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Which would fall first?
A. Anchor or B. Ball
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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
= + +
= ++
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Derivation of the kinematic equations
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Isolating from= +
And using this expression in
= + +
We have
=
+
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Derivation of the kinematic equations
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Kinematic equations for constant
acceleration
= +
= + +
=
+
= ++
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Graphical representation of
constant acceleration
= + +
= +
= constant
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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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= 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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( ) 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