successful cars have covered a distance of 500 miles.                                                                             

Velocity vs. Speed

• VELOCITY – change in DISPLACEMENT occurring over time– Includes both MAGNITUDE and DIRECTION– VECTOR

• SPEED – change in DISTANCE occurring over time– Includes ONLY MAGNITUDE– SCALAR

...321 dddd

t

dv

t

dv ...321 tttt

d is total displacement

...321 tttt

• In contrast, the average velocity is often computed using this formula:

Average Velocity/Speed Equations• As an object moves, it often undergoes changes in speed.

The average speed during the course of a motion is often computed using the following formula:

• The direction of velocity is the same as the direction of motion.

exampleJim gets on his bike and rides 300 meters west

in 60 seconds.

1.What is his average velocity?

2.What is his average speed?

example• Sally gets up one morning and decides to

take a three mile walk. She completes the first mile in 8.0 minutes, the second mile in 8.5 minutes, and the third mile in 9.0 minutes. What is her average speed?

total

totalavg t

dv

min12.0

min0.9min5.8min0.8

3 mimivavg

example

• In a drill during basketball practice, a player runs the length of the 30.-meter court and back. The player does this three times in 60. seconds. What is the average speed of the player during the drill?

example• The physics teacher walks 4 meters East, 2 meters

South, 4 meters West, and finally 2 meters North. The entire motion lasted for 24 seconds. Determine the average speed and the average velocity.

her average speed was 0.50 m/s and her average velocity of 0 m/s.

example• Use the diagram to determine the average speed and the average

velocity of the skier during these three minutes.

The skier has an average speed of(420 m) / (3 min) = 140 m/min

and an average velocity of(140 m, right) / (3 min) = 46.7 m/min, right

example• What is the coach's average speed and average

velocity?

The coach has an average speed of(95 yd) / (10 min) = 9.5 yd/min

and an average velocity of(55 yd, left) / (10 min) = 5.5 yd/min, left

Average speed is a measure of the distance traveled in a given period of time; Suppose that during your trip to school, you traveled a distance of 5 miles and the trip lasted 0.2 hours (12 minutes). The average speed of your car could be determined as

Instantaneous Speed - the speed at any given instant in time. For example, your speedometer tells the instantaneous speed. During your trip, there may have been times that you were stopped and other times that your speedometer was reading 50 miles per hour. Yet, on average, you were moving with a speed of 25 miles per hour.

Average Speed vs. Instantaneous Speed

hrmihours

miles

t

dv /25

2.0

5

Constant speed: the object will cover the same distance every regular interval of time

An object with a changing speed would be moving a different distance each second. The data tables below depict objects with constant and changing speed.

Constant speed vs. changing speed

In conclusion• Speed and velocity are kinematics quantities that have

distinctly different definitions. Speed, being a _______quantity, is the rate at which an object covers ___________. The average speed is the _____________ (a scalar quantity) per time ratio. Speed is ignorant of direction. On the other hand, velocity is a _________quantity; it is direction-aware. Velocity is the rate at which the position changes. The average velocity is the ______________ or position change (a vector quantity) per time ratio.

scalardistance distance

vector

displacement

Acceleration

• Definition: how fast the velocity is changing - change in VELOCITY over TIME• Change in speed (speed up or slow down)• Change in direction

• Equation:

• VECTOR• measured in velocity unit / time unit (m/s2, mi/hr, etc.)

t

va

∆ (delta) means change

∆v = vf - vi

t

vv

t

va if

According to the data, the velocity is changing over the course of time. In fact, the velocity is changing by a constant amount - 10 m/s - in each second of time. This is a case of constant acceleration.

Anytime an object's velocity is changing, the object is said to be accelerating; it has an acceleration

In a car, there are three controls that can create acceleration: gas petal, brake, steering wheel.

The Meaning of Constant Acceleration

• constant acceleration is when an accelerating object will change its velocity by the same amount each second.

Do not confuse constant velocity with constant acceleration

Acceleration is constant

Acceleration is changing

Example

Monty the Monkey accelerates uniformly from rest to a velocity of 9 m/s in a time span of 3 seconds. Calculate Monty's acceleration.

A child riding a bicycle at 15 meters per second accelerates at −3.0 meters per second2 for 4.0 seconds. What is the child’s speed at the end of this 4.0-second interval?

t

vva if

smv f /3

s

smvsm f

4

/15/3 2

The Direction of the Acceleration Vector

• The direction of the acceleration vector depends on whether the object is speeding up or slowing down or changing directions.– If an object is speeding up, then its acceleration is in

the same direction of its motion.– If an object is slowing down, then its acceleration is

in the opposite direction of its motion.– If an object is traveling east initially (vi), then goes

west (vf), its acceleration is west (same as vf)

Positive accelerations don't necessarily indicate an object speeding up, and negative accelerations don't necessarily indicate an object slowing down.

Example • The instant before a batter hits a 0.14-kilogram baseball,

the velocity of the ball is 45 meters per second west. The instant after the batter hits the ball, the ball’s velocity is 35 meters per second east. The bat and ball are in contact for 1.0 × 10–2 second. Determine the magnitude and direction of the average acceleration of the baseball while it is in contact with the bat.

t

vva if

vi = 45 m/s Wvf = 35 m/s Et = 2.0x10-2 sIf East is +Then vi = -45 m/s, vf = +35 m/s

232

/100.8100.2

)/45(/35sm

s

smsma

East

2. Describing Motion with Diagrams

1. Know how to interpret motion using Ticker Tape Diagrams

2. Know how to interpret motion using Vector Diagrams

Ticker Tape Diagrams• A common way of analyzing the motion of objects in

physics labs is to perform a ticker tape analysis. A long tape is attached to a moving object and threaded through a device that places a tick upon the tape at regular intervals of time - say every 0.10 second. As the object moves, it drags the tape through the "ticker," thus leaving a trail of dots. The trail of dots provides a history of the object's motion and therefore a representation of the object's motion.

Examples of ticker tape diagram

Example

• Oil drips at 0.4 seconds intervals from a car that has an oil leak.  Which pattern best represents the spacing of oil drops as the car accelerates uniformly from rest?

A.    .    .    .    .    .    .    .

B.    .    .         .                 .

C.    .  .  .  .      .      .      .

D.    .      . .    .   . .      . .    .

Example• A spark timer is used to record the position of a lab cart

accelerating uniformly from rest. Each 0.10 second, the timer marks a dot on a recording tape to indicate the position of the cart at that instant, as shown.

• The linear measurement between t = 0 second to t = 0.30 is 5.4 cm.

• Calculate the average speed of the cart during the time interval t = 0 second to t = 0.30 second.

• Vector diagrams are diagrams that depict the direction and relative size of a vector quantity by a vector arrow.

Vector Diagrams

Constant acceleration

Both velocity and acceleration change

3. Describing Motion with d-t Graphs

1.Understand The Meaning of Slope for a d-t Graph

2.Know The Meaning of Shape for a d-t Graph

3.Be able to Determine the Slope on a d-t Graph

The slope of the line on a d-t graph is equal to the speed/velocity of the object.

The Meaning of Slope for a d-t Graph

velocityslopet

dslope

x

yslope

• If the object is moving with a velocity of +4 m/s, then the slope of the line will be ______. If the object is moving with a velocity of -8 m/s, then the slope of the line will be _______. If the object has a velocity of 0 m/s, then the slope of the line will be ________.

+4 m/s

-8 m/s

0 m/s

Slope is speed/velocity in a d-t Graph

The Meaning of Shape for a d-t Graph

Slope is Constant, positive, velocity is constant, positive or speed is constant

Slope is increasing, positive, velocity is increasing, positive or speed is increasing. There is acceleration

As the slope goes, so goes the velocity

As the slope goes, so goes the velocity

Slow, Positive, Constant Velocity

Fast, Positive, Constant Velocity

Slow, Negative Constant Velocity

Fast, Negative, Constant Velocity

As the slope goes, so goes the velocity

Slope is negative, increasing (steeper)

velocity is negative, increasing (faster in negative direction)

Slope is negative, decreasing (flatter)

velocity is negative, decreasing (slower in negative direction)

Check Your Understanding• Use the principle of slope to describe the motion

of the objects depicted by the two plots below.

velocity is positive, increasing (faster in positive direction)

velocity is negative, increasing (faster in negative direction)

example

Describe the velocity of the object between 0-5 s and between 5-10 s.

The velocity is positive constant between 0-5 seconds

The velocity is zero between 5-10 seconds

exampleA cart travels with a constant nonzero acceleration along a straight line. Which graph best represents the relationship between the distance the cart travels and time of travel?

A B

C D

• Which graph best represents the motion of a block accelerating uniformly down an inclined plane?

example

• The displacement-time graph below represents the motion of a cart initially moving forward along a straight line. During which interval is the cart moving forward at constant speed?

example

Determining the Slope on a d-t Graph

1. Pick two points on the line and determine their coordinates.

2. Determine the difference in y-coordinates of these two points (rise).

3. Determine the difference in x-coordinates for these two points (run).

4. Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).

5. Make sure all your work has units.

Slope = -3.0 m/s

example

• Determine the velocity (i.e., slope) of the object as portrayed by the graph below.

The velocity (i.e., slope) is 4 m/s.

example

example• With the given d-t graph of Tom,

1. describe his motion during 0-5 s, 5-10 s, 10-12.5 s.

2. What is Tom’s total displacement in 12.5 seconds?

• The graph represents the relationship between distance and time for an object. What is the instantaneous speed of the object at  t  = 5.0 seconds?

example

• The graph below represents the displacement of an object moving in a straight line as a function of time. What was the total distance traveled by the object during the 10.0-second time interval?

example

4. Describing Motion with v-t Graphs

1. Know The Meaning of Slope for a v-t Graph

2. Be able to describe motion with given the Shape for a v-t Graph

3. Be able to Determining the Slope on a v-t Graph

4. Be able to Relate the Shape of d-t graph to the shape of v-t graph

5. Know the meaning of the Area on a v-t Graph

6. Be able to determine the Area on a v-t Graph

The Meaning of Slope for a v-t Graph

The slope of the line on a v-t graph is equal to the ACCELERATION of the object.

onacceleratislopet

vslope

x

yslope

The average velocity of constant acceleration can be determined by

2fi vv

v

velocity is constant, Slope is zero, Acceleration is Zero, Velocity is Increasing,

slope is positive, constant, acceleration is positive, constant.

Describe motion with given the Shape for a v-t Graph

The slope of the line on a velocity-time graph reveals the acceleration of the object.

Direction of velocity

• the velocity would be positive whenever the line lies in the positive region (above the x-axis) of the graph. Similarly, the velocity would be negative whenever the line lies in the negative region (below the x-axis) of the graph. And finally, if a line crosses over the x-axis from the positive region to the negative region of the graph (or vice versa), then the object has changed directions.

speeding up or slowing down?

• Speeding up means that the magnitude (or numerical value) of the velocity is getting large.

• Consider the graph at the right. The object whose motion is represented by this graph is ... (include all that are true):

a. moving in the positive direction. b. moving with a constant velocity. c. moving with a negative velocity. d. slowing down. e. changing directions. f. speeding up. g. moving with a positive acceleration. h. moving with a constant acceleration.

a, d and h

Example

Example

What is constant?

What is zero?

v is constant, positive. Slope is zero, a is zero.

What is constant?

What is zero?

v is constant, negative. Slope is zero, a is zero.

What is increasing?

What is constant?

v is increasing, slope is constant, a is constant. a and v are both positive.

v is decreasing, slope is constant, a is constant. a is neg. v is pos.

What is decreasing?

What is constant?

What is decreasing?

What is constant?

v is decreasing, slope is constant, a is constant. a is pos. v is neg.

What is increasing?

What is constant?

v is increasing, slope is constant, a is constant. a and v are both negative.

Check Your Understanding• Describe the motion, include the direction of motion (+ or

- direction), the velocity and acceleration and any changes in speed (speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C).

1.

A: + direction at a constant speed, zero acceleration

B: + direction, slowing down, negative acceleration

C: + direction, a constant speed slower speed than A, zero acceleration.

2.

3.

A: + direction, slowing down, constant negative acceleration

B: zero velocity, zero acceleration

C: - direction, speeding up, constant negative acceleration.

A: + direction, constant velocity, zero acceleration

B: + direction, slowing down, constant negative acceleration

C: - direction, speeding up, constant negative acceleration.

Point indicate changing direction

• Determining the Slope on a v-t Graph

A method for carrying out the calculation is1. Pick two points on the line and determine their

coordinates. 2. Determine the difference in y-coordinates for these two

points (rise). 3. Determine the difference in x-coordinates for these two

points (run). 4. Divide the difference in y-coordinates by the difference in

x-coordinates (rise/run or slope).

Check Your Understanding• Determine the acceleration (i.e., slope) of the object as

portrayed by the graph.

The acceleration (i.e., slope) is 4 m/s/s.

From 0 s to 4 s: slope = 0 m/s2 ; a = 0 m/s2

From 4 s to 8 s: slope = 2 m/s2 ; a = 2 m/s2

ExampleDescribe motion from 0 – 4 s, and 4 – 8 s with the given diagram.

Example• The velocity-time graph for a two-stage rocket is shown

below. Determine the acceleration of the rocket during the listed time intervals.

1. t = 0 - 1 second

2. t = 1 - 4 second

3. t = 4 - 12 second

+40 m/s/s

+20 m/s/s

-20 m/s/s

Relate d-t graph with v-t graph

• In d-t graph, the slope is velocity.• In v-t graph, the slope is acceleration.• Motion can be describe by both graphs.• Example: an object is moving with positive, constant

velocity.

dis

pla

cem

ent

time

velo

city

time

all graphs represent the same motiona

cce

lera

tion

time

example• Graph d-t and v-t graphs for an object moving with

constant, positive acceleration.

dis

pla

cem

ent

time

velo

city

time

Slope is velocity.Positive acceleration means increasing velocity in positive direction, which in turn means increasing slope, in positive direction.

Slope is acceleration.Positive acceleration means constant, positive slope.

all graphs represent same motion

acc

ele

ratio

n

time

acceleration is constant, positive.

Which pair of graphs represents the same motion of an object?

example

A

C

B

D

 

 Which pair of graphs represents the same motion?

example

A

C

B

D

Which pair of graphs represents the same motion of an object?

example

A

C

B

D

The meaning of the Area on a v-t Graph

For velocity versus time graphs, the area bound by the line and the axes represents the displacement.

A = b x hA = v x tA = displacement

A = ½ b x h A = ½ b x ( h1 + h2)Or

A = ½b2xh2 - ½ b1xh1

Determine the area of displacement• Determine the displacement (i.e., the area) of the object

during the first 4 seconds (Practice A) and from 3 to 6 seconds (Practice B).

120 m 90 m

example

• Determine the displacement of the object during the first second (Practice A) and during the first 3 seconds (Practice B).

5 m 45 m

example• Determine the displacement of the object during the time

interval from 2 to 3 seconds (Practice A) and during the first 2 seconds (Practice B).

25 m 40 m

• The graph below represents the velocity of an object traveling in a straight line as a function of time. Determine the magnitude of the total displacement of the object at the end of the first 6.0 seconds.

example

5. Free Fall and the Acceleration of Gravity

1. Understand the meaning of Free Fall

2. Know The Acceleration of Gravity

3. Be able to Represent Free Fall by Graphs

4. Be able to steer clear from The Big Misconception

What is Free Fall?

• Free-falling objects do not encounter air resistance.

Ticker tape trace for free fall

As an object free falls, its speed in increasing. The distance traveled during each second also increases.

• The acceleration for any object moving under the sole influence of gravity is called acceleration of gravity

• The symbol g is used to represent the acceleration of gravity.

• g = 9.81 m/s2, downward• The value of the acceleration of gravity (g) is different in

different gravitational environments. – On the moon, g = 1.6 m/s2

– On Mercury, g = 3.7 m/s2

Acceleration of gravity

• To accelerate at 9.81 m/s/s means to change the velocity by 9.81 m/s each second.

Time (s)Velocity

(m/s)

0 0

1 - 9.81

2 - 19.62

3 - 29.43

4 - 39.24

5 - 49.05

Acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path.

• If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern.

Representing Free Fall by Graphs

Velocity is increasing in neg. direction. Slope is increasing in negative direction.

Acceleration is constant, negative, slope is constant, negative. Velocity starts from zero, increase in neg. direction.

dis

pla

cem

ent

time

velo

city

time

time

velo

city

Upward: velocity is big, positive, decreasing, slope is constant (a = -9.8 m/s/s).

Top, velocity is zero. Slope remains the same (acceleration is still -9.8 m/s/s)

Downward: velocity increases in negative direction at the same constant rate of -9.81 m/s/s (slope remains the same), reaches the same speed as it started upward.

Graphs of up and down motion

time

pos

itio

n Upward: displacement increases, slope is positive, decreasing (velocity is positive, decreasing)

Top: slope = 0 (its velocity is zero)

Downward: displacement decreases, its slope increases in negative direction (velocity is negative, increasing)

• Does heavy object falls faster than lighter object? • NO. The acceleration of a free-falling object (on earth) is 9.81

m/s2. This value (known as the acceleration of gravity) is the same for all free-falling objects regardless of – how long they have been falling, – whether they were going up, down or go side ways.– How massive they were.

The Big Misconception

6. Describing Motion with Equations

1. Know The Kinematics Equations

2. Be able to apply Kinematics Equations to solve Problems

3. Be able to apply Kinematics Equations to solve problems in Free Fall

4. Be able to Relate Kinematics Equations and Graphs

• d - displacement t – time a – acceleration• - velocity vi - initial velocity vf - final velocity

Kinematics Equations

t

dv

t

vv

t

va if

2fi vv

v

tavv if

2)(2

1)( tatvd i

davv if 222

v

The strategy for solving problems:

1. Identify and list the given information in variable form.

2. Identify and list the unknown information in variable form.

3. Identify and list the equation that will be used to determine unknown information from known information.

4. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information.

5. Make sure your answer has proper unit

Example A• Ima Hurryin is approaching a stoplight moving with a

velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.)

vi vf ∆t a d

+30.0 m/s 0.00 m/s 15 m/s -8.00 m/s2 ?

v

davv if 222

a

vvd if

2

22

msm

smsmd 3.56

)/00.8(2

)/30()/0( 22

3 sig figs.

md 3.56

Example B• Ben Rushin is waiting at a stoplight. When it finally turns

green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.

vi vf ∆t a d

0.00 m/s 4.10 s 6.00 m/s2 ?

v

2)(2

1)( tatvd i

22 )10.4)(/00.6(2

1)10.4)(/00.0( ssmssmd

md 4.50 3 sig figs.

• A race car starting from rest accelerates uniformly at a rate of 4.90 meters per second2. What is the car’s speed after it has traveled 200. meters?

Example C

vi vf ∆t a d

0.00 m/s ? 4.10 s 4.90 m/s2 200. m

v

davv if 222

).200)(/90.4(2)/00.0( 22 msmsmv f

smv f /3.44

1. An object in free fall experiences an acceleration of -9.81 m/s2. ( downward acceleration.)

2. If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. (vi = 0)

3. If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematics equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s.

4. If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height.

Kinematics Equations and Free Fall

Example A• Luke Autbeloe drops a pile of roof shingles from the top of a

roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

t = 1.32 s

vi vf ∆t a d

0.00 m/s ? -9.81m/s2 -8.52m

v

up is positive, down is negative.

2)(2

1tad 2)(

2

1)( tatvd i

a

dt

2

)/81.9(

)52.8(22sm

mt

Example B• Rex Things throws his mother's crystal vase vertically

upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

d = 35.0 m

vi vf ∆t a d

26.2 m/s 0.00m/s 13.2m/s -9.81m/s2 ?

v

davv if 222

a

vvd if

2

22

)/00.81.9(2

)/2.26()/0( 22

sm

smsmd

Check Your Understanding1. An airplane accelerates down a runway at 3.20 m/s2 for

32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

3. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

d = 1720 m

a = 8.10 m/ s2

vf = -25.5 m/s

Kinematics Equations and Graphs• there are now two methods to solve problems involving

the numerical relationships between displacement, velocity, acceleration and time. – Using equations– Using graphs: the slope of the line on a velocity-time

graph is equal to the acceleration of the object and the area between the line and the time axis is equal to the displacement of the object.

• An object that moves with a constant velocity of +5.0 m/s for a time period of 5.0 seconds and then accelerates to a final velocity of +15 m/s over the next 5 seconds. What is the acceleration of the object? How far did the object travel in total of 10 seconds?

1. Using graphSlope = acceleration

t = 0 – 5 s: a = 0

t = 5 – 10 s: a = 2 m/s2

Area = displacement

t = 0 – 5 s: d = 25 m

t = 5 – 10 s: d = 50 m

dtotal = 75 m

Using Kinematics Equations and Graphs to solve problems

2. Kinematics equations:

vi vf ∆t a d1

5.0 m/s 5.0 m/s 5.0 m/s 5.0 s 0 ?

v

vi vf ∆t a d2

5.0 m/s 15 m/s 10. m/s 5.0 s ? ?

v

t = 0 s - 5 s

t = 5.0 s – 10. s

t

dv mssmtvd 25)0.5)(/0.5(1

t

dv mssmtvd .50)0.5)(/.10(2

sms

smsm

t

vva if /0.2

0.5

/0.5/.15

mmmdddtotal 75502521

practice1. A 747 jet, traveling at a velocity of 70. meters per second

north, touches down on a runway. The jet slows to rest at the rate of 2.0 meters per second2. Calculate the total distance the jet travels on the runway as it is brought to rest.

2. A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump?

3. A rocket initially at rest on the ground lifts off vertically with a constant acceleration of 2.0 × 101 meters per second2. How long will it take the rocket to reach an altitude of 9.0 × 103 meters?

4. A spark timer is used to record the position of a lab cart accelerating uniformly from rest. Each 0.10 second, the timer marks a dot on a recording tape to indicate the position of the cart at that instant, as shown.    The linear measurement between t = 0 second to t = 0.30 is 5.4 cm. Calculate the magnitude of the acceleration of the cart during that time interval.

5. Rennata Gas is driving through town at 25.0 m/s and begins to accelerate at a constant rate of -1.0 m/s2. Eventually Rennata comes to a complete stop.

a. Represent Rennata's accelerated motion by sketching a velocity-time graph. Use the velocity-time graph to determine this distance.

b. Use kinematic equations to calculate the distance that Rennata travels while decelerating.

Area = 313 m

vf2 = vi

2 + 2*a*d

(0 m/s)2 = (25.0 m/s)2 + 2 * (-1.0 m/s2)*d

313 m = d

6. Otto Emissions is driving his car at 25.0 m/s. Otto accelerates at 2.0 m/s2 for 5 seconds. Otto then maintains a constant velocity for 10.0 more seconds.

a. Represent the 15 seconds of Otto Emission's motion by sketching a velocity-time graph. Use the graph to determine the distance that Otto traveled during the entire 15 seconds.

b. Finally, break the motion into its two segments and use kinematic equations to calculate the total distance traveled during the entire 15 seconds.

Area = 500 m

For the 1st 5 second:d = vi*t + 0.5*a*t2 d = (25.0 m/s)*(5.0 s) + 0.5*(2.0 m/s2)*(5.0 s)2

d = 150 m

last 10 seconds:

d = vi*t + 0.5*a*t2

d = (35.0 m/s)*(10.0 s) + 0.5*(0.0 m/s2)*(10.0 s)2

d = 350 m + 0 md =350 m

distance = 150 m + 350 m = 500 m

7. Luke Autbeloe, a human cannonball artist, is shot off the edge of a cliff with an initial upward velocity of +40.0 m/s. Luke accelerates with a constant downward acceleration of -10.0 m/s2 (an approximate value of the acceleration of gravity).

a. Sketch a velocity-time graph for the first 8 seconds of Luke's motion.

b. Use kinematic equations to determine the time required for Luke Autbeloe to drop back to the original height of the cliff. Indicate this time on the graph.

vf = vi + a*tup

0 m/s = 40 m/s + (-10 m/s2)*tup

tup = 4.0 s

2*tup = 8.0 s

8. Chuck Wagon travels with a constant velocity of 0.5 mile/minute for 10 minutes. Chuck then decelerates at -.25 mile/min2 for 2 minutes.

a. Sketch a velocity-time graph for Chuck Wagon's motion. Use the velocity-time graph to determine the total distance traveled by Chuck Wagon during the 12 minutes of motion.

b. Finally, break the motion into its two segments and use kinematics equations to determine the total distance traveled by Chuck Wagon.

Area = 5.5 mi

first 10 minutes:

d = vi*t + 0.5*a*t2

d = (0.50 mi/min)*(10.0 min) + 0.5*(0.0 mi/min2)*(10.0 min)2

d = 5.0 mi

last 2 minutes:

d = vi*t + 0.5*a*t2

d = (0.50 mi/min)*(2.0 min) + 0.5*(-0.25 m/s2)*(2.0 min)2

d = 0.5 miThe total distance = 5.5 mi

9. Vera Side is speeding down the interstate at 45.0 m/s. Vera looks ahead and observes an accident that results in a pileup in the middle of the road. By the time Vera slams on the breaks, she is 50.0 m from the pileup. She slows down at a rate of -10.0 m/s2.

a. Construct a velocity-time plot for Vera Side's motion. Use the plot to determine the distance that Vera would travel prior to reaching a complete stop (if she did not collide with the pileup).

b. Use kinematics equations to determine the distance that Vera Side would travel prior to reaching a complete stop (if she did not collide with the pileup). Will Vera hit the cars in the pileup? That is, will Vera travel more than 50.0 meters?

Area = 101 m

vf2 = vi

2 + 2*a*d

(0 m/s)2 = (45.0 m/s)2 + 2 * (-10.0 m/s2)*d

101 m =d

10.Earl E. Bird travels 30.0 m/s for 10.0 seconds. He then accelerates at 3.00 m/s2 for 5.00 seconds.

a. Construct a velocity-time graph for Earl E. Bird's motion. Use the plot to determine the total distance traveled.

b. Divide the motion of the Earl E. Bird into the two time segments and use kinematics equations to calculate the total displacement.

Area = 488 m

first 10 seconds:

d = vi*t + 0.5*a*t2

d = (30.0 m/s)*(10.0 s) + 0.5*(0.0 m/s2)*(10.0 s)2

d =300 mlast 5 seconds:

d = vi*t + 0.5*a*t2

d = (30.0 m/s)*(5.0 s) + 0.5*(3.0 m/s2)*(5.0 s)2

d = 150 m + 37.5 m

d = 187.5 m

The total distance = 488 m

vocabulary• Vector• Scalar• Distance• Displacement• Speed • Average speed• Instantaneous speed• Velocity• Average velocity• Instantaneous velocity• Constant speed• Acceleration• Average acceleration• Constant acceleration

• Slope• Meaning of slope in p-t graph• Meaning of slope in v-t graph• Meaning of area in v-t graph• Free fall• Acceleration of gravity• Factors affect acceleration of

gravity• Kinematics equations