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Chapter 6A. Acceleration. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University. © 2007.

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chapter 6a acceleration

Chapter 6A. Acceleration

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

© 2007

slide2

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

Photo © Vol. 44 Photo Disk/Getty

objectives after completing this module you should be able to
Objectives: After completing this module, you should be able to:
  • Define and apply concepts of average and instantaneous velocity and acceleration.
  • Solve problems involving initial and final velocity, acceleration, displacement, and time.
  • Demonstrate your understanding of directions and signs for velocity, displacement, and acceleration.
  • Solve problems involving a free-falling body in a gravitational field.
uniform acceleration in one dimension
Uniform Acceleration in One Dimension:
  • Motion is along a straight line (horizontal, vertical or slanted).
  • Changes in motion result from a CONSTANT force producing uniform acceleration.
  • The cause of motion will be discussed later. Here we only treat the changes.
  • The moving object is treated as though it were a point particle.
distance and displacement

B

s = 20 m

A

Distance and Displacement

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

Distances is a scalar quantity (no direction):

Contains magnitude only and consists of a number and a unit.

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

distance and displacement1

B

D = 12 m, 20o

A

q

Distance and Displacement

Displacement is the straight-line separation 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)

distance and displacement2

D

8 m,E

x = -4

x = +8

12 m,W

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

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

x

D = 4 m, W

What is the distance traveled?

20 m !!

the signs of displacement

Examples:

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

The Signs of Displacement
  • Displacement is positive (+) or negative (-) based on LOCATION.

2 m

-1 m

-2 m

The direction of motion does not matter!

definition of speed

B

s

t

20 m

4 s

v = =

A

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

s = 20 m

v = 5 m/s

Not direction dependent!

Time t = 4 s

definition of velocity

s = 20 m

B

D=12 m

A

20o

Time t = 4 s

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

v = 3 m/s at 200 N of E

Direction required!

slide11

s2 = 300 m

s1 = 200 m

Avg. speed 8.33 m/s

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

Recall that average speed is a function only of total distance and total time:

start

Total distance: s = 200 m + 300 m = 500 m

Direction does not matter!

slide12

t = 60 s

xf= -100 m

x1= +200 m

xo = 0

Average velocity:

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

x0 = 0 m; xf = -100 m

Direction of final displacement is to the left as shown.

Note: Average velocity is directed to the west.

slide13

14 s

A

625 m

B

356 m

142 s

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

Total distance/ total time:

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

examples of speed

Orbit

2 x 104 m/s

Light = 3 x 108 m/s

Car = 25 m/s

Jets = 300 m/s

Examples of Speed
speed examples cont

Runner = 10 m/s

Glacier = 1 x 10-5 m/s

Snail = 0.001 m/s

Speed Examples (Cont.)
average speed and instantaneous velocity

s = 20 m

B

C

A

Time t = 4 s

Average Speed and Instantaneous Velocity
  • The averagespeed depends ONLY on the distance traveled and the time required.

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

the signs of velocity

-

+

+

-

+

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

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

average and instantaneous v

slope

x2

Dx

Dx

Displacement, x

x1

Dt

Dt

t1

t2

Time

Average and Instantaneous v

Average Velocity:

Instantaneous Velocity:

definition of acceleration

The direction of accel- eration is same as direction of force.

  • The acceleration is proportional to the magnitude of the force.
Definition of Acceleration
  • An acceleration is the change in velocity per unit of time. (A vector quantity.)
  • A changeinvelocity requires the application of a push or pull (force).

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

acceleration and force

F

a

2F

2a

Acceleration and Force

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

example of acceleration

Force

t = 3 s

+

v0= +2 m/s

vf= +8 m/s

Example 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.

the signs of acceleration

+

F

F

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

Choose + direction first. Then accelerationawill have the same sign as that of the force F —regardless of the direction of velocity.

a (-)

a(+)

slide24

Force

+

v1= +8 m/s

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

t = 4 s

v2= +20 m/s

Step 1. Draw a rough sketch.

Step 2. Choose a positive direction (right).

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

Step 4. Indicate direction of force F.

example 3 continued what is average acceleration of car

Force

+

t = 4 s

v2 = +20 m/s

v1= +8 m/s

Step 5. Recall definition of average acceleration.

Example 3 (Continued):What is average acceleration of car?
slide26

+

Force

E

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

vf= -5 m/s

vo= +20 m/s

Step 1. Draw a rough sketch.

Step 2. Choose the eastward direction as positive.

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

slide27

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

Choose the eastward direction as positive.

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

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

The change in velocity, Dv = vf - v0

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

slide28

vf - vo

tf - to

Dv

Dt

aavg= =

vo= +20 m/s

vf= -5 m/s

+

Force

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

E

-25 m/s

5 s

a =

Example 4: (Continued)

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

a = - 5 m/s2

slide29

C

D

B

A

+

a = - 5 m/s2

vf= -5 m/s

vo= +20 m/s

Force

E

Signs for Displacement

Time t = 0 at point A. What are the signs (+ or -) of displacement at B, C, and D?

At B,x is positive, right of origin

At C, x is positive, right of origin

At D, x is negative, left of origin

slide30

x = 0

C

D

B

A

+

a = - 5 m/s2

vf= -5 m/s

vo= +20 m/s

Force

E

Signs for Velocity

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

  • At B,v is zero - no sign needed.
  • At C, v is positive on way out and negative on the way back.
  • At D, v is negative, moving to left.
slide31

C

D

B

A

+

a = - 5 m/s2

vf= -5 m/s

vo= +20 m/s

Force

E

Signs for Acceleration

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

  • At B, C, and D, a = -5 m/s, negative at all points.
  • The force is constant and always directed to left, so acceleration does not change.
definitions
Definitions

Average velocity:

Average acceleration:

velocity for constant a
Velocity for constant a

Average velocity:

Average velocity:

Setting to = 0 and combining we have:

slide34

+

F

x

vf

vo

-2 m/s

5 m

Careful

t = 4 s

8 m/s

vo + vf

2

x = xo + t

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

2

= 5 m + (4 s)

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

slide35

(Continued)

F

+

x

vf

vo

-2 m/s

5 m

t = 4 s

8 m/s

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

2

x = 5 m + (4 s)

8 m/s - 2 m/s

2

x = 5 m + (4 s)

x = 17 m

constant acceleration
Constant Acceleration

Acceleration:

Setting to = 0 and solving for v, we have:

Final velocity = initial velocity + change in velocity

acceleration in our example

+

F

x

v

vo

-2 m/s

5 m

t = 4 s

8 m/s

Acceleration in our Example

What is the meaning of negative sign for a?

The force changing speed is down plane!

a = -2.50 m/s2

formulas based on definitions
Formulas based on definitions:

Derivedformulas:

For constant acceleration only

use of initial position x 0 in problems

0

0

0

0

Use of Initial Position x0in 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.

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

review of symbols and units
Review of Symbols and Units
  • Displacement (x, xo); meters (m)
  • Velocity (v, vo); meters per second (m/s)
  • Acceleration (a); meters per s2 (m/s2)
  • Time (t); seconds (s)

Review sign convention for each symbol

the signs of displacement1

2 m

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

-1 m

-2 m

The Signs of Displacement
  • Displacement is positive (+) or negative (-) based on LOCATION.
the signs of velocity1

-

+

+

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

-

+

The Signs of Velocity
  • Velocity is positive (+) or negative (-) based on direction of motion.
acceleration produced by force

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

F

a(-)

F

a(+)

Acceleration Produced by Force
  • Acceleration is (+) or (-) based on direction offorce(NOT based on v).

More will be said later on the relationship between Fand a.

problem solving strategy
Problem Solving Strategy:
  • Draw and label sketch of problem.
  • Indicate + direction and force direction.
  • 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.
slide45

+400 ft/s

v = 0

300 ft

vo

+

F

X0 = 0

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

Step 1. Draw and label sketch.

Step 2. Indicate + direction and F direction.

example cont

+400 ft/s

v = 0

300 ft

vo

+

F

X0 = 0

Example: (Cont.)

Step 3. List given; find information with signs.

Given:vo = +400 ft/s

v = 0

x = +300 ft

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

Find:a = ?; t = ?

slide47

x

+400 ft/s

v = 0

300 ft

vo

+

0

0

F

X0 = 0

-vo2

2x

-(400 ft/s)2

2(300 ft)

a = =

Continued . . .

Step 4. Select equation that contains aand not t.

2a(x -xo) = v2 - vo2

Initial position and final velocity are zero.

a = - 267 ft/s2

Why is the acceleration negative?

Because Force is in a negative direction!

acceleration due to gravity

g

W

Earth

Acceleration Due to Gravity
  • Every object on the earth experiences a common force: the force due to gravity.
  • This force is always directed toward the center of the earth (downward).
  • The acceleration due to gravity is relatively constant near the Earth’s surface.
gravitational acceleration
Gravitational Acceleration
  • In a vacuum, all objects fall with same acceleration.
  • Equations for constant acceleration apply as usual.
  • Near the Earth’s surface:

a = g = 9.80 m/s2 or 32 ft/s2

Directed downward (usually negative).

experimental determination of gravitational acceleration

Dt

y

Experimental Determination of Gravitational Acceleration.

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

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

experimental determination of gravity y 0 0 y 1 20 m

Dt

y

+

Acceleration of Gravity:

W

Experimental Determination of Gravity (y0 = 0; y = -1.20 m)

y = -1.20 m; t = 0.495 s

Acceleration ais negative because force W is negative.

sign convention a ball thrown vertically upward

UP = +

Sign Convention:A Ball Thrown Vertically Upward

a = -

v = 0

y = +

a = -

v = +

y = +

y = +

a = -

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

v = -

v = -

y = 0

a = -

y = 0

Release Point

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

y = -Negative

v= -Negative

a = -

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

Tippens

same problem solving strategy except a g
Same Problem Solving Strategy Except a = g:
  • Draw and label sketch of problem.
  • Indicate + direction and force direction.
  • 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.
slide54

+

a = g

Example 7:A ball is thrown vertically upward with an initial velocity of 30 m/s. What are its position and velocity after 2 s, 4 s, and 7 s?

Step 1. Draw and label a sketch.

Step 2. Indicate + direction and force direction.

Step 3. Given/find info.

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

vo = + 30m/sy = ?v = ?

vo= +30 m/s

finding displacement

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

+

a = g

0

vo= 30 m/s

Finding Displacement:

y = (30 m/s)t + ½(-9.8 m/s2)t2

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

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

finding velocity

+

a = g

vo= 30 m/s

Finding Velocity:

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

Substitute t = 2, 4, and 7 s:

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

example 7 cont now find the maximum height attained

+

a = g

vo = +96 ft/s

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

Displacement is a maximum when the velocity vf is zero.

To find ymaxwe substitute t = 3.06 s into the general equation for displacement.

y = (30 m/s)t + ½(-9.8 m/s2)t2

example 7 cont finding the maximum height

+

a = g

vo =+30 m/s

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

y = (30 m/s)t + ½(-9.8 m/s2)t2

t = 3.06 s

Omitting units, we obtain:

y = 91.8 m - 45.9 m

ymax = 45.9 m

summary of formulas
Summary of Formulas

DerivedFormulas:

For Constant Acceleration Only

summary procedure

Draw and label sketch of problem.

  • Indicate + direction and force direction.
  • List givens and state what is to be found.

Given: ____, _____, ______

Find: ____, _____

  • Select equation containing one and not the other of the unknown quantities, and solve for the unknown.
Summary: Procedure