AccelerationAcceleration• Is a change in velocity over Is a change in velocity over

time.time.

• Is usually described as Is usually described as “speeding up” or “slowing “speeding up” or “slowing down”, but ……down”, but ……

• If an objects direction If an objects direction changes its direction, it is changes its direction, it is accelerating even if it is accelerating even if it is speed is not changing.speed is not changing.

• Is a vector!Is a vector!

• If acceleration and velocity If acceleration and velocity are in same direction, the are in same direction, the object speeds up. Opposite object speeds up. Opposite directions, the object slows directions, the object slows down.down.

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

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

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))

• Values with an Values with an oo subscript means it is the subscript means it is the initial value. Values without it are final.initial value. Values without it are final.

• 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))

• Values with an Values with an oo subscript means it is the subscript means it is the initial value. Values without it are final.initial value. Values without it are final.

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

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

Indicate Indicate ++ 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)

Unkowns: ____, _____ 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+

Step 1. Draw and label sketch.

Step 2. Indicate + direction.

XX00 = = 00

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

300 ft

+400 ft/s

vo

v = 0

+

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 = 9.8 m/s9.8 m/s22 or approximately 10 or approximately 10 m/sm/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 = +

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 m/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 = 3.00 s = 3.00 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 = 3.00 = 3.00 ss

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

yy = 90.0 m – 45.0 = 90.0 m – 45.0 mm

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

ymax = 45.0 m