Motion in 1 Dimension

Positive displacement

Negative displacement

� Physics is all about describing motion. For now we are going to discuss

motion in 1 dimension, which means either along the x axis or the y axis.

� To describe an object’s motion, we must be able to specify its position (x)

at all times. An object’s position is the x or y coordinate of its location.

● Since we are discussing motion along the x-axis first, we will assign an

object’s initial position the label xi and its final position the label xf.

� When an object changes its position, it experiences a displacement.

● This is a vector that points from an initial position xi to a final

position xf.

● The magnitude is the shortest distance between the two positions.

● Mathematically, displacement is given by…

)( rightmovesobject

xxifpositiveisx if >∆

if xxx −≡∆

)( leftmovesobject

xxifnegativeisx if <∆The 3 bar

equal sign

means

mathematical definition.

● We will use the meter as our standard S.I. unit of measurement.

Do you remember

your

conversions?

● FROM THIS POINT ON, YOU REALLY NEED TO KNOW THOSE

METRIC CONVERSIONS!!!

� Now it’s time to realize that there is a HUGE difference between

distance traveled and displacement!!

As an example… A baseball player hitting a home run travels around the bases.

Questions… What is his initial

position?

What is his final

position?

What distance did he

run?

What is his

displacement?

1 meter = 100 centimeters

1 meter = 1000 millimeters

1 kilometer = 1000 meters

SAMPLE PROBLEM A A marble rolls alongside a meter

stick. It begins moving at the 98-

cm mark and stops on the 87-cm

mark. Calculate its displacement.

∆x = xf – xi = 0.87 – 0.98 = -0.11 m

SAMPLE PROBLEM B A “lazy susan” is spun 3 times. A container of parsley flakes

travels around a circular path having a radius of 6.0 cm. (a) What

was the distance traveled by the parsley? (b) What was the

parsley’s displacement?

a) d = 3 (2πr) = 6π(0.06) = 1.131 m

b) ∆x = 0 m

Graphing Position vs. Time (x vs. t)

� Important relationships can be discovered from graphs, so let’s plot a

sample position vs. time (x vs. t) graph to analyze.

4 Steps to Graphing Position vs. Time

● Step 1: Label the x-axis with Time (units)

Label the y-axis with Position (units)

Entitle the graph Position vs. Time

● Step 2: Determine an appropriate scale for each axis. The idea is to

make the graph as large as possible!

● Step 3: Plot the data neatly and connect the points with a smooth line.

● Step 4: Calculate the slope for each segment of the graph using the

equation below. Don’t forget units as this slope will now be a

meaningful quantity!!!

t

xgraphourforor

x

y

run

risem

∆

∆

∆

∆== :

SAMPLE (x vs. t) GRAPH

Step 1: Label, label, label!

Step 2: Determine an appropriate scale for each axis.

Step 3: Plot the data neatly and connect the points with a smooth line.

Step 4: Calculate the slope for each segment of the graph.

Slope 1 = ∆x / ∆t = 1200 m / 600 s = 2 m/s

Slope 2 = ∆x / ∆t = 0 m / 400 s = 0 m/s

Slope 3 = ∆x / ∆t = -800 m / 800 s = -1 m/s

What is significant about the slope’s units?

m/s. Indicates that the slope is telling you velocity.

Position (m) Time

(s)

0 0

200 100

400 200

600 300

800 400

1000 500

1200 600

1200 700

1200 800

1200 900

1200 1000

1100 1100

1000 1200

900 1300

800 1400

700 1500

600 1600

500 1700

400 1800

positiveistvsxofslope

or

positiveisxifpositiveisvx

.

∆

negativeistvsxofslope

or

negativeisxifnegativeisv x

.

∆

Average Velocity

v

� This slope is so important that it has been given a special name;

Average Velocity. Average velocity is the change in position with time.

● Since velocity depends on displacement, velocity is also a vector

quantity.

● meters per second (m/s) is the standard S.I. unit of measurement

● We can find between any two points by drawing a straight line

between them and calculating the slope, regardless of the graph’s

shape.

� Average Velocity is defined as:

graphtvsxofslopet

xvx .=

∆

∆≡

PRACTICE PROBLEM A NAME: ___________________________

A farmer collected the following data after lopping the head off one of his chickens. Set up a position

vs. time graph to help the farmer understand the motion of the bird. Label each straight-line segment

A − J.

Position (m) Time (s) 0 0

4 1

8 2

2 4

2 5

2 6

4 7

4 8

6 9

1 10

2 11

3 12

3 13

3 14

4 15

5 16

6 17

Calculate VAVG for the following time

intervals:

(0 – 2 sec): ∆x/∆t = 8m/2s = 4 m/s

(2 – 4 sec): ∆x/∆t = -6m/2s = -3 m/s

(1 – 6 sec): ∆x/∆t = -2m/5s = -0.4 m/s

(4 – 9 sec): ∆x/∆t = 4m/5s = 0.8 m/s

(4 – 6 sec): ∆x/∆t = 0m/2s = 0 m/s

(0 – 17 sec): ∆x/∆t = 6m/17s = 0.353 m/s

Describe the motion of the bird for each segment:

A:

B:

C:

D:

E:

F:

G:

H:

I:

J:

PRACTICE PROBLEM A

Velocity vs. Time Graphs

� Instead of graphing just position vs. time, we are also able to graph

velocity vs. time. This graph will show you how velocity changes

with time. Therefore, the rules that applied to a position vs. time

graph do not necessarily apply to a velocity vs. time graph.

∆x = positive ∆x = positive

v = increasing v = increasing

a = positive a = positive

∆x = positive ∆x = positive

v = constant v = decreasing

a = zero a = negative

time (s)

vel

oci

ty (

m/s

)

time (s)

vel

oci

ty (

m/s

)

time (s)

vel

oci

ty (

m/s

)

time (s)

vel

oci

ty (

m/s

)

Instantaneous Velocity

v

Why would not be useful if you wanted to know your

velocity the instant you noticed the flashing lights of a police car behind

you?

� We need to be able to specify our velocity just as precisely as we can our

position by knowing what is happening at a specific time.

● We could not actually do this mathematically until the late 1600’s

when Calculus gave us this ability.

● We can do this in Physics by taking smaller and smaller intervals of

time on our x vs. t graph.

� The limiting value of the ratio as approaches zero is called

Instantaneous Velocity.

● Notice that this is just the limit of the average velocity as time

approaches zero.

● We can interpret V graphically by drawing a line tangent to the

curve at a specific time. We can then calculate the slope of the

tangent line.

t

x

∆

∆ t∆

� Instantaneous Velocity is defined as:

� From here on, we use the term velocity to designate instantaneous

velocity.

SAMPLE PROBLEM An object moves along the x-axis as shown

in the position vs. time graph to the right.

(a) Determine the object’s displacement

during the time intervals t = 1 s to t = 3 s

and t = 4 s to t = 6 s. (b) Calculate the

average velocity during these two time

intervals. (c) Find the object’s instantaneous

velocity at t = 2.0 s and at 6.0 s.

∆x = xf – xi = 3m – 11m = -8 m

∆x = xf – xi = 5m – 2m = 3 m

v = ∆x / ∆t = -8 / 2 = -4 m/s

v = ∆x / ∆t = 3 / 2 = 1.5 m/s

v = ∆x / ∆t = -13 / 3.5 = 3.714 m/s

0

slope of tangent linelimt

xv

t∆ →

∆≡ =

∆

Instantaneous Velocity (Using Calculus)

SAMPLE PROBLEM

A person moves along the x-axis. Her position varies with time according to

the expression x = 2t2 - 4t where x is her instantaneous position in meters

while t represents the time at that position. The graph for this motion is

shown below. Find the instantaneous velocity at t = 0s, t = 1s, and t = 3s.

Step 1 = Take the derivative of the function:

x = 2t2 – 4t

v = 4t – 4

Step 2 = Plug in your value of time

v(0) = 4(0) – 4 = -4 m/s

v(1) = 4(1) – 4 = 0 m/s

v(3) = 4(3) – 4 = 8 m/s

� In everyday usage, the words speed and velocity are used

interchangeably. However, there is a HUGE difference!!!

EXAMPLE

During the 2008 Summer Olympics,

Michael Phelps swam four 50-m laps

in 1.52 minutes to win the men’s 200-m

Butterfly and set a new world record.

QUESTIONS:

What distance was covered?

4 x 50 m = 200 m

What was the swimmer’s displacement?

∆x = 0m

What was the swimmer’s average velocity?

V = ∆x/∆t = 0m / 1.52 min = 0 m/s

This is not very useful, so we still need to be able to quantify

“how fast” she swam.

distance

time

totals

total=

� This is where the term Average Speed comes in. Speed is a scalar

quantity so the direction is meaningless in describing it.

● Average Speed is given by:

� Instantaneous Speed is defined as the magnitude of the instantaneous

velocity. Your car’s speedometer measures instantaneous speed!!

� You need to very careful here! Average Speed IS NOT the magnitude of

Average Velocity because average speed is dependent on distance and

average velocity depends on displacement!

Acceleration

� We have already seen that velocity

can change while a particle moves.

� We can quantify changes in velocity as a function of time just as we could

with x(t).

� When velocity changes with time, a particle is undergoing Acceleration.

● Mathematically, acceleration is defined in the following way:

● As with velocity, when the motion is 1-Dimensional, we can simply

use +/- to indicate direction.

● meters per second2 (m/s

2) is the standard S.I. unit of measurement.

graphtvsvofslopet

va .=

∆

∆≡

SAMPLE PROBLEM A

Suppose a plane starts from rest (vi = 0 m/s) when ti = 0 s.

The plane accelerates down the runway and at tf = 29 s

attains a velocity of vf = + 260 km/h, where the plus sign

indicates the velocity points to the right. Determine the

acceleration of the plane.

260 km x 1000 m x 1 hr = 72.222 m/s

hr 1 km 3600s

a = (Vf – Vi) / t = (72.222 – 0) / 29 = 2.490 m/s2

SAMPLE PROBLEM B

A drag racer crosses the finish line, and the driver deploys a

parachute and applies the brakes to slow down. The driver

begins slowing down when t = 9.0 s and the car’s velocity is

v = 28 m/s. When t = 12.0 s, the velocity has been reduced

to v = 13 m/s. What is the acceleration of the dragster?

a = (Vf – Vi) / t = (13 – 28) / (12 - 9) = -5 m/s2

SAMPLE PROBLEM C

A glider’s velocity on a tilted air track increases from - 17.0 cm/s

at the time t = 0.33 s to - 55.0 cm/s at a time of t = 2.33 s. What

is the acceleration of the glider?

-17 cm/s = -0.17 m/s, -55 cm/s = -0.55 m/s

a = (Vf – Vi) / t = (-0.55 – (-0.17)) / 2 = 0.19 m/s2

1-D Motion w/ Constant Acceleration

� The following kinematic equations only work for situations

involving constant acceleration!!!

�

�

�

�

You need to have these memorized by

next class!

atvv if +=

xavv if ∆+= 222

2

2

1attvx i +=∆

tvvx fi )(2

1+=∆

Physics Name:_______________________

Kinematics SAMPLE PROBLEMS Date: __________ Period: ______

Directions: Solve the following problems in the space provided. Use the tables to

organize your data. If a variable is not given, record a −−−−−−−− in the block. If a variable is

unknown, record a ? in the block. Convert all measurements into units of meters,

seconds and meters per second. Be sure to show all work!

SAMPLE PROBLEM A

The doors open for a back−to−school sale and a crazed teen sprints

from her mom’s car. If the girl reaches a velocity of +4.6 m/s in

1.9 s, determine how far she ran to the store.

∆x = ½ (Vi + Vf)t

= 0.5 (0 + 4.6)1.9

= 4.37 m

SAMPLE PROBLEM B

Starting from rest, a cow reaches a velocity of +2.5 m/s in a time of 5.0 s.

A Calculate the cow’s average acceleration.

B By how much does the cow’s speed change each second?

C Calculate the bovine’s displacement during this time interval.

A. a = (Vf – Vi) / t = (2.5 – 0) / 5 = 0.5 m/s2

B. Increases + 0.5 m/s for every second of movement

C. ∆x = Vit + ½ at2 = 0.5(0.5)5

2 = 6.25 m

DATA TABLE

∆x vi vf a t

? 0 4.6 - 1.9

DATA TABLE

∆x vi vf a t

? 0 2.5 0.5 5

SAMPLE PROBLEM C

A man’s 2010 Porsche 911 reaches 60 mi/h with a constant

acceleration of + 8 m/s2. Assuming the car starts from rest,

how long does it take the car to reach this speed?

(Hint: 1 mile = 1609 m)

60 mi x 1609 m x 1 hr = 26.817 m/s

hr 1 mi 3600s

t = (Vf – Vi) / a = (26.817 – 0) / 8 = 3.352 s

SAMPLE PROBLEM D

Billy Bob is going to fetch some dinner. He catches sight of

a squirrel that he fancies and “takes off a runnin’.” Billy

Bob starts from rest, accelerates at the rate of + 1.5 m/s2 and

reaches a velocity of + 4.7 m/s before catching the critter.

Calculate Billy’s displacement for the hunt.

Vf2 = Vi

2 + 2a∆x

∆x = (Vf2 – Vi

2) / 2a = (4.7

2 – 0

2) / (2(1.5)) = 7.363 m

SAMPLE PROBLEM E

A motorcycle accelerates from 90 mi/h to 65 mi/h in 2.3 s.

Find the motorcycle’s acceleration and displacement

during this time interval.

90 mi x 1609 m x 1 hr = 40.225 m/s

hr 1 mi 3600s

65 mi x 1609 m x 1 hr = 29.051 m/s

hr 1 mi 3600s

a = (Vf – Vi) / t = (29.051 – 40.225) / 2.3 = -4.858 m/s2

∆x = Vit + ½ at2 = 40.225(2.3) + 0.5(-4.858)2.3

2 = 79.669 m

DATA TABLE

∆x vi vf a t

- 0 26.817 8 ?

DATA TABLE

∆x vi vf a t

? 0 4.7 1.5 -

DATA TABLE

∆x vi vf a t

? 40.225 29.051 ? 2.3

SAMPLE PROBLEM F

A car, initially traveling at 70.0 km/h, accelerates

at the constant rate of -1.50 m/s2. How far will the

car travel in 10.0 s?

70 km x 1000 m x 1 hr = 19.444 m/s

hr 1 km 3600s

∆x = Vit + ½ at2 = 19.444(10) + 0.5(-1.5)10

2 = 119.44 m

CHALLENGE PROBLEM

Watch Out for the Speed Limit! A car traveling at a constant speed of 45.0 m/s passes a trooper

hidden behind a billboard. One second after the speeding car

passes the billboard, the trooper sets out from the billboard to

catch it, accelerating at a constant rate of 3.00 m/s2. How long

does it take her to overtake the car?

∆xcar = 45 + 45t

∆xtrooper = Vit + ½ at2 = 0 + ½(3t

2) = 1.5t

2

1.5t2 = 45t + 45

1.5t2 - 45t – 45 = 0

USE QUADRATIC FORMULA

t = (-b ± √(b2 – 4ac)) / 2a = (-(-45) ± √(45

2 – 4(1.5 * -45))) / 2(1.5)

t = 30.968 s, -0.969 s

DATA TABLE

∆ x vi vf a t

? 19.444 -1.5 10

Freely Falling Objects

� All objects dropped near the earth’s surface fall with the same

constant acceleration in the absence of air resistance.

● Aristotle (384-322 B.C.) believed that heavier objects fell

faster than lighter ones.

● This fact was not fully accepted until the early 1600s. Galileo

(1564-1642) conducted experiments with balls on inclined

planes and drew conclusions about freely falling objects since a

freely falling ball is equivalent to a ball moving down a vertical

incline.

? What kinematic equation could be used to find the

acceleration in this experimental procedure?

● On August 2nd, 1971 David Scott released a hammer and a

feather on the surface of the moon. What was discovered?

2

2

1attvx i +=∆

� By definition, a freely falling object is any object moving freely

under the influence of gravity alone, regardless of its initial

motion. Objects thrown upward or downward and those

released from rest are all falling freely once they are released.

Any freely falling object experiences an acceleration directed

DOWNWARD, regardless of its initial motion.

� The value of g does change with increasing altitude, as we will

see when we study Newton’s Law of Universal Gravitation.

� At the earth’s surface (and for our discussions in this course) the

acceleration due to gravity is approximately - 9.80 m/s2.

gtvvif yy −=

ygvvif yy ∆−= 222

2

2

1gttvy

iy −=∆

� Our kinematic equations now become:

�

�

�

Name: ______________________________ Date: _______ Period: ______

Mr. Talboo - Physics

SAMPLE PROBLEM A

A stone is thrown downward from the top of a tall building with an initial speed of

5.00 m/s. What is the displacement of the stone after 3.00 s of free-fall? What is the

final velocity of the stone when it strikes the ground?

∆y = Vit – ½ gt2

= (-5)(3) - 1/2 (9.8)(32)

= -59.1 m

Vyf = Vyi – gt

= 5 – (9.8)(3)

= -24.4 m/s

SAMPLE PROBLEM B

You and a friend are going to toss a coin to determine who asks Mr. Talboo for help on

a kinematics problem. You toss the coin up with an initial velocity of 6.00 m/s. How

high does the coin go above the point of release? What is the total time the coin is in

the air before returning to its release point?

∆y = (Vf2 – Vi

2) / -2g

= (02 – 6

2) / -2(9.8)

= 1.837 m

t = (Vf – Vi) / -g

t = (-6 – 6) / -9.8

t = 1.224 s

DATA TABLE

∆y vyi vyf g t

? 5 ? 9.8 3

DATA TABLE

∆y vyi vyf g t

? 6 0 9.8 ?

SAMPLE PROBLEM C

A stone thrown from the top of a building is given an initial velocity of 20.0 m/s

straight upward. The building is 50.0 m high, and the stone just misses the edge of the

roof on its way down. Determine (a) the time at which the stone reaches its maximum

height, (b) the maximum height, (c) the total time that it is in the air and (d) the velocity

of the stone when it hits the ground.

t = (Vf – Vi) / -g

t = (0 – 20) / -9.8

t = 2.041 s

∆y = (Vf2 – Vi

2) / -2g

= (02 – 20

2) / -2(9.8)

= 20.408 m + 50 m = 70.408 m

∆y = Vit – ½ gt2

= (0) – ½ gt2

t = √((2∆y)/-g)

= √((2 * -70.408 )/-9.8)

= 3.791 s

+ 2.041 s

t = 5.832 s

Vyf = Vyi – gt

= 20 – (9.8)(5.832)

= -37.154 m/s

DATA TABLE

∆y vyi vyf g t