Chapter 6A. Chapter 6A. AccelerationAcceleration

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of Paul E. Tippens, Professor of PhysicsPhysics

Southern Polytechnic State Southern Polytechnic State UniversityUniversity© 2007

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. Photo © Vol. 44 Photo

Disk/Getty

Objectives: After completing Objectives: After completing this module, you should be this module, you should be able to:able to:• Define and apply concepts of Define and apply concepts of averageaverage and and

instantaneous instantaneous velocity and acceleration.velocity and acceleration.

• Solve problems involving initial and final Solve problems involving initial and final velocityvelocity, , accelerationacceleration, , displacementdisplacement, and , and timetime..

• Demonstrate your understanding of directions Demonstrate your understanding of directions and signs for velocity, displacement, and and signs for velocity, displacement, and acceleration.acceleration.

• Solve problems involving a free-falling body in a Solve problems involving a free-falling body in a gravitational fieldgravitational field..

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 cause of motion will be discussed later. The cause of motion will be discussed later. Here we only treat the changes.Here we only treat the changes.

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

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

Bs = 20 m

DistanceDistance ss 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.

(20 m, 40 mi/h, 10 (20 m, 40 mi/h, 10 gal)gal)

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.

(12 m, 300; 8 km/h, N)

A

BD = 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 DD is from the origin to is from the origin to the final position:the final position:

What is the What is the distancedistance traveled? traveled?20 m !!

12 m,W

D

D = 4 m, WD = 4 m, W

x8 m,E

x = +8

x = -4

The Signs of DisplacementThe Signs of Displacement• Displacement is positive (+) or Displacement is positive (+) or

negative (-) based on negative (-) based on LOCATIONLOCATION..

2 m

-1 m

-2 m

The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION.

Examples:

The direction of motion does not matter!The direction of motion does not matter!

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).

v = = s

t

20 m

4 s

v = 5 m/sv = 5 m/s

Not direction dependent!

A

Bs = 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.)

v = 3 m/s at 200 N of E

v = 3 m/s at 200 N of E

Direction required!

A

Bs = 20 m

Time t = 4 s

12 m

4 s

Dv

t

D=12 m

20o

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 8.33 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

mmxxff = -100 = -100 mm

0fx xv

t

xx00 = 0 m; x = 0 m; xff = -100 = -100 mm

100 m 01.67 m/s

60 sv

Direction of final Direction of final displacement is to displacement is to the left as shown.the left as shown.

Average velocity:

1.67 m/s, Westv

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

Example 2.Example 2. A sky diver jumps and falls A sky diver jumps and falls for 600 m in 14 s. After chute opens, he for 600 m in 14 s. After chute opens, he falls another 400 m in 150 s. What is falls another 400 m in 150 s. What is average speed for entire fall?average speed for entire fall?

625 m

356 m

14 s

142 s

A

B

600 m + 400 m

14 s + 150 sA B

A B

x xv

t t

1000 m

164 sv 6.10 m/sv 6.10 m/sv

Average speed is a Average speed is a function function onlyonly of total of total distance traveled and the distance traveled and the total time required.total time required.

Average speed is a Average speed is a function function onlyonly of total of total distance traveled and the distance traveled and the total time required.total time required.

Total distance/ total time:Total distance/ total time:

Examples of SpeedExamples of Speed

Light = 3 x 108 m/s

Orbit

2 x 104 m/s

Jets = 300 m/s Car = 25 m/s

Speed Examples (Cont.)Speed Examples (Cont.)

Runner = 10 m/s

Snail = 0.001 m/s

Glacier = 1 x 10-5 m/s

Average Speed and Average Speed and Instantaneous VelocityInstantaneous Velocity

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

The instantaneous velocity is the magn-itude and direction of the speed 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

The Signs of VelocityThe Signs of Velocity

First choose + direction; then v is positive if motion is with that direction, and negative if it is against that direction.

First choose + direction; then v is positive if motion is with that direction, and negative if it is against that direction.

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

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

-+

-++

Average and Average and Instantaneous Instantaneous vv

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:

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 of AccelerationExample of Acceleration

The wind changes the speed of a boat from 2 m/s to 8 m/s in 3 s. Each second the speed changes by 2 m/s.

Wind force is constant, thus acceleration is constant.Wind force is constant, thus acceleration is constant.

+

vf = +8 m/s v0 = +2 m/s

t = 3 s

Force

The Signs of AccelerationThe Signs of Acceleration• Acceleration is positive (Acceleration is positive (++) or negative ) or negative

((--) based on the ) based on the direction direction ofof force force..• Acceleration is positive (Acceleration is positive (++) or negative ) or negative

((--) based on the ) based on the direction direction ofof force force..

Choose + direction Choose + direction first. Then first. Then accelerationacceleration aa will will have the have the same signsame sign as that of the as that of the force F force F —regardless of the —regardless of the direction of velocity.direction of velocity.

Choose + direction Choose + direction first. Then first. Then accelerationacceleration aa will will have the have the same signsame sign as that of the as that of the force F force F —regardless of the —regardless of the direction of velocity.direction of velocity.

F

F

+a (-)

a(+)

Average and Average and Instantaneous Instantaneous aa

v

t

v2

v1

t2t1

v

t

time

slope

2 1

2 1avg

v v va

t t t

2 1

2 1avg

v v va

t t t

( 0)inst

va t

t

( 0)inst

va t

t

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.

+

v1 = +8 m/s

t = 4 s

v2 = +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.

2 1

2 1avg

v v va

t t t

2 1

2 1avg

v v va

t t t

20 m/s - 8 m/s3 m/s

4 sa

3 m/s, rightwarda

+

v1 = +8 m/s

t = 4 s

v2 = +20 m/s

Force

Example 4:Example 4: A wagon moving east at A wagon moving east at 20 20 m/sm/s encounters a very strong head-wind, encounters a very strong head-wind, causing it to change directions. After causing it to change directions. After 5 s5 s, , it is traveling west at it is traveling west at 5 m/s5 m/s. What is the . What is the average acceleration?average acceleration? (Be careful of (Be careful of signs.)signs.)

Step 1. Draw a rough sketch.Step 1. Draw a rough sketch.

+ Force

Step 2. Choose the eastward direction as Step 2. Choose the eastward direction as positive.positive.

vo = +20 m/s

vf = -5 m/s

Step 3. Label given info with + and - Step 3. Label given info with + and - signs.signs.

E

Example 4 (Cont.):Example 4 (Cont.): Wagon moving east at Wagon moving east at 20 20 m/sm/s encounters a head-wind, causing it to encounters a head-wind, causing it to change directions. Five seconds later, it is change directions. Five seconds later, it is traveling west at traveling west at 5 m/s5 m/s. What is the average . What is the average acceleration?acceleration?

Choose the eastward direction as Choose the eastward direction as positive.positive.

Initial velocity, Initial velocity, vvoo == +20 m/s, east (+)+20 m/s, east (+)

Final velocity, Final velocity, vvff = -5 m/s, west (-) = -5 m/s, west (-)

The change in velocity, The change in velocity, vv = = vvff - - vv00

vv = (-5 m/s) - (+20 m/s) = -25 m/s = (-5 m/s) - (+20 m/s) = -25 m/s

Choose the eastward direction as Choose the eastward direction as positive.positive.

Initial velocity, Initial velocity, vvoo == +20 m/s, east (+)+20 m/s, east (+)

Final velocity, Final velocity, vvff = -5 m/s, west (-) = -5 m/s, west (-)

The change in velocity, The change in velocity, vv = = vvff - - vv00

vv = (-5 m/s) - (+20 m/s) = -25 m/s = (-5 m/s) - (+20 m/s) = -25 m/s

Example 4: Example 4: (Continued)(Continued)

aaavgavg == =vv

tt

vvff - - vvoo

ttff - t - too

aa = =-25 m/s-25 m/s

5 s5 s

a = - 5 m/s2

a = - 5 m/s2

Acceleration is directed to left, west (same as F).

+ Force

vo = +20 m/s vf = -5 m/s

E

v = (-5 m/s) - (+20 m/s) = -25 m/s

Signs for Signs for DisplacementDisplacement

Time t = 0 at point Time t = 0 at point AA. What are the . What are the signs (+ or -) signs (+ or -) of of displacementdisplacement at at BB, , CC, and , and DD??At B, x is positive, right of origin

At C, x is positive, right of originAt D, x is negative, left of origin

+ Force

vo = +20 m/s vf = -5 m/s

E a = - 5 m/s2

A BCD

Signs for VelocitySigns for Velocity

What are the signs (+ or -) of velocity What are the signs (+ or -) of velocity at points B, C, at points B, C, and D? and D? At At B,B, vv is is zerozero - no sign - no sign

needed.needed. At At CC, , vv is is positive positive on way out on way out and and negativenegative on the way back. on the way back.

At At DD, , vv is is negativenegative, moving to , moving to left.left.

+ Force

vo = +20 m/s vf = -5 m/s

E a = - 5 m/s2

A BCD

x = 0

What are the signs (+ or -) of acceleration at points B, C, and D?

The force is constant and always directed to left, so acceleration does not change.

At At B, C, and DB, C, and D, , aa = -5 m/s, = -5 m/s, negativenegative at all points. at all points.

Signs for AccelerationSigns for Acceleration

+ Force

vo = +20 m/s vf = -5 m/s

E a = - 5 m/s2

A BCD

DefinitionsDefinitions

Average velocity:

Average acceleration:

2 1

2 1avg

x x xv

t t t

2 1

2 1avg

x x xv

t t t

2 1

2 1avg

v v va

t t t

2 1

2 1avg

v v va

t t t

Velocity for constant Velocity for constant aa

Average velocity:

Average velocity:

Setting to = 0 and combining we have:

0

0

favg

f

x xxv

t t t

0

0

favg

f

x xxv

t t t

0

2f

avg

v vv

0

2f

avg

v vv

00 2

fv vx x t

00 2

fv vx x t

Example 5:Example 5: A ball A ball 5 m5 m from the bottom of an from the bottom of an incline is traveling initially at incline is traveling initially at 8 m/s8 m/s. . FourFour secondsseconds later, it is traveling down the later, it is traveling down the incline at incline at 2 m/s2 m/s. How far is it from the . How far is it from the bottom at that instant?bottom at that instant?

x = xo + tvo + vf

2= 5 m + (4 s)

8 m/s + (-2 m/s)

2

5 m

x

8 m/s

-2 m/st = 4

s

vo

vf

+ F

CarefuCarefull

x = 5 m + (4 s) 8 m/s - 2 m/s

2x = 17 mx = 17 m

x = 5 m + (4 s)

8 m/s + (-2 m/s)

2

5 m

x

8 m/s

-2 m/s

t = 4 s

vo

vf

+ F(Continued)(Continued)

Constant AccelerationConstant Acceleration

Acceleration:Acceleration:

Setting tSetting too = 0 and solving for = 0 and solving for v,v, we we have:have:

Final velocity = initial velocity + change in Final velocity = initial velocity + change in velocityvelocity

0fv v at 0fv v at

0

0

favg

f

v vva

t t t

0

0

favg

f

v vva

t t t

Acceleration in our Acceleration in our ExampleExample

a = -2.50 m/s2

a = -2.50 m/s2

What is the What is the meaning of meaning of negative sign for negative sign for aa??

0fv va

t

0fv v at

5 m

x

8 m/s8 m/s

-2 m/st = 4 s

vo

v+ F

2( 2 m/s) ( 8 m/s)2 m/s

4 sa

The force changing The force changing speed is down speed is down plane!plane!

Formulas based on definitions:Formulas based on definitions:

DerivedDerived formulasformulas:

For constant acceleration onlyFor constant acceleration only

210 0 2x x v t at

210 0 2x x v t at 21

0 2fx x v t at 210 2fx x v t at

00 2

fv vx x t

00 2

fv vx x t

0fv v at 0fv v at

2 20 02 ( ) fa x x v v 2 2

0 02 ( ) fa x x v v

Use of Initial Position Use of Initial Position xx00 in in Problems.Problems.

If you choose the origin of your x,y axes at the point of the initial position, you can set x0 = 0, simplifying these equations.

If you choose the origin of your x,y axes at the point of the initial position, you can set x0 = 0, simplifying these equations.

210 0 2x x v t at 21

0 0 2x x v t at

210 2fx x v t at 21

0 2fx x v t at

00 2

fv vx x t

00 2

fv vx x t

2 20 02 ( ) fa x x v v 2 2

0 02 ( ) fa x x v v

0fv v at 0fv v at

The The xxoo term is very term is very useful for studying useful for studying problems involving problems involving motion of two motion of two bodies.bodies.

00

00

00

00

Review of Symbols and Review of Symbols and UnitsUnits

• Displacement ( (x, xx, xoo); meters (); meters (mm))

• Velocity ( (v, vv, voo); meters per second (); meters per second (m/sm/s))

• Acceleration ( (aa); meters per s); meters per s22 ( (m/sm/s22))

• TimeTime ( (tt); seconds (); seconds (ss))

• Displacement ( (x, xx, xoo); meters (); meters (mm))

• Velocity ( (v, vv, voo); meters per second (); meters per second (m/sm/s))

• Acceleration ( (aa); meters per s); meters per s22 ( (m/sm/s22))

• TimeTime ( (tt); seconds (); seconds (ss))

Review sign convention for each Review sign convention for each symbolsymbol

The Signs of DisplacementThe Signs of Displacement

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

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

The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION.

2 m2 m

-1 m-1 m

-2 m-2 m

The Signs of VelocityThe Signs of Velocity

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

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

First choose + direction; then velocity v is positive if motion is with that + direction, and negative if it is against that positive direction.

+

-

-

+

+

Acceleration Produced by Acceleration Produced by ForceForce

• Acceleration is (Acceleration is (++) or () or (--) based on ) based on direction ofdirection of forceforce ((NOTNOT based on based on vv).).

• Acceleration is (Acceleration is (++) or () or (--) based on ) based on direction ofdirection of forceforce ((NOTNOT based on based on vv).).

A push or pull (force) is necessary to change velocity, thus the sign of a is same as sign of F.

FF aa((-)-)

FF aa(+)(+) More will be said later on the relationship between F and a.

Problem Solving Strategy:Problem Solving Strategy: Draw and label sketch of problem.Draw and label sketch of problem.

Indicate Indicate ++ direction and direction and forceforce direction. direction.

List givens and state what is to be found.List givens and state what is to be found.

Given: ____, _____, _____ (x,v,vo,a,t)

Find: ____, _____ Select equation containing one and

not the other of the unknown quantities, and solve for the unknown.

Example 6:Example 6: A airplane flying initially at A 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?

300 ft

+400 ft/s

vo

v = 0+ F

Step 1. Draw and label sketch.

Step 2. Indicate + direction and F F direction.

XX00 = = 00

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

300 ft

+400 ft/s

vo

v = 0

+ F

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

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

vv = 0 = 0xx = +300 = +300 ftft

Find:Find: aa = ?; t = ?; t = ?= ?

List t = ?, even List t = ?, even though time was not though time was not asked for.asked for.

XX00 = = 00

Step 4.Step 4. Select equation that contains aa and not tt.

300 ft

+400 ft/s

vo

v = 0

+ F

x

2a(x -xo) = v2 - vo

2

0 0

a = = -vo

2

2x

-(400 ft/s)2

2(300 ft) aa = - 267 = - 267 ft/sft/s22

aa = - 267 = - 267 ft/sft/s22

Why is the acceleration negative?Why is the acceleration negative?

Continued . . Continued . . ..

Initial position and Initial position and final velocity are final velocity are zero.zero.

XX00 = 0 = 0

Because Force is in a negative Because Force is in a negative direction!direction!

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 ft/s or 32 ft/s22

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

Experimental Experimental Determination of Determination of Gravitational Gravitational Acceleration.Acceleration.

The apparatus consists of a The apparatus consists of a device which measures the device which measures the time required for a ball to fall time required for a ball to fall a given distance.a given distance.

Suppose the height is 1.20 Suppose the height is 1.20 m and the drop time is m and the drop time is recorded as 0.650 s. What recorded as 0.650 s. What is the acceleration due to is the acceleration due to gravity?gravity?

yy

tt

Experimental Determination Experimental Determination of Gravity of Gravity (y(y00 = 0; y = -1.20 = 0; y = -1.20 m)m)

yy

tt

y = -1.20 m; t = y = -1.20 m; t = 0.495 s0.495 s 21

0 02 ; 0y v t at v

2 2

2 2( 1.20 m)

(0.495 s)

ya

t

29.79 m/sa Acceleration of Gravity:

Acceleration Acceleration aa is is negativenegative because force because force WW is is negative.negative.

Acceleration Acceleration aa is is negativenegative because force because force WW is is negative.negative.

++

WW

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 = +

TippensTippens

• 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 = -

Same Problem Solving Same Problem Solving Strategy Except Strategy Except aa = g = g::

Draw and label sketch of problem.Draw and label sketch of problem.

Indicate Indicate ++ direction and direction and forceforce direction. direction.

List givens and state what is to be found.List givens and state what is to be found.

Given: ____, _____, a = - 9.8 m/s2

Find: ____, _____ Select equation containing one and

not the other of the unknown quantities, and solve for the unknown.

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 m/s30 m/s. . What are its position and velocity after What are its position and velocity after 2 s2 s, , 4 s4 s, and , and 7 s7 s??

Step 1. Draw and label a sketch.

a = g

+

vo = +30 m/s

Step 2. Indicate + direction and force direction.Step 3. Given/find info.

a = -9.8 ft/s2 t = 2, 4, 7 s

vo = + 30 m/s y = ? v = ?

Finding Finding Displacement:Displacement:

a = g

+

vo = 30 m/s

0

y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8 m/sm/s22))tt22

Substitution of t = 2, 4, and Substitution of t = 2, 4, and 7 s will give the following 7 s will give the following values: values:

y = 40.4 m; y = 41.6 m; y = -30.1 my = 40.4 m; y = 41.6 m; y = -30.1 m

210 0 2y y v t at

Step 4. Select equation that contains y and not v.

Finding Velocity:Finding Velocity:Step 5. Find v from equation that contains v and not x:

Step 5. Find v from equation that contains v and not x:

Substitute t = 2, 4, and 7 Substitute t = 2, 4, and 7 s:s:

v = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/sv = +10.4 m/s; v = -9.20 m/s; v = -38.6 m/s

a = g

+

vo = 30 m/s

0fv v at 0fv v at

230 m/s ( 9.8 m/s )fv t

Example 7: (Cont.)Example 7: (Cont.) Now Now find the find the maximum heightmaximum height attained:attained:

Displacement is a Displacement is a maximum when the maximum when the velocity velocity vvff is zero. is zero. a =

g

+

vo = +96 ft/s

230 m/s ( 9.8 m/s ) 0fv t

2

30 m/s; 3.06 s

9.8 m/st t

To find To find yymaxmax we we substitute substitute tt = 3.06 s = 3.06 s into the general into the general equation for equation for displacement.displacement.y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8

m/sm/s22))tt22

Example 7: (Cont.)Example 7: (Cont.) Finding the Finding the maximum maximum height:height:

y = y = (30 m/s)(30 m/s)tt + + ½½(-9.8 (-9.8 m/sm/s22))tt22

a = g+

vo =+30 m/s

tt = 3.06 = 3.06 ss

212(30)(3.06) ( 9.8)(3.06)y

yy = 91.8 m - 45.9 = 91.8 m - 45.9 mm

Omitting units, we obtain:Omitting units, we obtain:

ymax = 45.9 m

Summary of FormulasSummary of Formulas

DerivedDerived FormulasFormulas:

For Constant Acceleration Only

210 0 2x x v t at

210 0 2x x v t at 21

0 2fx x v t at 210 2fx x v t at

00 2

fv vx x t

00 2

fv vx x t

0fv v at 0fv v at

2 20 02 ( ) fa x x v v 2 2

0 02 ( ) fa x x v v

Summary: ProcedureSummary: Procedure

Draw and label sketch of problem.Draw and label sketch of problem.

Indicate Indicate ++ direction and direction and forceforce direction.direction.

List givens and state what is to be List givens and state what is to be found.found.Given: ____, _____, ______

Find: ____, _____ Select equation containing one and

not the other of the unknown quantities, and solve for the unknown.

CONCLUSION OF CONCLUSION OF Chapter 6 - AccelerationChapter 6 - Acceleration