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.

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

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

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

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

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

Review of Symbols and Review of Symbols and UnitsUnits

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

• Velocity ( (vvff, v, 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 ( (vvff, v, 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 m/s400 m/s lands on a carrier deck and lands on a carrier deck and stops in a distance of stops in a distance of 300 m.300 m. What is What is the acceleration?the acceleration?

300 m

+400 m/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 m

+400 m/s

vo

v = 0

+ F

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

Given:Given: vvoo = +400 = +400 m/sm/s vv = 0 = 0

xx = +300 = +300 mm

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 m/s)2

2(300 m)

aa = - 267 = - 267 m/sm/s22

aa = - 267 = - 267 m/sm/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!

CONCLUSION OF CONCLUSION OF Chapter 6 - AccelerationChapter 6 - Acceleration