1. Chapter 1 Motion in a straight line
Displacement vs Distance Average Velocity
INSTANTANEOUS VELOCITY
Acceleration
the acceleration of gravity and falling objects
10/19/2011
Norah Ali Almoneef

2. 1-2 Displacement vs Distance Average Velocity
Displacement is a vector that points from an object’s initial position to its final position
and has a magnitude that equals the shortest distance between the two positions.
_Only depends on the initial and final positions
Independent of actual paths between the initial and final positions
Distance is a scalar
Depends on the initial and final positions as well as the actual path between them
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Norah Ali Almoneef

3. Displacement
This type of x(t) plot shows the position of an object at any time, e.g.,
x (m)
Position at t=3 s, x(3) = 1 m 3
t (s) 4
Displacement between t=1 s and t=5 s
Dx = 1.0 m m = -1.0 m -3
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Norah Ali Almoneef

4. What is the distance traveled by the train ? Displacement =
Given the train’s initial position and its final position what is the displacement of the train?
What is the distance traveled by the train ?
Displacement =
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Norah Ali Almoneef

5. Example: A boy travels from D to A,A to B .B to C.C to D
Displacement from D to D ( which are initial and final points ) = 0
Distance traveled = = 24 m
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6. Example : Distance = 4 m + 3 m =7 m Displacement = 5 m 10/19/2011
Norah Ali Almoneef

7. Speed and Velocity
The average speed being the distance traveled divided by the time required to cover the distance:
How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m /s?
Distance = s x m / s = m
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Norah Ali Almoneef

8. Speed Speed can be defined in a couple of ways:
How fast something is moving
The distance covered in a certain amount of time
The rate of change of the position of an object
Units for speed are: miles / hour (mi/hr)
kilometers / hour (km/hr)
feet / second (ft/s)
This is the standard unit meters / second (m/s)
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Norah Ali Almoneef

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11. example
A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m. Determine the velocity of the particle when t = 4 s.
At t = 4 s,
the velocity = 3 (4)2 – 6(4) = 24 m/s
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Norah Ali Almoneef

12. example
What is
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13. example
From A to B
What is A B
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17. 1-3 INSTANTANEOUS VELOCITY
Instantaneous velocity – is how fast an object is moving at a particular instant.
The position of a particle moving on the x axis is given by
x = t – 2.1t2. What is its instantaneous velocity at t = 3.5 seconds?
v = – (2)(2.1)t
v = – (2)(2.1)(3.5) = -5.5 m/s
example
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Norah Ali Almoneef

18. 1 - 4 Acceleration Instantaneous acceleration
Acceleration: is a rate at which a velocity is changing.
Instantaneous acceleration
= dv / dt = d2 x / d t2
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Norah Ali Almoneef

19. Example
A car’s velocity at the top of a hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocity of 26 m/s. What is the acceleration of the car?
The car is increasing its velocity by 8 m/s for every second it is moving.
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Norah Ali Almoneef

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21. Instantaneous Acceleration
Suppose a particle is moving in a straight line so that its position is given by the relationship x = (2.10 m/s2)t m. Find its instantaneous acceleration at 5 seconds.
v = dx / dt = (3)(2.1)t2
a = dv / dt = (2)(3)(2.1)t at t= 5s
a = (2)(3)(2.1)(5) = 63 m/s2
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Norah Ali Almoneef

22. example
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23. example OR Total distance travelled
A bullet train starts from rest from a station and travels along a straight horizontal track towards another station. The graph in fig. shows how the speed of the train varies withtime over the whole journey. Determine: (a) the total distance covered by the train,
(b) the average speed of the train.
A ) Total distance travelled
Speed / ms-1
Time 40 2 12 16
OR Total distance travelled
= ‘area under the graph’
= (1/2)( )(40) = 520 m
Average speed = (total distance) / (total time ) = 520 / 16 = ms-1
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Norah Ali Almoneef

24. example
: a car is traveling 30 m/s and approaches 10 m from an intersection when the driver sees a pedestrian and slams on his brakes and decelerates at a rate of 50 m/s2.
(a) How long does it take the car to come to a stop?
(b) how far does the car travel before coming to a stop?
vf -vi= a t, where vo= 30 m/s, v = 0 m/s, and a = -50 m/s2
t = (0 -30)/(-50) = 0.6 s
Δx= vit + ½ a t2= (30)(0.6) + ½(-50)(0.6)2= = 9 m
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25. 1.5 finding the motion of an object
Equations of Kinematics for Constant Acceleration
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26.                                
example
  How long does it take a car going 30 m/sec to stop of it decelerates at 7 m/sec2?
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27. example
-  A car starting from rest attains a speed of 28 m/sec in 20 sec.  Find the acceleration of the car and the distance it travels in this time.
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29. 1. Velocity & acceleration are both vectors
1. Velocity & acceleration are both vectors. Are the velocity and the acceleration always in the same direction?
NO!!
If the object is slowing down, the acceleration vector is in the opposite direction of the velocity vector!
2. Velocity & acceleration are vectors.
Is it possible for an object to have a zero acceleration and a non-zero velocity?
YES!!
If the object is moving at a constant velocity, the acceleration vector is zero!
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Norah Ali Almoneef

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31. Examples :
1 ) What is the acceleration of a car that increased its speed from 10 m/s to 30 m/s in 4 seconds?
a = (30 m/s – 10 m/s) ÷ 4s
= 20 m/s ÷ 4s
= 5 m/s2
2)the same car now slows down back to 10 m/s in 5 seconds. What is his acceleration?
a = (10 m/s – 30 m/s) ÷ 5s
= (- 20 m/s) ÷ 5s
= - 4 m/s2
Means slowing down
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32. Graphical Analysis
*deduce from the shape of a speed-time graph when a body is:
(i) at rest
(ii) moving with uniform speed
(iii) moving with uniform acceleration
(iv) moving with non-uniform acceleration
Velocity
(i) at rest
(ii) moving with uniform speed
(iii) moving with uniform acceleration
(iv) moving with non-uniform acceleration
(ii)
(iii)
(iv)
(i)
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Time

33. example
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34. A bus stopped at a bus-stop for 10 seconds before accelerating to a velocity of 15 m/s in 4 seconds and then at a constant speed for the next 9 seconds. How does the graph look like? How far did the bus go in this 23 seconds?
Distance travelled in first 10 seconds is zero
Distance travelled in the next 4 seconds is
= ½ x 4 x 15 = 30 m
Distance travelled in the final 9 seconds is
= 9 x 15 = 135 m
Total distance travelled = 165 m
Velocity /m/s
(ii) 15
(iii)
(i) 10 14 23
Time/s
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36. 1.6 the acceleration of gravity and falling objects
Objects thrown straight up
The acceleration of a falling object is due to the force of gravity between the object and the earth.
Galileo showed that falling objects accelerate equally, neglecting air resistance.
Galileo found that all things fall at the same rate.
On the surface of the earth, in a vacuum, all objects accelerate towards the surface of the earth at 9.8 m/s2.
The acceleration of gravity (g) for objects in free fall at the earth's surface is 9.8 m/s2. ( down ward )
g actually changes as we move to higher altitudes
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37. For free fall Equations of Kinematics for Constant Acceleration
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39.   A ball is dropped from a tall building and strikes the ground 4 seconds later.  A ) what velocity does it strike the ground B ) what distance does it fall?
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40. g = -
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41. example
How high can a human throw a ball if he can throw it with initial velocity 90 m / h?.
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42. example
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43. example
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44. Notice in free fall
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45. Word clues to numbers for problem solving
“free-fall”  acceleration due to gravity
a=9.8m/s2, down
“at rest”  not moving
v=0
“dropped”  starts at rest and free-fall
vi=0 and a=9.81m/s2, down
“constant velocity”  no acceleration
a=0
“stops”  final velocity is zero
vf=0
10/19/2011
Norah Ali Almoneef

46. Summary 2. Time interval: 1.Displacement: 2. Average velocity:
3.Instantaneous velocity:
4. Average acceleration:
5.nstantaneous acceleration:
10/19/2011
Norah Ali Almoneef