Motion in One Dimension

Speedtraveloftime

traveledcetandisSpeedAverage

Mr. Wolf drives his car on a long

trip to a physics store.

Given the distance and time

data for his trip, plot a graph of

his distance versus time.

Graphing Distance Vs. Time

∆X

∆Y

Slope = 64 mph

Kinematics

• Kinematics - using mathematics to

describe how objects are moving

examples: (Vertical) Falling brick, (Horizontal) Moving

Truck on a flat road or a flat ramp

• One Dimensional - Rectilinear motion,

moving in a straight line path, no rotation

Reference Point

• A zero location used to determine position.

Sometimes called the origin.

01

2

4

Position - x

Displacement• Definition of Displacement:

the shortest distance between the two positions.

• Unit: meter (m)

• Represented as Δx

Position and Displacement

Xi

1

2

0

1

2

4

Initial Position- The location where the physics

problem starts

Δx = xf - xi

Displacement vs. Distance

• Displacement is the shortest distance

between start and finish.

• Distance is the total path length taken to

arrive at point B from point A.

Displacement vs. Distance cont.

Xi

Xf

Speed vs. Velocity• The Sears Tower in Chicago is 443 m tall. Joe

wants to set the world’s stair climbing record and

runs all the way to the roof of the tower. If Joe’s

average upward speed is 0.60 m/s, how long will it

take Joe to climb from street level to the roof of the

Sears Tower?

TimeElapsed

anceDistspeedAverage

The Sears Tower in Chicago is 443m tall. Joe

wants to set the world’s stair climbing record and

run all the way to the roof of the tower. If Joe’s

average upward speed is 0.60 m/s, how long will it

take Joe to climb form street level to the roof of the

Sears Tower?

740

2.0

Speed vs. Velocity

Time

anceDistSpeed

Distance = 443 m

Speed = 0.60 m/s

Speed

anceDistTime 443 m

Speed vs. Velocity cont.

t

x

tt

xxv

TimeElapsed

ntDisplacemeVelocityAverage

if

if

sm

s

m60.0

476,1

866

t

x

tt

xxv

if

if

Now Joe climbs to the roof of the Sears

and then back down in 1,476 sec. What

is his..

a) average speed?

b) average velocity?443 m

Xi = 0Xf = 0

sm

m0

sec)0476,1(

)00(

The Peregrine falcon is the fastest of

flying birds (and, as a matter of fact, is

the fastest living creature). A falcon

can fly 1.73 km downward in 25 s.

What is the average velocity of a

peregrine falcon in m/s?

-69

0.5

The peregrine falcon is the fastest of

flying birds (and, as a matter of fact, is

the fastest living creature). A falcon can

fly 1.73 km downward in 25 s. What is

the average velocity of a peregrine

falcon?

Instantaneous Velocity• Instantaneous velocity is how fast the car is

moving and the direction it is moving at any

instant during the motion.

Frame of Reference• The object or position from which

movement is determined.

• The most common frame of reference

is…. the earth.

Position-Time GraphsLet’s plot the position and time for Joe Driver.

0m, 0s 25m, 1s 50m, 2s 75m, 3s 100m,

4s

• Note that every second, the

position increases by 25 m.

0

25

50

75

100

0 1 2 3 4

Time, s

Me

ters

, m

•What is the average velocity?

•Is the velocity constant?

•What is the slope of the line?

X

t

X

t

The slope of position-time

graph = the velocity

Acceleration• Acceleration is when an object changes its

velocity over a certain time.

• Average acceleration:

• Average acceleration points in the same

direction as the change of velocity.

t

v

tt

vva

timeElapsed

velocityinChange

if

if

Velocity-Time Graphs

Let’s plot the velocity and time for Joe Driver.

0m, 0s 10m, 1s

V=10 m/s

20m, 2s

V=10 m/s

30m, 3s

V=10 m/s

40m, 4

V=10 m/s

The velocity is constant in this case.

Since it is not changing (V = 0),

the acceleration = zero.

-15

-10

-5

0

5

10

15

0 1 2 3 4

Time,s

Velo

city,

m/s

Velocity-Time Graphs

What if Joe Driver increases his velocity?

0m, 0s 5m, 1s

V=10 m/s

20m, 2s

V=20 m/s

45m, 3s

V=30 m/s

80m, 4s

V=40 m/s

The velocity is changing. The

acceleration can be determined from

the slope of the velocity-time graph.

a =V

t=

40-0

4-0

During the 1st second, the velocity

increases from 0 to 10 m/s, during

the next second it increases another

10 m/s to 20 m/s, and so on.

=10 m/s

s

0

10

20

30

40

0 1 2 3 4

Velo

city m

/s

Time, s0 1 2 3 4

0

10

20

30

40

Example:

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 ti=9.0 s and the car’s

velocity is vi=+28 m/s. When t = 12 s the

velocity of the car is v = +13 m/s. What is

the average acceleration?

What is the average

acceleration?-5.0

1.0

Equations for Kinematics for

Constant Acceleration

• 5 Variables: x = displacement

a = acceleration

v = final velocity at time t

vi = initial velocity at time ti

t = time elapsed since ti = 0

One Dimensional Motion With

Constant Acceleration

If acceleration is constant (constant net force)

and assuming the time interval starts at zero

seconds. t1 0

t

vva

if

av v

t

2 1

solve for v2

* tavv if

Example: A golf ball rolls up a hill toward a

putt-putt hole.

a. If it starts with a velocity of 2.0

m/s and accelerates at a

constant rate of -0.5 m/s2, what

is its velocity after 2.0 sec?

b. If the acceleration occurs for

6.0 s, what is its final velocity?

In 1934, the wind speed on Mt.

Washington in New Hampshire reached a

record high. Suppose a very sturdy glider

is launched in this wind, so that in 45.0 s

the glider reaches the speed of the wind.

If the glider undergoes a constant

acceleration of 2.29 m/s2, what is the

wind’s speed? Assume that the glider is

initially at rest.

tavv if

If acceleration is constant and ...

set them equal and solve for

vv v

and vx

t

( )1 2

2

x

xv v

t

( )1 2

2*

When is not known . . .

substitute the first equation into the second.

v2

v v a t2 1 xv v

t

( )1 2

2into

x v t at 112

2*

When is not known . . .t

av v

tt

v v

a

2 1 2 1

sub. the time expression into the second Equation

tv v

a

2 1into

* axvv if 222

t

vvx

fi

2

Equations for Kinematics for

Constant Acceleration

tvvx

axvv

attvx

atvv

fi

if

i

if

2

1

2

2

1

22

2

Applications of the Equations of

Kinematics

• Starting point for all problems:

1. Make a drawing

2. Decide (+) direction

3. Write down givens

4. Write down unknown

5. Write down equation(s) that will be used

6. Solve

Example Problem 1A VW Beetle goes from 0 to 60 mph with an

acceleration of +2.35 m/s2. (A) How much

time does it take for the Beetle to reach

this speed? (B) A top-fuel dragster can go

from 0 to 60 mph in 0.600 seconds. Find

the acceleration of the dragster (m/s2)

Example Problem 2A jetliner, traveling northward, is landing with

a speed of 69 m/s. Once the jet touches

down, it has 750 m of runway in which to

reduce its speed to 6.1 m/s. (A) Compute

the average acceleration of the plane

during landing. (B) How long did it take

the plane to accomplish this?

Example Problem 3A car is traveling at a constant speed of 33

m/s on a highway. At the instant this car

passes an entrance ramp, a second car

enters the highway from the ramp. The

second car starts from rest and has a

constant acceleration. What acceleration

must it maintain so that the two cars meet

for the first time at the next exit, which is

2.5 km away?

Freely Falling Bodies

• In the absence of air resistance, all bodies

fall at the same rate.

• Galileo Galilee discovered this property

over 400 years ago.

• Idealized Motion is called free-fall.

• Acceleration due to gravity is:

g = 9.80 m/s2 or 32.2 ft/s2

Equations for Kinematics for

Constant Acceleration (a = -g)

tvvx

axvv

attvx

atvv

i

i

i

i

2

1

2

2

1

22

2

Before

tvvy

gyvv

gttvy

gtvv

i

i

i

i

2

1

2

2

1

22

2

After

How far does it fall?

• Galileo predicted that all objects

regardless of mass, fall at the same rate.

In his experiments he also discovered a

relationship between the distance covered

during the first second of travel and each

consecutive time interval of one second.

V0 = 0

Example Problem 4A stone is dropped from rest from the top of

a tall building. After 3 seconds of free fall,

what is the displacement of the stone?

What is the velocity of the stone after 3

seconds?

Example Problem 5The John Hancock Center in Chicago is the

tallest building in the United States in which

there are residential apartments. The Hancock

Center is 343 m tall. Suppose a resident

accidentally causes a chunk of ice to fall from

the roof. What would be the velocity of the ice

as it hits the ground? Neglect air resistance.

tvvy

gyvv

gttvy

gtvv

i

i

i

i

2

1

2

2

1

22

2

Up and Down Problems

• When solving for a problem

in which the object travels

up and then down, it is

helpful to break the problem

into two parts. The up

motion and the down

motion.

Up and Down Problems

• The Up motion.

• The objects initial velocity is

not zero.

• The objects final velocity is

zero.

• a = -g

Up and Down Problems• The Down Motion

• The initial velocity is zero.

• The final velocity is not zero. (In the case where it returns to the starting position, the final velocity is equal in magnitude to the starting velocity.)

• a = -g

• tdown = tup (if returning to starting position)

Example Problem 6

A rock is tossed up into the air with an initial

velocity of 15 m/s. How high does the

rock travel? How long is it in the air? How

fast is it traveling when it reaches its

starting position?

Example Problem 7A young girl is standing on a chair which is 1

meter above the ground when she tries to

toss an apple to her friend in a tree house.

She tosses up the apple with a velocity of

5 m/s. The apple doesn’t make it all the

way to the tree house. As the apple falls

back down, the girl misses the apple and it

continues to hit the ground. How fast was

the apple traveling right before impact?

How long was the apple in the air?

Graphical Analysis of Kinematics

• Reading graphs is another important part

of Physics.

• Information regarding the motion of an

object can be obtained by either finding

the slope of the graph or by finding the

area under the curve of a graph.

Graphing

In a graph of Position vs. Time, the velocity of the object can be obtained from finding the slope of the graph. Since the displacement covered is 8 m and it took 2 seconds, then the average velocity is 4 m/s.

In the above graph, we can find the average velocity of

the object by finding the slopes at different points in

the motion.

1. v = 2 m/s

2. v = 0 m/s

3. v = -1 m/s

Changing Velocity

When the velocity of the object is changing, the position vs. time graph is a curve. To find the instantaneous velocity, draw a tangent line to the point in question and find the slope.

Po

sitio

n c

m

Velocity vs. Time

• When you have a graph of velocity vs.

time, then the slope gives the acceleration

of the object.

• Finding the area under the curve for a

certain time interval gives the

displacement of the object.

The average acceleration of the graph is given by the

slope, 6 m/s2.

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Converted by:

Arzelo D. Rivas

http://www.zephyrosglumph.wordpress.com/