Kinematics in One Kinematics in One DimensionDimension

The CheetahThe Cheetah: A cat that is built for speed. Its strength : A cat that is built for speed. Its strength and agility allow it to sustain a top speed of over 100 and agility allow it to sustain a top speed of over 100 km/h. Such speeds can only be maintained for about km/h. Such speeds can only be maintained for about ten seconds.ten seconds.

Distance and Distance and DisplacementDisplacement

Distance is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below:

Distance is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below:

A

Bd = 20 m

DistanceDistance dd is a is a scalarscalar quantity (no quantity (no direction):direction):Contains Contains magnitudemagnitude only and consists of a only and consists of a numbernumber and a and a unit.unit.

Distance and Distance and DisplacementDisplacement

DisplacementDisplacement is the straight-line separation is the straight-line separation of two points in a specified direction.of two points in a specified direction.DisplacementDisplacement is the straight-line separation is the straight-line separation of two points in a specified direction.of two points in a specified direction.

A vector quantity:

Contains magnitude AND direction, a number, unit & angle.

A

BΔs = 12 m, 20o

Distance and Distance and DisplacementDisplacement

• For motion along x or y axis, the For motion along x or y axis, the displacementdisplacement is determined by the x or y coordinate of its is determined by the x or y coordinate of its final position. Example: Consider a car that final position. Example: Consider a car that travels 8 m, E then 12 m, W.travels 8 m, E then 12 m, W.

• For motion along x or y axis, the For motion along x or y axis, the displacementdisplacement is determined by the x or y coordinate of its is determined by the x or y coordinate of its final position. Example: Consider a car that final position. Example: Consider a car that travels 8 m, E then 12 m, W.travels 8 m, E then 12 m, W.

Net displacement Net displacement ΔxΔx is from the is from the origin to the final origin to the final position:position:

What is the What is the distancedistance traveled? traveled?d = 20

m !!

12 m,W

Δx

Δx = 4 m, WΔx = 4 m, W

x8 m,E

x = +8

x = -4

Definition of SpeedDefinition of Speed

• SpeedSpeed is the distance traveled per is the distance traveled per unit of time (a scalar quantity).unit of time (a scalar quantity).

• SpeedSpeed is the distance traveled per is the distance traveled per unit of time (a scalar quantity).unit of time (a scalar quantity).

vs = = d

t

20 m

4 s

vs = 5 m/svs = 5 m/s

Not direction dependent!

A

Bd = 20 m

Time t = 4 s

Definition of VelocityDefinition of Velocity

• VelocityVelocity is the displacement per is the displacement per unit of time. (A vector quantity.)unit of time. (A vector quantity.)

• VelocityVelocity is the displacement per is the displacement per unit of time. (A vector quantity.)unit of time. (A vector quantity.)

Direction required!

A

Bs = 20 m

Time t = 4 s

Δx=12 m

20o

= 3 m/s at 200 N of E

= 3 m/s at 200 N of E

Average Speed and Average Speed and Instantaneous VelocityInstantaneous Velocity

The instantaneous velocity is the magn-itude and direction of the velocity at a par-ticular instant. (v at point C)

The instantaneous velocity is the magn-itude and direction of the velocity at a par-ticular instant. (v at point C)

The The averageaverage speedspeed depends depends ONLYONLY on the distance traveled and the on the distance traveled and the time required.time required.

The The averageaverage speedspeed depends depends ONLYONLY on the distance traveled and the on the distance traveled and the time required.time required.

A

Bs = 20 m

Time t = 4 s

C

Example 1.Example 1. A runner runs A runner runs 200 m, east,200 m, east, then changes direction and runs then changes direction and runs 300 m, 300 m, westwest. If the entire trip takes . If the entire trip takes 60 s60 s, what is , what is the average speed and what is the the average speed and what is the average velocity?average velocity?

Recall that Recall that average average speedspeed is a function is a function onlyonly of of total total distancedistance and and total total timetime::Total distance: Total distance: ss = 200 m + 300 m = 500 = 200 m + 300 m = 500 mm 500 m

60 s

total pathAverage speed

time

Avg. speed= 8 m/s Direction does not Direction does not matter!matter!

startstart

ss11 = 200 = 200 mm

ss22 = 300 = 300 mm

Example 1 (Cont.)Example 1 (Cont.) Now we find the Now we find the average velocity, which is the average velocity, which is the net net displacement displacement divided by divided by timetime. In this . In this case, the direction matters. case, the direction matters.

xxoo = 0 = 0

t t = 60 = 60 ssxx11= +200 = +200

mmxx = -100 = -100 mm

xxoo = 0 m; = 0 m; xx = -100 = -100 mm Direction of final Direction of final

displacement is to displacement is to the left as shown.the left as shown.

Average velocity:

Note: Average velocity is directed to the Note: Average velocity is directed to the west.west.

Definition of AccelerationDefinition of Acceleration

An An accelerationacceleration is the change in is the change in velocity per unit of time. (A velocity per unit of time. (A vectorvector quantity.)quantity.)

A A changechange inin velocityvelocity requires the requires the application of a push or pull (application of a push or pull (forceforce).).

A formal treatment of force and acceleration A formal treatment of force and acceleration will be given later. For now, you should will be given later. For now, you should know that:know that:•The direction of

accel- eration is same as direction of force.

•The acceleration is proportional to the magnitude of the force.

Pulling the wagon with twice the Pulling the wagon with twice the force produces twice the force produces twice the acceleration and acceleration is in acceleration and acceleration is in direction of force.direction of force.

Pulling the wagon with twice the Pulling the wagon with twice the force produces twice the force produces twice the acceleration and acceleration is in acceleration and acceleration is in direction of force.direction of force.

Acceleration and ForceAcceleration and Force

F a

2F 2a

Example 3 (No change in direction):Example 3 (No change in direction): A A constant force changes the speed of a car constant force changes the speed of a car from from 8 m/s8 m/s to to 20 m/s20 m/s in in 4 s4 s. What is . What is average acceleration?average acceleration?

Step 1. Draw a rough sketch.Step 1. Draw a rough sketch. Step 2. Choose a positive direction (right).Step 2. Choose a positive direction (right).Step 3. Label given info with + and - Step 3. Label given info with + and - signs.signs.Step 4. Indicate direction of force F.Step 4. Indicate direction of force F.

+

vo = +8 m/s

t = 4 s

v = +20 m/s

Force

Example 3 (Continued):Example 3 (Continued): What is average What is average acceleration of car?acceleration of car?

Step 5. Recall Step 5. Recall definition of average definition of average acceleration.acceleration.

+

vo = +8 m/s

t = 4 s

v = +20 m/s

Force

Graphical AnalysisGraphical Analysis

slope:slope:

velocity:velocity:

acceleratioacceleration:n:

x, (m)x, (m)

Position vs time graph (velocity)Position vs time graph (velocity)

v, (m/s)v, (m/s)

velocity vs time graph (acceleration)velocity vs time graph (acceleration)

Graphical AnalysisGraphical Analysis

x

tt

xx

22

xx

11

tt22tt11

2 1

2 1avg

x x xv

t t t

2 1

2 1avg

x x xv

t t t

( 0)inst

xv t

t

( 0)inst

xv t

t

x

t

Time

slope

Dis

pla

cem

en

t,

x

Average Average Velocity:Velocity:

Instantaneous Instantaneous Velocity:Velocity:

Uniform Acceleration Uniform Acceleration in One Dimension:in One Dimension:

• Motion is along a straight line (horizontal, Motion is along a straight line (horizontal, vertical or slanted).vertical or slanted).

• Changes in motion result from a CONSTANT Changes in motion result from a CONSTANT force producing uniform acceleration.force producing uniform acceleration.

• The velocity of an object is changing by a The velocity of an object is changing by a constant amount in a given time interval.constant amount in a given time interval.

• The moving object is treated as though it The moving object is treated as though it were a point particle.were a point particle.

Average velocity for constant a:

setting to = 0

combining both equations:

For constant acceleration:

Formulas based on definitionsFormulas based on definitions::

DerivedDerived formulasformulas:

For constant acceleration onlyFor constant acceleration only

Example 6:Example 6: An airplane flying initially at An airplane flying initially at 400 ft/s400 ft/s lands on a carrier deck and lands on a carrier deck and stops in a distance of stops in a distance of 300 ft.300 ft. What is What is the acceleration?the acceleration?

Δx = 300 ft

vo = 400 ft/sv = 0

+

Step 1. Draw and label sketch.

Step 2. Indicate + direction

Example: Example: (Cont.)(Cont.)

+

Step 3.Step 3. List given; find information with signs.

Given:Given: vvoo = 400 ft/s = 400 ft/s - initial velocity of - initial velocity of airplaneairplane vv = 0 = 0 - final velocity after - final velocity after

traveling traveling ΔΔxx = +300 ft = +300 ft

Find:Find: aa = ? = ? - acceleration of airplane- acceleration of airplane

Δx = 300 ft

vo = 400 ft/sv = 0

Step 4.Step 4. Select equation that contains aa and not tt.v 2 - vo

2 = 2aΔx0

a = = -vo

2

2x

-(400 ft/s)2

2(300 ft)

aa = - 300 = - 300 ft/sft/s22

aa = - 300 = - 300 ft/sft/s22

Why is the acceleration negative?Why is the acceleration negative?Because Force is in a negative direction Because Force is in a negative direction which means that the airplane slows downwhich means that the airplane slows down

Given:Given: vvoo = +400 ft/s = +400 ft/s

vv = 0 = 0

ΔΔxx = +300 = +300 ftft

Example 5:Example 5: A ball A ball 5.0 m5.0 m from the bottom of from the bottom of an incline is traveling initially at an incline is traveling initially at 8.0 m/s8.0 m/s. . Four seconds Four seconds (4.0 s)(4.0 s) later, it is traveling later, it is traveling down the incline at down the incline at 2.0 m/s2.0 m/s. How far is it . How far is it from the bottom at that instant?from the bottom at that instant?

5.0 m

Δx

8.0 m/s

-2.0 m/s

t = 4.0 s

+

Given:Given: d d = 5.0 m= 5.0 m - distance from initial position of the ball

vvoo = 8.0 m/s= 8.0 m/s - initial velocity velocity

v v = -2.0 m/s = -2.0 m/s - final velocity after - final velocity after tt = 4.0 s = 4.0 s

Find:Find: x x = ? = ? - - distance from the bottom of the inclinedistance from the bottom of the incline

x = 17.0 mx = 17.0 m

Given:Given: d d = 5.0 m= 5.0 m

vvoo = 8.0 m/s= 8.0 m/s - initial velocity velocity

v v = -2.0 m/s = -2.0 m/s - final velocity after - final velocity after tt = 4.0 s = 4.0 s

Find:Find: x x = ? = ? - - distance from the bottom of the inclinedistance from the bottom of the incline

Solution: Solution: wherewhere

Acceleration Due to Acceleration Due to GravityGravity

• Every object on the earth Every object on the earth experiences a common force: experiences a common force: the force due to gravity.the force due to gravity.

• This force is always directed This force is always directed toward the center of the toward the center of the earth (downward).earth (downward).

• The acceleration due to The acceleration due to gravity is relatively constant gravity is relatively constant near the Earth’s surface.near the Earth’s surface.

• Every object on the earth Every object on the earth experiences a common force: experiences a common force: the force due to gravity.the force due to gravity.

• This force is always directed This force is always directed toward the center of the toward the center of the earth (downward).earth (downward).

• The acceleration due to The acceleration due to gravity is relatively constant gravity is relatively constant near the Earth’s surface.near the Earth’s surface.

Earth

Wg

Gravitational AccelerationGravitational Acceleration

• In a vacuum, all objects fall In a vacuum, all objects fall with same acceleration.with same acceleration.

• Equations for constant Equations for constant acceleration apply as acceleration apply as usual.usual.

• Near the Earth’s surface:Near the Earth’s surface:

• In a vacuum, all objects fall In a vacuum, all objects fall with same acceleration.with same acceleration.

• Equations for constant Equations for constant acceleration apply as acceleration apply as usual.usual.

• Near the Earth’s surface:Near the Earth’s surface:

aa = g = - = g = -9.80 m/s9.80 m/s22 or -32 or -32 ft/sft/s22

Directed downward (usually Directed downward (usually negative).negative).

Sign Convention:Sign Convention: A Ball Thrown A Ball Thrown

Vertically Vertically UpwardUpward

• Velocity is positive (+) Velocity is positive (+) or negative (-) based or negative (-) based on on direction of motiondirection of motion..

• Velocity is positive (+) Velocity is positive (+) or negative (-) based or negative (-) based on on direction of motiondirection of motion..

• Displacement is positive Displacement is positive (+) or negative (-) (+) or negative (-) based on based on LOCATIONLOCATION. .

• Displacement is positive Displacement is positive (+) or negative (-) (+) or negative (-) based on based on LOCATIONLOCATION. .

Release Point

UP = +

• Acceleration is (+) or (-) Acceleration is (+) or (-) based on direction of based on direction of forceforce (weight). (weight).

y = 0

y = +

y = +

y = +

y = 0

y = -NegativeNegative

v = +

v = 0

v = -

v = -

v= -NegativeNegative

a = -

a = -

a = -

a = -

a = -a = -

Example 7:Example 7: A ball is thrown vertically A ball is thrown vertically upward with an initial velocity of upward with an initial velocity of 30.0 m/s30.0 m/s. . What are its position and velocity after What are its position and velocity after 2.00 2.00 ss, , 4.00 s4.00 s, and , and 7.00 s7.00 s? Find also the ? Find also the maximum height attainedmaximum height attained

a = g

+

vo = +30.0 m/s

Given: Given: a a = -= -ΔΔ9.8 m/s9.8 m/s22

vvoo = 30.0 m/s= 30.0 m/s

t t = 2.00 s; 4.00 s; 7.00 s= 2.00 s; 4.00 s; 7.00 sFind:Find:

ΔΔyy = ? – displacement = ? – displacement

v v = ? - final velocity= ? - final velocity

After those three “times”After those three “times”

ΔΔy y = ? – maximum height= ? – maximum height

Given: Given: a a = -9.8 m/s= -9.8 m/s22; ; vvoo = 30.0 m/s= 30.0 m/s

t t = 2.00 s; 4.00 s; 7.00 s= 2.00 s; 4.00 s; 7.00 s

Solutions: Solutions:

For For tt = 2.00 s: = 2.00 s:

For For tt = 4.00 s: = 4.00 s:

For For tt = 7.00 s: = 7.00 s:

Given: Given: a a = -9.8 m/s= -9.8 m/s22; ; vvoo = 30.0 m/s= 30.0 m/s

t t = 2.00 s; 4.00 s; 7.00 s= 2.00 s; 4.00 s; 7.00 s

Solutions: Solutions:

For For tt = 2.00 s: = 2.00 s:

For For tt = 4.00 s: = 4.00 s:

For For tt = 7.00 s: = 7.00 s:

Given: Given: a a = -9.8 m/s= -9.8 m/s22; ; vvoo = 30.0 m/s= 30.0 m/s

t t = 2.00 s; 4.00 s; 7.00 s= 2.00 s; 4.00 s; 7.00 s

Solutions: Solutions:

For maximum height, For maximum height, v v = 0 (the ball = 0 (the ball stops at maximum height):stops at maximum height):

Experiment 10Experiment 10 Uniformly Accelerated Motion Uniformly Accelerated Motion

(Acceleration due to Gravity) 39 (Acceleration due to Gravity) 39 (06A)(06A)