Chapter 2 Unit Vector

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Chapter 2Unit Vector

• Vector addition by components
• Vector addition by graphing

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Displacement (Vector)
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Speed Velocity

• Average and Instantaneous
• Average Speed
• Distance/ Elapsed time
• (How fast it moves, but no direction)
• Units and dimensions

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Average Velocity
a)
m/s
b)
m/s
Time to go to Florida from Atlanta
Average Velocity?
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Instantaneous Velocity

• Smaller and smaller gives smaller and
  smaller
• As ? 0 it gives the limiting value
• (In general instantaneous values are given unless
  noted otherwise.)
• What if changes ( push the accelerator
  pedal)

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Acceleration

• Rate of change of velocity is acceleration

Average acceleration
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Equation of Kinematics

• Constant acceleration (straight line- motion)
• Say X0 and t0 0

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Constant acceleration
Velocity increases
What is the average velocity?
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If starts at
and with constant acceleration.
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If you do not know t
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Summary
Table 2.1 Equations of Kinematics for Constant
Acceleration
Equation Variables Variables Variables
number Equation x a v v0 t
(2.4) vv0at -----
(2.7) x(1/2)(v0v)t -----
(2.8) xv0t(1/2)at2 -----
(2.9) v2v022ax -----
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Application of the Equations

• Draw the situation
• Pick ve and ve direction and co-ordinates
• Write the known values (with signs)
• Look for hidden information
• Enough information, what is unknown?
• If two objects, are they connected?

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Free FALL
g 9.80 m/s2 or 32.2 ft/s2 (on earth surface)
Gallileos test
Leaning tower (better on moon)
Air filled tube
Evacuated tube
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Example 10 A Falling Stone
After 3.0 sec what is the displacement
Known v00 ag (i.e.) t3.0
yv0t(1/2)at2 vv0at v2v022ay
Which eqn to use?
v00 m/s t3.0 sec
yv0t(1/2)at2 0(1/2)(-9.8)9(-44.1m)
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What is the velocity after 3.0 sec?
vv0at v2v022ay
Can use both----why? Which is easier?
v -9.8m/s23sec -29.4ms -ve sign indicate
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Example 12 How high the coin reach, if it is
tossed up with an initial speed of 5.00 m/s?
At the highest point the velocity should be zero.
v05.0 m/s, a-9.8 m/s2 v0.0 m/s, y?
v2v022ay
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How long will it take to reach the top ?
v0, v05, a -9.8
Which eqn ?
How much time it takes to come to the same level
as it was thrown ?
v05.0 a-9.8 y0 t?
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t0 or
i.e.t0 or
Why two values? Explain each answer.
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symmetry
No air resistanceacceleration is the same for
the motion in both
directions
Velocity vector changes continuously Acceleration
does not. Time to reach the highest point
time to reach the ground At
same height, speed is the same.

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Bullet reaches the ground with the same speed
irrespective of which direction it was fired
first.
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Show the speed of the bullet when it comes down
to the same height. What is the velocity if shot
up with initial velocity 30 m/s under gravity
when it reaches the same level? Known
information v030, y0,
Lets say
150 m
a-g
Which eqn ?
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Harder way How high it reaches before changing
direction? v0 30, a-g, v?(hidden
information) y?(needs to
find)
Which eqn ?
v2v022ay 0(30)22(-9.8)y y(3030)/(29.8)4
5.92 m
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For the downward motion,
v2v022ay 02(9.8)(3030)/(29.8)
(why ve ?) 3030v30 m/s (again /-)
Find the time to reach max height and then find
the speed when it reaches the same level.
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What is the velocity of the bullet when it
reaches the ground if we start from the
mountain? v? v0 30, a-9.8, y-150
(why ve ?)
v2v022ayv2(30)22(-9.8)(-150)
(30)22.94103 3.84103 v61.97 m/s
(which sign?)
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If you drop down with 30 m/s from the mountain,
what is the speed? v0 -30, g-9.8, y-150
v2v022ay (30)22(-9.8)(-150)
(30)22.94103 3.84103 v61.97 m/s
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Example
tt
t0 v0
2v0
vv0at 0v0at
v2v0at1 02v0at1
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Answer-- 2t, 4y
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Graphical analysis
x0 and t0
Slope
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Example 16 Bicycle Trip
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1st segment
2nd segment
3rd segment
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If velocity changes (i.e. acceleration)
(assume v00 for simplicity)
Slope of the tangent
Instantaneous velocity
vv0at
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Conceptual questions 2
REASONING AND SOLUTION The buses do not have
equal velocities. Velocity is a vector, with both
magnitude and direction. In order for two
vectors to be equal, they must have the same
magnitude and the same direction. The direction
of the velocity of each bus points in the
direction of motion of the bus. Thus, the
directions of the velocities of the buses are
different. Therefore, the velocities are not
equal, even though the speeds are the same.
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Conceptual question 4
REASONING AND SOLUTION Consider the four
traffic lights 1, 2, 3 and 4 shown below. Let
the distance between lights 1 and 2 be x12, the
distance between lights 2 and 3 be x23, and the
distance between lights 3 and 4 be x34.
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The lights can be timed so that if a car travels
with a constant speed v, red lights can be
avoided in the following way. Suppose that at
time t 0 s, light 1 turns green while the rest
are red. Light 2 must then turn green in a time
t12, where t12  x12/v. Light 3 must turn green
in a time t23 after light 2 turns green, where
t23  x23/v. Likewise, light 4 must turn green
in a time t34 after light 3 turns green, where
t34  x34/v. Note that the timing of traffic
lights is more complicated than indicated here
when groups of cars are stopped at light 1. Then
the acceleration of the cars, the reaction time
of the drivers, and other factors must be
considered.
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Conceptual question 5
REASONING AND SOLUTION The velocity of the car
is a vector quantity with both magnitude and
direction The speed of the car is a scalar
quantity and has nothing to do with direction.
It is possible for a car to drive around a track
at constant speed. As the car drives around the
track, however, the car must change direction.
Therefore, the direction of the velocity changes,
and the velocity cannot be constant. The
incorrect statement is (a).
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Conceptual question 14
REASONING AND SOLUTION The magnitude of the
muzzle velocity of the bullet can be found (to a
very good approximation) by solving Equation 2.9,
With v0 0 m/s that is
where a is the acceleration of the bullet and x
is the distance traveled by the bullet before it
leaves the barrel of the gun (i.e., the length of
the barrel).
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Since the muzzle velocity of the rifle with the
shorter barrel is greater than the muzzle
velocity of the rifle with the longer barrel, the
product ax must be greater for the bullet in the
rifle with the shorter barrel. But x is smaller
for the rifle with the shorter barrel, thus the
acceleration of the bullet must be larger in the
rifle with the shorter barrel.
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Problem 4
REASONING The distance traveled by the Space
Shuttle is equal to its speed multiplied by the
time. The number of football fields is equal to
this distance divided by the length L of one
football field.
SOLUTION The number of football fields is
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Problem 8
REASONING AND SOLUTION Let west be the positive
direction. The average velocity of the
backpacker is
.
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Combining these equations and solving for xe
(suppressing the units) gives
The distance traveled is the magnitude of xe , or
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Problem 8 solution (2)
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t1t2t
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Problem 10
Reasoning The definition of
average velocity is given by Equation 2.2 as
Average velocity  Displacement/(Elapsed time).
The displacement in this expression is the total
displacement, which is the sum of the
displacements for each part of the trip.
Displacement is a vector quantity, and we must be
careful to account for the fact that the
displacement in the first part of the trip is
north, while the displacement in the second part
is south.
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Solution According to Equation 2.2, the
displacement for each part of the trip is the
average velocity for that part times the
corresponding elapsed time. Designating north as
the positive direction, we find for the total
displacement that
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Where tNorth and tSouth denote, respectively,
the times for each part of the trip. Note that
the minus sign indicates a direction due south.
Noting that the total elapsed time is tNorth
tSouth we can use Equation 2.2 to find the
average velocity for the entire trip as follows
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But
Therefore, we have that
The plus sign indicates that the average velocity
for the entire trip points north.
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Problem 19
REASONING AND SOLUTION
xv0t(1/2)at2
Uniform acceleration, starts from rest.
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Problem 23
REASONING We know the initial and final
velocities of the blood, as well as its
displacement. Therefore, Equation 2.9
can be used to find the acceleration of the
blood. The time it takes for the blood to reach
it final velocity can be found by using Equation
2.7
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SOLUTION a.) The acceleration of the blood is
b.) The time it takes for the blood, starting
from 0 cm/s, to reach a final velocity of 26
cm/s is
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Problem 28
REASONING AND SOLUTION The speed of the car at
the end of the first (402 m) phase can be
obtained as follows v12 vo2 2a1x1
The speed after the second phase (3.50 ? 102 m)
can be obtained in a similar fashion. v22 v022
2a2x2
v296.9 m/s
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Problem 39
REASONING The initial velocity and the elapsed
time are given in the problem. Since the rock
returns to the same place from which it was
thrown, its displacement is zero (y 0 m). Using
this information, we can employ Equation 2.8
yv0t(1/2)at2 to determine the acceleration a
due to gravity.
SOLUTION Solving Equation 2.8 for the
acceleration yields
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Problem 44
REASONING AND SOLUTION
a.
The minus is chosen, since the diver is now
moving down. Hence, v-7.9 m/s.
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b. The diver's velocity is zero at his highest
point. The position of the diver relative to the
board is
The position above the water is 3.0 m 0.17 m
3.2 m/s
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Problem 48
t
t
t00
v025.0 m/s
t01.2 sec
v0?
vv0at 0v0-9.8(t-1.2) v0 9.8(t-1.2)
9.8(2.55-1.2) 9.81.3513.23 m/s
vv0at 025-9.8t t25/9.82.55 sec
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Problem 51
v00
75 m
What is the time for stone to hit the water(log)?
5.0 m/s
x?
If the log is moving with constant velocity
during this time, the log should have moved.
yv0t(1/2)at2 (-75)(1/2)(-9.8)t2 t2275/9.815
.3 sec
xvt53.919.5 m
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Problem 55
REASONING AND SOLUTION The balls pass at a time
t when both are at a position y above the ground.
Applying Equation 2.8 to the ball that is
dropped from rest, we have
(1)
Note that we have taken into account the fact
that y  24 m when t  0 s in Equation (1). For
the second ball that is thrown straight upward,
Equating Equations (1) and (2) for y yields
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Thus, the two balls pass at a time t, where
The initial speed v02 of the second ball is
exactly the same as that with which the first
ball hits the ground. To find the speed with
which the first ball hits the ground, we take
upward as the positive direction and use Equation
2.9 v2v022ay . Since the first ball is dropped
from rest, we find that
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Thus, the balls pass after a time
At a time t  1.11 s, the position of the first
ball according to Equation (1) is
which is 6.0m below the top of the cliff.