d change of distance s = step 1: variables step 3: put in...

Name: _____________________

Period: _____________________

cstephenmurray.com Copyright © 2008, C. Stephen Murray Legal copying of this worksheet requires written permission.

Unit 6:1

Speed

Speed

Speed is how fast something is moving. Precisely, it is how far an object travels in a certain amount

of time. The standard metric units are meters per second (m/s), but any units of distance divided by

time will work (like miles per hour [mph] or cm per sec [cps], etc).

S =

Speed equal change of distance (distanced traveled)

divided by change of time.

Change of Distance (in meters)

Change of Time (in seconds)

Speed (in meter/sec)

∆D

∆T

Where ∆D = Dfinal − Dinitial

Ex. A plane flies 200 meters in 5 sec. Calculate its speed.

Step 1: Variables S = ________

∆D = 200 m

∆T = 5 sec

Step 2: Formula

Step 3: Put in numbers and solve

Step 4: Check units

S = 40 m/sec D

ST

∆=

∆

200

5

40

DS

T

S

∆= =

∆

=

Speed is proportional to distance:

A faster object goes farther, in the same amount of time.

Speed is indirectly proportional to time:

A faster object travels the same distance in less time.

Each dot represents an object’s position at regular time intervals (time is constant).

Measuring Speed Initial Position Final Position 25 m

Distance Traveled

0:05.0 Elapsed Time

5 sec 0:00.0

To measure speed you must measure

the distance traveled and the elapsed

time.

Measure distance in meters using a

meter stick or measuring tape.

Measure time with a stopwatch or

with photogates.

Photogates (which start and stop when

an object breaks beams of light) are a

very accurate and precise method of

measuring time.

2 5 m5 m /s

5 sec

DS

T

∆= = =

∆

100m in 10sec

200m in 10sec

1

10010m/s

10

DS

T

∆= = =

∆

2

20020m/s

10

DS

T

∆= = =

∆

Doubling the distance,

doubles the speed.

200m in 20sec

200m in 10sec 2

20020m/s

10

DS

T

∆= = =

∆

1

20010m/s

20

DS

T

∆= = =

∆

Doubling the time,

halves the speed.

Constant Speed

A slower object can travel the same distance as a faster object, it

just takes more time. A fast object travels the same distance faster.

If an object moves at constant speed,

it travels the same amount of distance

each second. Notice that there is

equal space between each dot.

Why we use change of distance:

A tree 4 m away

for 2 sec has a

speed of zero

— it hasn’t moved.

That’s why we

have to use ∆D

(change of distance) instead of

distance (D).

An object has to be moving to

have speed.

Physics Explains Mathematics: If ∆T = 0 (in S = ∆D/∆T), then an

object is in two places at once,

which is impossible. This is why

dividing by zero is undefined: it

makes no physical sense!

Fast object

Slow object

Name: _____________________

Period: _____________________


Unit 6:1

1. Speed

2. Distance Traveled

3. Elapsed Time

4. ∆

5. Constant Speed

A. How far an object moves between two positions.

B. When an object covers equal amounts of time each second.

C. The rate of how fast an object travels a particular distance.

D. How many seconds it takes for an event to occur.

E. Delta: means “change of”.

True or false (and why): “A fast car goes farther.”

Can a slow object travel as far as a fast object?

Explain.

Why do we have to use change of distance (∆D) instead of just

distance (D)?

A bike moves 50 m in 10 seconds.

Calculate the speed of the bike.

Step 1: Variables:

S =

∆D =

∆T =

Step 2: Formula:

Step 3: Plug in numbers and solve:

Step 4: Give answer with units:

1. Slow speed

2. Fast speed

3. Photogate

4. Directly Proportional

5. Indirectly Proportional

A. An object that travels a long distance quickly.

B. Can travel a long distance, but requires a lot of time.

C. Uses a beam of light to start and stop a timer.

D. One quantity increases as another quantity increases.

E. One quantity decreases as another quantity increases.

_____ 5 mm/sec

_____ 10 inches

_____ 50 m/s2

____ 20 meters/sec

____ 228 meters

____ 8 minutes

____ 15 ft/min

____ 78 sec

____ 6 Newtons

Mark these as Speed, Distance, Time, or Other

A car travels 60 m/s for 10 secs.

Calculate how far it traveled.

Step 1: __________

Step 2: __________

Step 3: ______________________

Step 4: ______________________

On holiday, a family travels from Meyerville (10 miles away)

to Sprytown (70 miles away), in 3 hours. Find their speed.

Step 1: __________

Step 2: __________

Step 3: ______________________

Step 4: ______________________

A car travels 200 miles in 4 hours.

Calculate the car’s speed.

Step 1: Variables:

S =

∆D =

∆T =

Step 2: Formula:

Step 3: Plug in numbers and solve:

Step 4: Give answer with units:

____ Distance is constant and time increases.

____ Time is constant and distance decreases.

____ Time is constant and distance increases.

____ Distance is constant and time decreases.

Will Speed Increase or Decrease?

1. Is the above motion at constant speed?

2. Why or why not?

3. Each dot = 1 sec. How long did it take to go 15 m?

4. Calculate the object’s speed.

5. How would the dots change if it were moving faster?

start

Name: _____________________

Period: _____________________


Name: _____________________

Period: _____________________


Unit 6:2

Velocity and Acceleration

Example: A person

walks 4 m/s—speed (no direction).

Speed vs. Velocity Velocity is speed with direction.

20 m/s

north

20 m/s

west Same speed; different

velocities because they

have different directions.

Scalars vs. Vectors

Remember: Speed is a Scalar; Velocity is a Vector.

Vectors require direction;

Scalars only need magnitude (how big).

Vectors require magnitude (how much) and direction, often

vectors can cancel each other out (not acceleration, though).

12 m/s west Magnitude Direction

Speed: 12 m/s.

Velocity: 12 m/s west. Velocity changes when direction changes.

Ex. A plane starts at rest and ends up going

200 m/s in 10 secs. Calculate its acceleration.

Step 1: Variables Vi = 0 m/s (at rest)

Vf = 200 m/s

T = 10 sec

a = _________

Step 2: Formula

V

aT

∆=

∆


Step 4: Add units

a = 20 m/s2

200 0

10

20020

10

f iV VVa

T T

a

−∆ −= = =

∆ ∆

= =

Acceleration

Acceleration is how fast

you change velocity OR

how much the velocity

changed in a certain

amount of time.

An object accelerates

when it changes speed

OR changes direction!

a =

Acceleration equal change of velocity

divided by change of time.

Change of Velocity (in meters/sec)

Change of Time (in seconds)

Acceleration

(in m/s2)

∆V

∆T

, so, final initial

final initial

V VV V V a

T

−∆ = − =

∆

Ex. A race car starts at 40 m/s slows to 10 m/s

in 5 seconds. Calculate the car’s acceleration.

Step 1: Variables Vi = 40 m/s

Vf = 10 m/s

T = 5 sec

a = _________

Step 2: Formula

V

aT

∆=

∆


Step 4: Add units

a = –6 m/s2

10 40

5

306

5

f iV VVa

T T

a

−∆ −= = =

∆ ∆

−= = −

Neg. means

slowing

down

Negative

acceleration

means an object

is slowing down

OR speeding up

in the negative

direction.

Slowing down

is also called

“deceleration”.

Finding ∆V.

∆ always = final – initial.

∆V = Vfinal – Vinitial OR

Final velocity – Initial velocity.

If ∆V is positive the object is

speeding up.

If ∆V is negative the object is

slowing down (see below).

Distance and Acceleration

Pos. means

speeding

up

Measuring Acceleration

To measure an object’s

acceleration you need to

measure the object’s

velocity before and after

the acceleration.

If the object starts at rest

you know that Vi = 0m/s.

If the object stops

you know that Vf = 0m/s.

Points are equal distance, so velocity is constant.

Since the velocity is constant, the initial and final velocity

are equal and the acceleration equals zero.

The distance between the points is increasing, so velocity

is increasing. The object is accelerating: traveling faster

each second and covering more distance every second.

An object that is accelerating will travel farther each second.

4 m

1 sec

4 m /s

i

in it ia l

DV

T

V

∆= =

∆

=

8 m

1 se c

8 m /s

f

fin a l

DV

T

V

∆= =

∆

=

2

8 4

2

42m/s

2

f i

initial

V Va

T

V

− −= =

∆

= =

Constant Speed—Equal Distance Positive Acceleration—Increasing Distance

Accelerates

for 2 seconds

So ∆T = 2 sec 4 m in 1 sec

Measure Vf

(Final Velocity)

8 m in 1 sec

Measure Vi

(Initial Velocity)

Measure ∆T

(Time it took to Accelerate)

Name: _____________________

Period: _____________________


Unit 6:2

Mass, Time, Distance, Velocity, or Acceleration?

___ 2 hrs

___ 3 m/s

___ 6 mph/sec

____5 sec

____9 mph

____12 m

____8 kg

____4 m/s2

____1 in

___A bike goes 25 m/s toward

main street.

___A person walks 4 mph.

___A plane flies 200 m/s.

___A bird flies 100 mph due

south.

Speed (S) or Velocity (V)

___ 40 mph toward Dallas.

___ 3 m/s2 to the left.

___ 10 meters up the hill.

___ 12 meter per sec2.

___ Direction matters.

___ No direction is needed

Scalar (S) or Vector (V)

A dragster’s top acceleration is 60 m/s2. If it starts from rest at the

starting line, how fast will it be going after 3 seconds?

Variables:

Formula:

Solve:

A person starts running from 2 m/s to 6 m/s in 2 seconds.

Calculate the person’s acceleration.

Variables:

Formula:

Solve:

A car travels 30 m in 5 seconds. After accelerating for 3 seconds,

it travels 20 m in 2 seconds. Calculate the car’s acceleration.

1) Find Vi.

2) Find Vf.

3) Calculate a.

A plane stops from 250 mph in 25 seconds.

Calculate the planes acceleration.

Variables:

Formula:

Solve:

10 m/s

10 m/s

Accelerating? Yes, No, or Maybe?

___ At constant velocity.

___ Going 5 m/s then going 3 m/s.

___ A car going around a corner.

(see graphic at right).

___ At constant speed.

___ Stopping.

___ A car at rest.

Object A accelerates at 10 m/s2; Object B accelerates at 5 m/s2.

___ Which one will go faster?

___ Which one will take more time to reach a high speed?

___ If they start at rest, which one will reach 40 m/s first?

___ Which one goes farther (longer distance)?

___ Which one will be 100m away sooner?

Object A

Object B

Object C

Choose which of the above applies to the following

____ Constant speed.

____ Positive acceleration.

____ At constant velocity.

____ Accelerating.

____ Decelerating.

____ Acceleration = 0.

____ Distance increases

____ Starts at rest.

____ Is stopping.

____ Constant direction.

____ Negative acceleration.

____ Vi = Vf

Object D

Give what you know for the following: (Vi, Vf, or a)

An object at constant velocity.

An object that is stopping.

An object that accelerates from rest.

An object at rest.

Name: _____________________

Period: _____________________


Name: _____________________

Period: _____________________


Unit 6:3

Graphing Linear Motion

Conventions: X-axis (horizontal): Independent or manipulated variable.

Y-axis (vertical): Dependent or responsive variable.

Meaning of Slope Changes The slope of a position vs. time graph is speed. The slope of a velocity vs. time graph

is acceleration. Yet for some graph, the slope has no physical meaning.

Position vs. Time

Graphs

Graphing Variables

A Position vs. Time graph shows where an object is at a particular time. The slope of a position vs. time

graph shows the speed of an object. A steeper line shows faster speed. A downward line means negative

speed (moving left or coming back).

A steeper line = a faster speed.

306m/s

5LineA

DS

T

∆= = =

∆

303m/s

10LineB

DS

T

∆= = =

∆

Object B travels 30 m in 10 seconds.

Line B shows slow positive speed.

Position vs. Time

0

5

10

15

20

25

30

35

0 1 2 3 4 5 6 7 8 9 10 11 12

Time (sec)

Po

sit

ion

(m

)

Line A fa

st spe

ed

slow sp

eed Line B

negative speed Line D

Starting position (t = 0)

no speed Line C

Object C stays 15 m away.

Line C shows a speed of zero.

00m/s

10LineC

DS

T

∆= = =

∆

Object D travels –20 m in 10 seconds.

Line D shows slow negative speed.

202m/s

10LineD

DS

T

∆ −= = = −

∆

Object A travels 30 m in 5 seconds.

Line A shows fast positive speed.

To figure out what the

slope of a graph means:

divide the y-axis units by

the x-axis units to find the

units for the slope.

Scientists have rules for choosing which variable is graphed on which axis. This allows scientists to

understand how an experiment was conducted just by reading the graph.

Independent

vs. Dependent The independent vari-

able is not affected by

the changing depend-

ent variable. The de-

pendent variable

changes as the inde-

pendent variable

Manipulated

vs. Responsive Sometimes it is hard to

determine which is the

independent variable. In

these cases, the variable

that you are manipulating

(varying) will graphed on

the x-axis.

Velocity vs. Time

Dep

end

ent

vari

ab

le

Vel

oci

ty (

in m

/s)

Time (in sec)

Independent variable

Acceleration vs. Force

Res

po

nsi

ve v

ari

ab

le

Acce

lera

tio

n (

in m

/s2)

Force (in N)

Manipulated variable

The above object’s acceleration

changes (responds) as the force is

changed (manipulated).

This graph shows the change of acceleration

over time which is undefined.

Acceleration vs. Time

Acce

lera

tio

n

(in

m/s

2)

Time (in sec)

23m/s

m/s ?s

rise ySlope

run x

∆= = = = =

∆

Velocity vs. Time

Vel

oci

ty (

in m

/s)

Time (in sec)

This graph shows the change of velocity

over time which is acceleration.

2m/sm/s acceleration

s

rise ySlope

run x

∆= = = = =

∆

Slope = −acceleration

Time (as in “a particular

moment in time”) is always an

independent variable (x-axis)

because nothing stops time.

Time does not change with

speed; speed changes over time.

Duration (how long it takes) can be

dependent (y-axis). Ex. The period

of a spring (how long it takes to

move back and forth) changes as

more mass is added. Mass is inde-

pendent, not period of time.

The slope of

this graph

means nothing.

The manipulated variable is the

one you are changing in your ex-

periment and is often the experi-

mental variable.

Meaning of Slope

units of y-axis

units of x-axis

rise

run=

=

Name: _____________________

Period: _____________________


Unit 6:3

When was the object moving at 150 m/s? ______________________

How fast is the object going after 10 seconds? __________________

What was the initial velocity of the object? _____________________

How much speed does it gain in the first 5 seconds? ______________

Find the slope of the graph (must show work) ___________________

What does the slope you just found stand for? __________________

1. Linear

2. Responsive variable

3. Independent variable

4. Dependent variable

5. Slope

6. Manipulated variable

A. Vertical axis (y) variable.

B. The variable you change.

C. Any straight line graph.

D. Measure of how steep a line is.

E. The variable on the horizontal axis (x-axis).

F. What changes because you change something.

Position vs. Time

0

20

40

60

80

100

120

0 2 4 6 8 10 12

Time (sec)

Po

sit

ion

(m

)

What does the slope of this line show? ________________________

How much time does it take Object A to travel 100m? ___________

How much time does it take Object B to travel 100m? ___________

Which Object (A or B) has the faster velocity? _________________

Object C starts where? ________ Object C ends where? _________

Which line shows negative speed? ___________________________

Which line shows positive speed? ___________________________

Which line shows an object at rest? __________________________

What is Object D’s initial position? __________________________

Which is the independent variable? ___________________________

Which is the dependent variable? _____________________________

Where was the object at 4 seconds? ___________________________

Where did the object begin? _________________________________

Find the slope of the graph (must show work)

What does the slope you just found stand for? ___________________

The slope of this graph means:

Which segment shows:

Increasing velocity:

Constant velocity:

Positive acceleration:

Negative acceleration:

Speeding up:

Slowing down:

Position vs. Time

Time

Po

siti

on A

B C D

Which segments shows:

At rest:

Fast speed:

Slow speed:

Going backwards:

Going forward:

Negative speed:

Speed equals zero:

Position vs. Time

0

2

4

6

8

10

12

14

16

18

0 1 2 3 4 5 6

Time (sec)

Po

siti

on (

m)

Velocity vs. Time

Time

Vel

oci

ty

A B

C D

Circle the Independent Variable

A. Time or Acceleration

B. Velocity or Time

C. Time or Position

Circle the Manipulated Variable for these Graphs

A. Force on an object or Acceleration of the object?

B. Period of a Spring or Mass hung from the spring?

C. Number of batteries or Brightness of a bulb?

Velocity vs. Time

0

50

100

150

200

250

300

350

0 1 2 3 4 5 6 7 8 9 10 11

Time (secs)

Ve

loc

ity

(m

/s)

A B

C

D

Name: _____________________

Period: _____________________



Name: _____________________

Period: _____________________

Step 1: Calculate slope (m) with two good points

(where the line hits “cross hairs” [see graph]).

The slope (the tilt) tells you the rate of

change of y, not x. In this case slope tells us the

change of velocity which is acceleration (notice the

units: “m/s2”). More slope (more tilt) would mean more

acceleration, the velocity would change faster.

Step 2: Find the y-intercept (b)

(where the line crosses the y-axis ).

b = 2 m/s

The y-intercept (the vertical shift) tells you

the initial condition of the object: this object’s initial

velocity = 2 m/s (velocity at 0 sec).

Step 3: Find what the x and y variables are

for this graph.

y-axis = velocity = v (in m/s)

x-axis = time = t (in sec)

Why is this step so important? If you leave x and y

in the linear equation it is easy to be confused when putting

in numbers. Which one is time? Which one is velocity?

If you change your variables there will be no confusion.

x and y will be different for each graph!

Step 4: Put all of the above into the linear equation

to find the equation for this particular line.

Step 5: Use the linear equation to solve problems. You now

have a formula for the object depicted on the graph.

Given any x or y you can now solve for the other.

y = mx + b y - axis variable

y-intercept

slope

x - axis variable

Must be for the same point

24 m/s4 m/s

1 s

rise ym

run x

∆= = = =

∆

Velocity vs. Time

0123456789

1011

0 0.5 1 1.5 2 2.5 3

Time (sec)V

elo

cit

y (

m/s

)

Another “good point”

y – intercept

∆y =

8 – 4

= 4 m/s

∆x = 1.5 – 0.5

= 1 s

y = mx + b

v = 4t + 2 THIS LINE

Any line y = v

x = t

m = 4 m/s2

b = 2 m/s

The linear equation is the form of ANY straight line.

The linear equation is just a formula and like any other formula

you can solve for any unknown given the other variables.

For example, if you are given x and y for a point and the

y-intercept (b), you could solve for the slope of the line.

v = 4t + 2

20 = 4t + 2

20—2 = 4t

18 = 4t

t = 18/4

t = 4.5 sec

Example: When will the object graphed above

be going 20 m/s?

Solution: use the linear equation for this line.

v = 20 m/s

t = _____

The object will be going 20 m/s at 4.5 seconds.

(Notice this is a point beyond the graph. This is

known as extrapolation. “Extra” = outside.)

Velocity vs. Time

0123456789

1011

0 0.5 1 1.5 2 2.5 3

Time (sec)

Velo

cit

y (

m/s

)

The x variable for this graph is time, t.

The y variable

for this graph

is velocity, v.

The Linear Equation


Name: _____________________

Period: _____________________

1. m, b, x, or y?

A. ____vertical axis.

B. ____Slope

C. ____y-intercept

D. ____horizontal axis

E. ____Dependent variable.

F. _____Gives initial condition.

G._____Independent variable

H._____Rate of change of y.

I. _____Are constants.

J. _____Are variables.

2. Write the equation for slope.

3. Write the equation that defines a line.

Po

siti

on

(in

m)

Time

(in sec)

Vel

oci

ty

(in

m/s

)

Time

(in sec)

Acce

lera

tio

n

(in

m/s

2)

Time

(in sec)

A B C

4. Use the graphs above to answer the following.

A. What is the y variable for graph C?

B. What is the x variable for graph B?

C. What is y for graph A?

D. What is x for graph B?

E. In the linear equation what is y for graph B?

Position vs. Time

-4

-2

0

2

4

6

8

10

12

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time (sec)

Po

sit

ion

(m

)

Graph A Velocity vs. Time

0

1

2

3

4

5

6

7

8

0 1 2 3 4 5 6 7 8

Time (sec)

Ve

loc

ity

(m

/s)

Graph B

5. Use the two graphs below to answer the following questions.

A. What is the y variable for Graph B?

B. What is the x variable for Graph A?

C. What is the y-intercept for Graph A?

D. What is the y-intercept for Graph B?

E. Over time, what changes in Graph A?

F. So, what does the slope of Graph A show?

G. Over time what changes in Graph B?

H. So, what does the slope of Graph B show?

6. Use Graph A above to answer the following questions. 7. Use Graph B above to answer the following questions.

A. On the above graph, calculate the line’s slope.

B. Put a square around the y-intercept.

C. Write the linear equation variables for this line:

m =

b =

y =

x =

D. Write the linear equation

for this line:

E. Seconds would go into what part of this linear equation?

F. How fast is the object going after 10.5 seconds?

G. What is the initial velocity of the object?

A. On the above graph, calculate the line’s slope.

B. Put a square around the y-intercept.

C. Write the linear equation variables for this line:

m =

b =

y =

x =

D. Write the linear equation

for this line:

E. Meters would go into what part of this linear equation?

F. At what time will the object be at 15 meters?

G. What is the initial position of the object?

Linear Equation— p2


Name: _____________________

Period: _____________________

12 in

1ft

Conversions are how we change units. 1 foot

equals 12 inches: the amount is the same,

but how we express the amount is different.

12 in = 1 ft

1 Equality

2 Conversion Factors

12 in

1 ft

Example 1: Convert 35 m/s to ft/sec.

Step 1: Write what you are given as a fraction with one unit

on top and one unit on bottom.

35 m

1 sec

To do a conversion you need a con-

version factor. Conversion factors

come from equalities. Since any-

thing divided by itself is 1, a conver-

sion factor also equals 1. Any equal-

ity can make two conversion factors.

Conversion Factors

How To Do Conversions

If you need to perform multiple conversions, you can either do each

conversion independently or in one long chain.

Convert: 560 hours to weeks.

560 hr 1 days 1 w eeks

1 24 hr 7 days

560 w eeks 3.33 w eeks

24(7)

= =

Convert: 560 hours to weeks.

Chaining:

560 hr 1 days = 23.33days

1 24 hr

23.33 days 1 weeks= 3.33 weeks

1 7 days

One conversion at a time:

Both ways will give the same answer, but once you master the

chaining method, you will find it easier and less prone to mistakes.

OR 1 ft

12 in

Follow these steps exactly and you will be able to perform any conversion.

35 m 3.3 ft

1 sec 1 m

Since we know 3.3 ft = 1 m.

35 m ft

1 sec m

NO NUMBERS YET!

Notice: m’s are diagonal.

Step 2: In parenthesis and WITHOUT NUMBERS, write the units

you want to get rid of diagonal from itself. In the other

part of the fraction write what you’re converting to.

Step 3: Put numbers into the parenthesis so that the top

equals the bottom.

Step 4: Cancel out the units BUT NOT THE NUMBERS!

Step 5: Do the math. Multiply the numbers if they are both

on top. Divide if the second one is on the bottom.

35 m 3.3 ft

1 sec 1 m

m’s cancel because m/m = 1

35 m 3.3 ft 35(3.3) ft115.5 ft/sec

1 sec 1 m 1 sec

= =

35 m/s (given) becomes

Multiple Conversions

50 / 60 0.83 mi/min= =

Ex. 2: Convert 50 mi/hr to mi/min.

50 miStep 1:

1 hr

50 mi hrStep 2:

1 hr min

50 mi 1 hrStep 3:

1 hr 60 min

50 mi 1 hrStep 4:

1 hr 60 min

50 mi 1 hr 50 miStep 5:

1 hr 60 min 60 min

=

hr’s are diagonal

from each other

put in #s

since hr/hr = 1

write as a fraction

60 on

bottom

means ÷

If you have a single unit, just write it over 1.

15 ft

115 ft becomes

Why no numbers? Because most mistakes are made by

assuming that you will multiply or divide by some

number. Let the units guide you NOT the numbers.

Conversions


Name: _____________________

Period: _____________________

2. Find the mistakes in each of the following and write a corrected version underneath.

4 mph 5,280 ft

1 1 mi

52.2 m 1 min

1 sec 60 sec

82 years 320 days

1 1 year

1 ft42 in

12 in

1 2 in

1

A. Ex. 12 in B. 6 m/sec C. 4 sec D. 19 mph E. 3.7 meters

1. Prepare these numbers for conversion.

16 m 1 m

1 sec 3.3 ft

=

220 sec 1 min

1 60 sec

=

4. Do the following conversions. Given: 1 in = 2.54 cm;

3.3 ft = 1 m; 12 in = 1 ft; 5,280 ft =1 mi (mile)

A. Convert 3.5 miles to feet

B. Convert 6 ft to meters

C. Convert 2.5 weeks to days

D. Convert 2500 seconds to minutes

E. Convert 18 m/sec to m/min

F. Convert 60 mph (miles) to m/hr (meters)

5. Convert 120 m/min to m/hour.

6. There are 1,000,000 micrometers (µm) per meter.

How many meters is 48,000 µm?

7. A. Convert 15 in/min to feet per min

B. Using the above answer, convert to feet per second.

8. A. Convert 540 cm/min to cm/sec

B. Convert to inches per second.

9. Convert 12 mph (miles) to m/s (meters).

6 4

1 3

=

Conversions— p2

3. Perform the following functions (do the math).

A. B. C. D.

A. B. C. km km

1 1

=

km 1

1 km

=

D. E. F. m sec

sec min

=


Name: _____________________

Period: _____________________

Position (x)

Distance (D)

Displacement (∆x)

Position is where you are relative to a reference point. xi is the initial position. xf is the final position.

Position, distance, and displacement are all measured in meters, but they have different physical meanings.

Distance is how far you have traveled between two positions. Distance is always positive.

Displacement is the straight line distance between the initial and final positions. ∆x = xf − xi.

Displacement can be positive or negative.

0 1 2 3 4 5 -1 -2 -3 -4 -5

in meters in meters

Final position:

xf = −4 m

Initial position:

xi= 3 m

Reference

point

Distance: D = 7 meters

Displacement: ∆x = −7 m = -4 −3

In this example the displacement and distance

are the same amount, but the displacement is

negative, because they moved to the left.

Vertical Displacement (∆y)

An object that travels a circular path

and ends up at its starting point has a

distance equal to the circumference

of the circle: D = 2πr.

Yet the displacement is zero because

it ended up where it started: its ini-

tial and final positions are the same. xi

r

D = 2πr

∆x = 0 m

xf

But what if an object turns around? The distance trav-

eled would continue to increase, but the displacement

would begin to decrease as the final position became

closer to the initial position. If it were to return to its

initial position, its displacement would be zero.

D1 = 6 m

0 1 2 3 4 5 -1 -2 -3 -4 -5

in meters in meters

D2 = 3 m ∆x

Total Distance: D = 9 meters

Displacement: ∆x = 3 meters

xi

xf

D1 = 4 m

D2 =

3 m ∆x =

5 m

(since

32 +

42 =

52 )

Remember that displace-

ment is the straight line

distance between the

initial and final positions.

In some cases you may

need to use Pythagorean

theorem to find ∆x:

A2 + B2 = C2.

–∆y

If an object moves up

or down we use ∆y,

not ∆x. Remember

that down is negative,

so a falling object will

have a negative

displacement.

Initial position

∆x and ∆y

When an object moves at an angle we can find

both the x and y displacements independently. ∆y is just like ∆x except it is up or down..

+∆y means the final position is above the initial.

−∆y means the final position is below the initial.

initial

final

+∆x

−∆y

In this example, the

object has a positive

x-displacement

(because it moved

to the right) and

a negative

y-displacement

(because it fell).

Position, Distance, Displacement


Name: _____________________

Period: _____________________

0 1 2 3 4 5 -1 -2 -3 -4 -5

in meters in meters

A B C D

1. Use the number line at the right to answer the

following questions.

2.

40 m

30 m

I II

III

A. What is the position of letter A? xA =

B. What is the position of letter C? xC =

C. What is the distance from A to C?

D. What is the distance from D to A?

E. What is the displacement from D to A?

A. If II is the reference point, what is the position of the

car at I?

B. What is the total distance the car traveled? D =

C. What is the car’s first displacement from I to II?

D. What is the total displacement of the car from

I to III: ∆x =

3. A. What is the curved distance from a to c?

B. What is ∆x from a to c?

C. What is the curved distance from c to a?

D. What is ∆x from c to a?

E. What is the distance 1 time around the circle?

F. What is the displacement 1 time around?

4. A ball is thrown horizontally from the top of a 7 m tall ledge.

A. What is its vertical displacement during the fall? ∆y =

B. What is its horizontal displacement? ∆x =

C. What is the total displacement (straight line) from start to finish?

A. From D to E: ∆x = ∆y =

B. From A to M: ∆x = ∆y =

C. From B to O: ∆x = ∆y =

D. Draw this path: D to B to J to L:

i. ∆x = ii. ∆y = iii. Dtotal =

E. What is the total displacement (straight line) from B to P?

5. The grid at the right is 1 m between each of the horizontal and vertical rows.


Name: _____________________

Period: _____________________

Distance (D) is how far an

object has traveled. Displace-

ment (∆x) is how far an object

has moved from its original

position. Displacement can

be positive or negative.

Velocity is how fast an object changes position. Vi = initial velocity; Vf is final velocity.

Displacement (∆x) in m

–∆y

Vertical Displacement

If an object moves up

or down we use ∆y,

not ∆x. Remember

that down is negative

(as is moving to the

left for ∆x).

An object that travels a

circular path and ends

up at its starting point

has a distance equal to

the circumference (2πr),

but no displacement.

Velocity (v) in m/s

V is – if moving

to the left V is + if moving

to the right

When an object turns around v = 0 m/s.

V is – when an object moves down. V is + when an

object moves up.

Acceleration (a) in m/s2 Acceleration is how fast an object changes velocity. The kinematic equations work only

with constant acceleration. Acceleration can be positive, negative, or zero.

Time (t) in sec Time is always elapsed time, not a point in time. Also, time in any other units other than

seconds must be converted first.

A positive acceleration occurs

when an object speeds up in the

positive direction or slows

down in the negative direction.

A negative acceleration occurs

when an object speeds up in the

negative direction or slows

down in the positive direction.

Vi is – a is + Vf is 0

Vi is +

Vf is 0

a is –

start stop

r

D = 2πr

∆x = 0 m

Choosing an Equation Just as with any other word problem, first write a variable list

from the given information, including your unknown. Then

choose an equation which includes these variables.

Example 1: An object moves 12 m to the left in

4 seconds. If its initial velocity was 5 m/s to the

right, what is the acceleration of the object?

Variables:

∆x = –12 m

(moves left)

t = 4 sec

Vi = 5 m/s

(right is +)

a = _____

Vf is not in this list

2112 5(4) (4)

2

112 20 (16)

2

a

a

− = +

− = +

2

12 20 (8 )

32 8

4m/s

a

a

a

− = +

− =

= −

Kinematic Equations

With these five equations you

are able to calculate for any

unknown in linear motion.

Example 2: An object at rest ends up moving

20 m/s to the right after traveling 80 meters

to the right. How much time did this take?

Variables:

Vi = 0 m/s

(at rest)

Vf = 20 m/s

∆x = 80 m

t = _____

“a” is not in

this list.

1( )

2

180 (0 20)

2

180 (20)

2

80 10

8 sec

i fx v v t

t

t

t

t

∆ = +

= +

=

=

=

2

2

2 2

1( )

2

1( ) ( )

2

1( ) ( )

2

(2 )

i f

f i

i

f

f i

x v v t

v va

t

x v t a t

x v t a t

v v a x

∆ = +

−=

∆ = +

∆ = −

= + ∆

“a” is not used

“∆x” is not used

“vf” is not used

“vi” is not used

“t” is not used

Big Trick: Figure out which variable is not used in your variable list,

then chose the equation that is also not using this variable. Remember

that your unknown is still in your list, you just don’t know its value yet.

If your unknown is not in the equation, you can’t solve for it.

Variables:

∆x = 50 m

t = 10 sec

a = 2 m/s2

Vf = _____

Vi is not used

in our list.

21( ) ( )

2f

x v t a t

∆ = −

Vf is on this list:

it is the unknown.

So choose this equation

because it does not use “Vi”

and has all of your variables.

“a” is

not used “Vf” is

not used 21

( ) ( )2

ix v t a t

∆ = +

Kinematic Equations (R)


Name: _____________________

Period: _____________________

2. A person swims to the other end of a 20 m long pool

and back. What is their displacement?

3. A rock falls 15 m.

Is this vertical or horizontal motion?

What is the displacement of the rock?

4. A car moving 12 m/s stops in 3 seconds.

Vf =

5. You throw a rock into the air and catch it as it returns.

What is the displacement of the rock?

1. ∆x, ∆y, t, vi, vf, or a?

___ 2 sec

___ 3 m/s

___ 6 m right

____ How far...

____ 4 m/s2

____ How fast...

____How long did

it take?

____How high...

6. Choose the correct kinematic equation for the following:

Variables:

a = 2 m/s2

Vi = 6 m/s

Vf = −6 m/s

∆x = ____

What’s missing from the list:

So use this equation:

Variables:

a = 4 m/s2

t = 10 s

Vf = −2 m/s

∆x = ____



Variables:

a = −3m/s2

Vi = 6 m/s

Vf = −12 m/s

t = ____



Variables: Equation and Solve:


Variables: Equation and Solve: Variables: Equation and Solve:

7. In 10 seconds a car accelerates 4m/s2 to 60 m/s. How fast

was the car going before it accelerated?

8. A object moving 2 m/s experiences an acceleration of 3m/s2

for 8 seconds. How far did it move in that time?

9. An object at rest starts accelerating. If it travels 40 meters

to end up going 20 m/s, what was its acceleration?

10. A model rocket climbs 200 m in 4 seconds. If was moving

10 m/s to begin with, what is its final velocity?

11. A car stops in 120 m. If it has an acceleration of –5m/s2,

how long did it take to stop?

12. An object drops 20 m from a cliff. If it started at rest and is

going 20 m/s just before it hits the ground, what is its accel-

eration?

Kinematic Equations— p2


Name: _____________________

Period: _____________________

“Free-fall” is the expression we use for

any object falling in our earth’s gravi-

tational field with only gravity acting

on it: it is falling freely.

On the earth the acceleration due to

gravity (“g”) is 9.8 m/s2. Because we

usually call “up” the positive direction,

g is given a negative value.

Not all falling objects are in

freefall. Parachutes, balloons,

and airplanes all have air resis-

tance or buoyancy slowing

them down: a ≠ –9.8 m/s2.

Special Situations

For objects in freefall:

a = g = –9.8 m/s2

Without air resistance light and heavy objects

fall at the same rate. This can be proven in a

vacuum chamber when all of the air is

removed. On the moon,

Apollo 15 astronauts

showed this by drop-

ping a feather

and a hammer at the

same time. They

hit the ground at the same time.

The moon has no atmosphere so it is a

vacuum. It has gravity, but no air resistance.

Freefall

vi =

5 m/s

vf = −vi

= −5 m/s

Returns to initial position: ∆y = 0, and vf = −vi .

If an object comes back to its

starting position then ∆y = 0 m

and vf = −vi.

Examples: “Back to the

ground”; “back to your hand”;

“from ground to ground.”

Example 1. An object is dropped from 40 m.

How fast is it going at the bottom?

2 2

f i

2

f

2

f

f

v = v + 2a∆y

v = 0 +2(-9.8)(-40)

v = 784

v = 784 = 28±

Because it is going down we

choose the negative: vf = –28 m/s

Variables:

Dropped so: vi = 0 m/s

Falling so: ∆y = –40 m

a = –9.8 m/s2

vf = ______

(t is not used)

Example 2. An object is thrown up into the air going 8 m/s.

How long does it take for it to get back to the ground?

f iv = v + at

-8 = 8 + (- 9 .8 t)

-16 = -9 .8 t

-16t = 1 .63 sec

-9 .8=

Notice that mass is

not in the equation,

meaning two objects

of different mass

will hit the ground

at the same time!

Variables:

vi = 8 m/s

Because it comes back

to its original position:

∆y = 0 m

vf = –8 m/s

a = –9.8 m/s2

t = ______

Because we have all of the variables,

we choose the easiest equation.

If the object’s final

position is at the

top, then y is + and

vf = 0 m/s.

Examples: “How

high does it go?”;

“find maximum

height.”

Final position at top: Vf = 0 m/s.

−∆y

vi =

0 m/s

Dropped objects

begin at rest and go

down, so ∆y is −

and vi = 0 m/s.

Examples: “is

dropped”; “pushed

off a ledge”; “sitting

on a cliff.”

Dropped objects: ∆y is −; vi = 0 m/s.

Because a = g, very little information is needed to be able to solve a freefall problem, but

often you must use your everyday knowledge to pull additional information out of a problem.

2

2

2 2

1( )

2

( )

1( ) ( )

2

1( ) ( )

2

(2 )

i f

f i

i

f

f i

y v v t

v v at

y v t a t

y v t a t

v v a y

∆ = +

= +

∆ = +

∆ = −

= + ∆

“a” is not used

“∆y” is not used

“vf” is not used

“vi” is not used

“t” is not used

Vertical Kinematic Equations

The kinematic

equations become

the vertical kinematic

equations just by

putting ∆y in for ∆x.

Choose the correct

equation by deciding

which variable is not

used in your problem.

vi =

5 m/s

vf =

0 m/s

∆y = 0

+∆y


Name: _____________________

Period: _____________________

3. What do we call any space that has no air?

4. If the two objects at the right are dropped

in a vacuum, which will hit the ground first?

5. What if there is air resistance?

8. An object is dropped from a 15 m ledge. How fast it is

moving just before it hits the ground?

10. A ball is thrown 24 m/s into the air. How high does it go? 11. A rock falls off a cliff and falls for 3 secs. How high was

the cliff? Variables: Equation and Solve:


12. An object is thrown up into the air going 9 m/s. How fast

is it going 2 seconds later?

13. An object is thrown 16 m/s straight up from a 7 m tall

cliff. How much time does it take to hit the ground

below?

2. Freefall? Yes or No?

_____ An airplane.

_____ A volleyball hit over a net.

_____ Paper floating down.

_____ A ball rolling off a table.

_____ A person jumping.

6. “An object is thrown 3 m/s from the ground and it lands on the ground.”

vi = _______; vf = _______; a = _______; ∆y = _______.

7. “An object is thrown into the air going 80 m/s. How high does it go?”

vi = _______; vf = _______; a = _______;

20 kg

5 kg

Freefall— p2


9. A person throws tennis ball 6 m/s straight up. How long

does it take for it to come back to their hand?

A

The Ground

VA = 12m/s

a = ____

VB = ____

a = ____

VE = ____

a = ____

VD = – 4m/s

a = ____

VC = ____

a = ____

1. Fill in the missing information.

B D

C

E



Name: _____________________

Period: _____________________

7) The tape timers at the left show 4 objects moving to the

right. The dots show the positions of the objects each

second. Which objects apply to the following?

Linear Motion In Class Test Review

A. Centimeters or megameters?

B. Micrometers or millimeters?

C. Kilometers or megameters?

D. Centimeters or millimeters?

2) Convert 18 m/s to meters per min.

1) Circle the bigger one:

3) An object moves 120 m in 15 seconds.

Calculate the object’s speed.

4) An object moves 18 m/s. How long does

it take the object to move 154 m?

This is NOT the homework!!!




____ Accelerating.

____ Decelerating.




____ Is stopping.



____ Vi = Vf

Object A

Object B

Object C

Object D

8) A car begins at a stop sign. It ends up going 100 m in 6.5 seconds. Find the car’s acceleration.

Variables: Equation and solve:

9) +, –, or 0?

A. ____Acceleration of an object that is moving to the left and speeding up?

B. _____Acceleration of an object that is moving up and slowing down?

C. _____Velocity of an object that is moving to the right?

D. _____Displacement of an object that ends at its starting position?

E. _____Acceleration of an thrown object at the top of its path?

F. _____Displacement of an object moving to the left?

10) What is the acceleration of a full bottle of water dropped from a desk? An empty bottle?

11) When an object is dropped or thrown into the air, what is its acceleration?

12) An object dropped from a 4 m tall roof. ∆y = _____ and vi = _____.

13) An object is thrown 10 m/s into the air. How high does it go? vi = _____; a = _____; and vf = _____.

14) A person throws a ball into the air at 6 m/s from the ground. When it comes back, vf = _____ , a = _____, and ∆y = _____ .

15) “Sitting on the dock of the bay, wasting time” with my sister. I get bored and push her off the 2 m dock. How fast is she moving

when she belly flops into the water? (And more importantly how badly is she going to hurt me when she catches me?)

Variables: Equation: Solve:

16) What is the velocity of the stop sign in the car’s frame of reference?

17) What is the motorcyclist’s velocity relative to the car?

5) Speed or velocity:

A person walks 0.5 m/s to the east.

6) Scalar or vector:

A car is moving 30 m/s.

30 m/s

20 m/s


Name: _____________________

Period: _____________________

18) In the graphic above, the car is at constant speed between the first two positions and between the last two positions.

Between the middle two positions it is accelerating. Calculate its acceleration.

Use the three motion graphs below to answer the following questions.

19) What does the slope of the graphs below tell us: Graph 1: __________; Graph 2: __________; Graph 3: __________.

20) Transfer the following graphs. Each vertical square is 1 m; each horizontal square is 1 sec.

Velocity vs. Time

Time

0


Time

Acc

eler

atio

n

0

Position vs. Time

Time

Po

siti

on

Vel

oci

ty A C

B

In Class Review— p2

Position vs. Time

0

20

40

60

80

100

120

140

160

180

200

220

240

0 5 10 15 20 25 30 35 40 45 50Time (sec)

Po

sit

ion

(m

)


-5

-4

-3

-2

-1

0

1

2

3

4

5

0 5 10 15 20 25 30 35 40 45 50

Time (sec)

Accele

rati

on

(m

/s2)

Velocity vs. Time

-5

-4

-3

-2

-1

0

1

2

3

4

5

0 5 10 15 20 25 30 35 40 45 50

Time (sec)

Velo

cit

y (

m/s

)

21) Use the graph at the right to answer the following.

A. Give the linear equation for the graph at the right.

B. Where is the object on the graph at 4.2 seconds?

C. What does the y-intercept tell us about this object?

D. What is the speed of the graph?

E. Transfer the position graph to the velocity and acceleration

graphs below.

0:07.0 0:09.0

24 m

0:00.0 0:04.0

12 m

Constant

speed Constant

speed

Accelerating


Name: _____________________

Period: _____________________

Instantaneous Speed—Velocity

at a particular instant. How fast

something is moving at a particular

point in time. This is what your

speedometer reads.

Home

Banalville

Gastin

Porkerville

Pulchritude

Destiny

80 mi

1.3 hr

60 mi

0.75 hr

Fuel Stop —30 min

50 miles

1.4 hr

75 miles

1 hr

110 mi

1.5 hr

Stop at “Fatties Finest

Foods” —1.2 hr

Sightseeing—30 min

Instantaneous Vs. Average Speed

Average Speed—The average velocity over an en-

tire distance. Average velocity is found from total

distance divided by the total time.

Total Distance (in meters)

Total Time (in seconds)

Average Speed

(in meter/sec)

totalave

total

Dv

t=

The diagram shows a person’s

circuitous journey. During any

trip your speed does not stay

constant due to different speed

limits, traffic, stops, etc. To

find the average speed between

any two points, you need total

distance and total time between

those two points.

Instantaneous Speed—Velocity

at a particular instant. How fast

something is moving at a particular

point in time. This is what your

speedometer reads.

Home

Banalville

Gastin

Porkerville

Pulchritude

Destiny

80 mi

1.3 hr

60 mi

0.75 hr

Fuel Stop —30 min

50 miles

1.4 hr

75 miles

1 hr

110 mi

1.5 hr

Stop at “Fatties Finest

Foods” —1.2 hr

Sightseeing—30 min

Instantaneous Vs. Average Speed

Average Speed—The average velocity over an en-

tire distance. Average velocity is found from total

distance divided by the total time.

Total Distance (in meters)

Total Time (in seconds)

Average Speed

(in meter/sec)

totalave

total

Dv

t=

The diagram shows a person’s

circuitous journey. During any

trip your speed does not stay

constant due to different speed

limits, traffic, stops, etc. To

find the average speed between

any two points, you need total

distance and total time between

those two points.

Name: _____________________

Period: _____________________

7) The tape timers at the left show 4 objects moving to the

right. The dots show the positions of the objects each

second. Which objects apply to the following?

Linear Motion In Class Test Review

2) Convert 18 m/s to meters per min.1) Convert 6 ft/s to m/s

3) An object moves 120 m in 15 seconds.

Calculate the object’s speed.

4) An object moves 18 m/s. How long does

it take the object to move 154 m?




____ Accelerating.

____ Decelerating.




____ Is stopping.



____ Vi = Vf

Object A

Object B

Object C

Object D

8) A car begins at a stop sign. It ends up going 100 m in 6.5 seconds. Find the car’s acceleration.

Variables: Equation and solve:

9) +, –, or 0?

A. ____Acceleration of an object that is moving to the left and speeding up?

B. _____Acceleration of an object that is moving up and slowing down?

C. _____Velocity of an object that is moving to the right?

D. _____Displacement of an object that ends at its starting position?

E. _____Acceleration of an thrown object at the top of its path?

F. _____Displacement of an object moving to the left?

10) What is the acceleration of a full bottle of water dropped from a desk? + or -11) When an object is dropped or thrown into the air, what is its acceleration? + or -12) An object moves from rest to 4 m away. ∆x = _____ and vi = _____.

13) What are two ways a velocity can change?14) What does the slope of this velocity vs. time graph mean?15) A shopping cart is going 4.0 m/s. It undergoes -5.0 m/s2 of acceleration for 4 seconds. How fast is it going afterwards?


16) If our velocity is positive and our acceleration is negative what is happening?17) An object at rest accelerates for 6 seconds. Afterwards it is going 60 m/s. How far it traveled in this time?


5) Speed or velocity:

A person walks 0.5 m/s to the east.

6) Scalar or vector:

A car is moving 30 m/s.

Name: _____________________

Period: _____________________

18) In the graphic above, the car is at constant speed between the first two positions and between the last two positions.

Between the middle two positions it is accelerating. Calculate its acceleration.

Use the three motion graphs below to answer the following questions.

19) What does the slope of the graphs below tell us: Graph 1: __________; Graph 2: __________; Graph 3: __________.

20) In Graph 1, which letter has the highest velocity.

Velocity vs. Time

Time


Time

Acc

eler

atio

n

Position vs. Time

Time

Po

siti

on

Vel

oci

ty A C

B

In Class Review— p2

Position vs. Time

0

20

40

60

80

100

120

140

160

180

200

220

240

0 5 10 15 20 25 30 35 40 45 50Time (sec)

Po

sit

ion

(m

)

21) Use the graph at the right to answer the following.

A. Give the linear equation for the graph at the right.

B. Where is the object on the graph at 4.2 seconds?

C. What does the y-intercept tell us about this object?

D. What is the speed of the graph?

0:07.0 0:09.0

24 m

0:00.0 0:04.0

12 m

Constant

speed Constant

speed

Accelerating

22. How fast does the big car seem to be moving to a person looking from the little car (in the car’s frame of reference)?

25 m/s

10 m/s

23. A race car is going at 32 m/s but crashes into a wall. The crash lasts 4 seconds. Calculate the car's deceleration. *Remember if an object is slowing down or stoping what the acceleration will be.Variables: Equation: Solve:

rgollahon

Line

rgollahon

Line

rgollahon

Line

rgollahon

Line


Name: _____________________

Period: _____________________

d change of distance s = step 1: variables step 3: put in...

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