1stle lecture 05 - r2 motion with constant acceleration(a)

8/11/2019 1stLE Lecture 05 - R2 Motion With Constant Acceleration(a)

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Which would fall first?

A. Anchor or B. Ball

1


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Before 1600s

2

Heavier objects

fall faster than the

lighter ones

Aristotle


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3

LOOK OUT FOR FALLING OBJECTS!

In 1600s

Galileos inclined plane

all objects fall at

the same rate


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August 2, 1971

4

David Scott, performed

the following experiment

in the vacuum of space;

using geologists hammerand a falcons feather hit

Reminder: Play video


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August 2, 1971

5

David Scott, performed

the following experimentin the vacuum of space.

A geologists hammer

and a falcons feather hit

the lunar surface at thesame time.

I told you so


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Objectives

6

Solve problems involving motion withconstant acceleration and free falling bodies


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Lecture 5:

Motion with constant acceleration

= +

= + +

=

+

= ++


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Derivation of the kinematic equations

For constant acceleration,=

Let = 0and = .

Let = , initial velocity at time = 0and= , final velocity at time = .

=

=

Rearranging,

= +

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The average velocity is given by=

For a constant acceleration,

=+

2

Using the above expressions, we have

= + +

= ++

9



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Isolating from= +

And using this expression in

= + +

We have

=

+

10



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Kinematic equations for constant

acceleration

= +

= + +

=

+

= ++

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Graphical representation of

constant acceleration

= + +

= +

= constant

12

0

0


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Guides in problem solving

Determine if problem assumes constant .

When is not specified, it can well be taken as 0.

When stated that object is initially at rest, = .

You need at least two valuesto solve any of thefour equations. Find which pair is given and use

appropriate form of equation.

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Approach in problem solving

Write the givenand draw the situation.

Determine what was asked.

Determine the right expression to solve theproblem.

To checkif your answer is right or wrong:

See if the value you got makes sense.

Check the units.

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Sample Problem:

A motorcyclist heading east accelerates after he passes

a signpost. His acceleration is constant at 4.0m/s2. At

time t = 0 he is 5.0m east of the signpost, moving east at

15m/s.

(a) Find his position and velocity at time t = 2.0s.

(b) Where is the motorcyclist when his velocity is 25m/s?

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= + +1

2

= 5.0 + (15/)(2.0) + 12

(4.0/2)(2.0)

=

= +

= 15/ + (4.0/2)(2.0)

= /

To solve for the position x at time t = 2.0s;

Then, for velocity at time t = 2.0s:

(a) Find his position and velocity at time t = 2.0s.

Given: ax = 4.0m/s2 v0 = 15m/s

x0 = 5.0m t = 2.0s


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17

= 2

+ 2( 0)

= 0 +

2 02

2

= 5.0 +(25/)2 (15/)2

2(4.0/2)

=

(b) Where is the motorcyclist when his velocity is 25m/s?

Solving for x and substituting the known values:

Note that time is not given here but we know vx, v0x,axand x0,

therefore we can use:

Evaluate: Do our results make sense?

Positive acceleration; velocity is increasing

Given: ax = 4.0m/s2 v0 = 15m/s

x0 = 5.0m vx= 25m/s


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Special case: free fall

All bodies at a particular

location fall with the same

downward acceleration

regardless of size and weight

Neglect air resistance

Distance of fall is smaller

compared to the radius of the

earth

Ignore effects due to the

earths rotation

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Acceleration due to gravity, Constant acceleration of a free-falling body

= 9.81 m/s2= 981 cm/s2= 32 ft/s2

= +

= + +1

2

=

+ 2

= +

+

2

=

= +

=

= +

+

19

S l P bl


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Sample Problem:

You throw a ball vertically upward from the roof of a tall

building. The ball leaves your hand at a point even with the

roof railing with the speed of 15.0m/s; the ball is in free fall.

On its way back down, it just misses the railing. Find:

(a) the position and velocity of the ball 1.00s and 4.00s after

leaving your hand;

(b) the velocity when the ball is 5.00m above the railing;

(c) the maximum height reached and the time at which it is

reached; and

(d) the acceleration of the ball at its maximum height.

Given: v0y= 15.0m/s

=

= +

=

= +

+


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21

( ) Th iti d l it ti t ft th b ll


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(a) The position y and velocity vya time t after the ball

leaves your hand are given by:

= + (. /)

(. /)

= . / (. /)

= +

=

(1)

(2)

(3)

(4)

To get the position & velocity at time t = 1.0s, we substitute

it to eqs. 2 and 4:

We use the same equation to get position & velocity for t = 4.00s

( = . ) = +.

= . = .

( = . ) = +. /

= . = . /


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(b) The y-velocity vyat any point y is given by:

=

When the ball is 5.00m above the origin, y = +5.00m,substituting this to eq. 6:

(5)

(6)

= (. /) . / . = /

We get two values of vybecause the ball passes

through the point y = +5.00m twice!

= . /

= (. /)(. /)


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(c) Just at the instant when the ball reaches the highest point;

it is momentarily at rest and vy=0. The maximum height ycan

then be solve using:

=

= (. /)

(. /)

(d) At highest point, the acceleration is

still ay = -g = -9.8m/s2

=

=

= .


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Seatwork- solve problems in your

notebooks

- write the answers only inyour bluebook

- indicate the date

August 20, 20141. Blah?

2. Blah blah!

3. Blah blah blah!

4. Blah blah blah blah!

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Seatwork: Young and Freedman, 2.21

An antelope moving with constant acceleration covers

the distance between two points 70.0 m apart in 7.00 s.

Its speed as it passes the second point is 15.0 m/s.

1) What is its speed at the first point?

2) What is its acceleration?

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Hint: Given: x0x = 70.0m (assign x0=0)

t = 7.0s vx = 15m/s

= +

= + +

=

+

= ++

Equations for motion with constant acceleration:


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Seatwork: Young and Freedman, Prob. 2.39

A flea can jump straight up to a height of 0.440 m,

3) what is the initial speed as it leaves the ground?4) how long is it in the air?

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Given: y = 0.440m y0= 0

g = 9.81m/s2 for SW3: vy = 0

=

= +

=

= ++

Equations for motion with constant acceleration:Equations for motion with constant acceleration:

multiply t*2


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Seatwork

- seatwork will be checked atthe end of the class

- if wrong, write the correct

answer

- in checking: place the score

above the checkers name

- the checker must sign under

his/her name & studentnumber

August 20, 20141. Blah?

2. Blah blah!

3. Blah blah blah!

4. Blah blah blah blah!

Score: 3/4Checked by:

(signed)

Albert Einstein Jr.

(2013-24601)

X Bleh!!!

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AnswersAugust 14, 2014

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Answers to Seat work

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1-2) a = 0.2m/s2

(increases)3-4) a = -0.2m/s2(decreases)

5-6) a = -0.3m/s2 (increases)

7-8) a = 0.4m/s2 (decreases)

Negative slope

-negative acceleration


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a

Parabolic, reverses

concavity in the

middle

v=0 in the middle

(slope of x-t),

+ andvelocity on

other side

a

1stle lecture 05 - r2 motion with constant acceleration(a)

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