T.Merfat Al-Zumia
‫بسم هللا الرحمن الرحيم‬
Chapter (2) :Motion in one dimension
8- Motion under a constant velocity:
Velocity = ∆x/∆t
X (m)
(
)
*
But we know the line equation: Y= slop X + Co **
By comparing equation * with **:
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-The condition when the particle moves under a
constant velocity:
So a Motion under a constant :
1- Velocity means
2- Speed means
A constant speed
(magnitude)
A constant speed
(magnitude)
And on a straight line
(direction)
And even on a straight
or a curved line
(no direction)
T.Merfat Al-Zumia
9- Motion under a constant acceleration:
1- A First motion’s law :
From the average acceleration
(
)
(1)
We know the line equation is Y= slop X + co
=
slop
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2- A second motion’s law:
From the average velocity equation
(
)
(
(
)
)
(
)
From equation (1)
(
)
(
(
)
)
T.Merfat Al-Zumia
(
)
(
(
(
)
)
(
)
)
(2)
3- A Third motion’s law :
From the average velocity
equation
From
and the average
acceleration equation
From
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From two columns
(
)
(
)
(3)
We now have all the equations we need to solve
constant-acceleration problems.
(
)
(1)
(
)
(2)
(3)
T.Merfat Al-Zumia
Galileo dropped two balls of different weights
from the top of building
the two balls will reach the ground at the same
time. He proved that when objects of different
weights are dropped at the same height and time,
they take the same amount of time to fall to the
ground (ignoring air resistance).
T.Merfat Al-Zumia
Example (2.7) :
A jet lands on an aircraft carrier at (63 =m/s).
1-What is its acceleration (assumed constant) if it
stops in 2.0 s ?
2-If the plane touches down at position xi = 0, what
is the final position of the plane?
3-if the displacement is 75m find the maximum
initial speed with which the plane can land and still
come to rest
T.Merfat Al-Zumia
10- The thrown up and freely falling:
1-Means Acceleration Due to Gravity
2-The acceleration of a freely falling body is called the
acceleration due to gravity, g= 9.80 m/s2
3-The acceleration due to gravity is directed downward,
toward the center of the earth.
Maximum height
Note :
Vyf = 0
yf= value
Vyi = 0
yi = value
Vy,falling = -Vy,thrown
ay,falling = - ay,thrown
(9.8)
(-9.8)
tfalling = tthrown
Thrown up
t = t thrown
ay= - 9.8
Freely falling
t = t falling
ay= 9.8
Vyi = value
yi = 0
Vyf = value
yf= 0
yi,falling = -yf,thrown
T.Merfat Al-Zumia
The equation for motion in one dimension:
In x-axis
Xi ≠ 0
(
Thrown up
a = -g
Vyf = 0 yi = 0
Freely falling
a = g Vyi = 0 yf = 0
)
(
)
)
Also
In x-axis But
Xi = 0
(
)
(
)
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Example (2.12):
A stone thrown from the top of a building
is given an initial velocity of 20.0 m/s
straight upward. The building is 50.0 m
high, and the stone just misses the edge of
the roof on its way down, as shown. Using
tA = 0 as the time the stone leaves the
thrower’s hand at position (A)
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determine:
1- the time at which the stone
reaches its maximum height?
2-the maximum height
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3-the time at which the stone returns to the height from
which it was thrown,
4- the velocity of the stone at this instant
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6- the velocity of the stone at t = 5.00s
5- the position of the stone at t = 5.00s
Best wishes , T.Merfat Al-Zumia,
T.Merfat Al-Zumia