Transcript Slide 1

Ch.3 Kinemtics In 2-D
Displacement
 r  r  r0
Displacement vector is in the
x-y plane (not in x or y
direction)
r  r0 r
v

t  t0
t
r
v = limit
t
t  0
1
Velocity Components In 2-D
Tangent to the path
(instantaneous velocity)
Average acceleration
v  v0 v
a

t  t0
t
Instantaneous Acceleration
a = limit
t  0
v
t
2
Motion Only In X Direction
x – motion equations
x
ax
vx
v0x
t
vx= v0x+axt
(3.3a)
x=(1/2)(v0x+vx)t
(3.4a)
x=v0xt+(1/2)axt2
(3.5a)
vx2=v0x2+2axx
(3.6a)
3
Motion In Y Direction
y component
y
ay
vy
v0y
t
vy= v0y+ayt
(3.3b)
y=(1/2)(v0y+vy)t
(3.4b)
y=v0yt+(1/2)ayt2
(3.5b)
vy2=v0y2+2ayy
(3.6b)
4
2 D Motion
5
x Component
x
Variable
Displacement
y Component
y
ax
Acceleration
ay
vx
Final velocity
vy
v0x
Initial velocity
v0y
Elapsed time
t
t
vx= v0x+axt
(3.3a)
vy= v0y+ayt
(3.3b)
x=(1/2)(v0x+vx)t
(3.4a)
y=(1/2)(v0y+vy)t (3.4b)
x=v0xt+(1/2)axt2
(3.5a)
y=v0yt+(1/2)ayt2
(3.5b)
vx2=v0x2+2axx
(3.6a)
vy2=v0y2+2ayy
(3.6b)
6
Example 1
x-Direction Data
x
ax
vx
? +24m/s ?
2
y-Direction Data
v0x
t
y
ay
vy
+22m/s
7s
?
12m/s2
?
v0y
14 m/s
t
7s
1 2
1
x  v0t  axt  (22 m / s )( 7 s )  (24 m / s 2 )( 7  7 s 2 )  740 m
2
2
vx  v0 x  axt  (22m / s)  (24m / s2 )  (7s)  190m / s
7
1
1
2
y  v0 y t  a y t  (14 m / s)  (7 s)  (12 m / s 2 )( 7  7 s 2 )  390 m
2
2
vy  v0 y  ayt  14m / s  (12m / s 2 )  7s  98m / s
v  v x  v y  (190)  (98)  210m / s
2
2
2
2
8
Projectile Motion
• 2-D motion
motion under
undergravity
gravity
• What
What isisthe
the acceleration
accelerationininXXdirection
direction
? ?
• What
What isisthe
theacceleration
accelerationininy ydirection
direction
? ?
ax= 0
ay = - 9.8 m/s2
Garden Horse – watering a plant
a Plane dropping a bomb
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Example 2: Falling Package
The time to hit the ground ?
y
ay
-1050 -9.8
2(1050)
t
 14.6s
 9.8
vy v0y
?
0
t
?
y = v0y t +(1/2) ay t2 = ½ (-9.8) t2
i.e., -1050 = ½ (-9.8) t2
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Velocity Of The Package When It
Hits The Ground ?
vx = v0x + ax t = v0x = 115 m/s
vy = v0y + ay t = (-9.8 m/s2) 14.6 s
= -143 m/s (Why Negative ?)
v2 = (115)2+ (-143)2
v = 184 m/s
vx
t an 
θ
vy
R
vx
vy
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Shoot A Bullet Straight Up In A
Moving Car
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Example 5: Height Of A Kickoff
v0x = v0 cosθ = 22 m/s cos 40.0o = 17m/s
For y direction
y
ay
H=? -9.8
v0y = v0 sinθ = 22 m/s sin 40.0o =14m/s
v0y
vy
t
14
0
?
v2 = v02 + 2ay
0 = (14)2 = 2*(-9.8)H
H = 14*14 /(2*9.8)= 10 m
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Time Of Flight ( Time In The Air)
Time until it hits the ground
y = v0yt + (½) ay t2
0 = 14 t + (½) (-9.8) t2
i.e., t(14 -4.9t) = 0 sec
t = 0 sec, or t= 14/4.9 = 2.9 sec
(why two answers?)
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Time To Reach The Maximum Height
y
ay
v0y
H=? -9.8 14
vy
t
0
?
vy = v0y +ay t
0 = 14 -9.8t
t = 14/9.8 = 1.45 sec
2t = 2.9 (time of flight) sec
Symmetry: time of flight = twice the time to reach the top --- Why?
RANGE vx
R = v0x t + (½ )ax t2
= 17 *2.9 = 49 m
t an 
θ
vy
R
vx
vy
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Same Height Same Speed
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Relative Velocity
vPG = vPT + vTG
= 2.0 m/s + 9.0 m/s
= 11.0 m/s
vPG = Velocity of Passenger relative to Ground
(vPG = - vGP)
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Check your understanding 4
Three
moving
aa straight
Three cars
Three
cars
carsA,
A,
A,B,
B,
B,C,
C,
C,are
are
are
moving
movingalong
along
along
straight
a
line,
are
line, Relative
straight
Relative
line, velocities
velocities
Relative velocities
are given.
given.are given.
vvAB
AB
1.
1.
2.
2.
3.
3.
4.
4.
vvAC
AC
AC
70
??m/s
40
m/s
70
m/s 40
40m/s
m/s
m/s
???m/s 50
50
m/s
30
50m/s
m/s
m/s
60
20
m/s
60 m/s
m/s 20
20m/s
m/s
m/s
-50
???m/s
-50 m/s
m/s -60
vvCB
CB
CB
CB
30m/s
m/s
30
30
m/s
-20
-20m/s
m/s
m/s
-20
-20
m/s
40
40
???m/s
m/s
10
10m/s
m/s
m/s
10
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Crossing A River
vBS = vBW + vWS
R2 = (vBW)2 + (vWS)2
vWS
vBW
vBS
R
vWS
tan  
v BW
 vWS
  t an 
 v BW
1




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10. Crossing A River
Width of the river = 1800 m
vBS = vBW + vWS
= 4.0 m/s + 2.0 m/s
= 4.5 m/s
v BW
t an 
 2.0
vWS
θ = tan-1(2) = 63º
width
tim e to cross 
v BW
1800

 450 s
4.0
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11. Approaching Intersection
vAG = 25.0 m/s
vBG = 15.8 m/s
vAB = vAG + vGB
vAB = 25.0 m/s
v AB  (25)  (15.8 )  29.6 m / s
2
2
+ 15.8 m/s
tanθ = 15.8/25
Θ = tan-1(15.8/25) = 32.4º
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Raindrops On The Car
vRC = vRG + vGC
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Concepts: Circus Clowns
No air resistance
vox = 4.6 m/s
v0y = 10.0 m/s
v0  (4.6)  (10 )
2
2
Θ = tan-1(v0y/v0x)
= tan-1(10/4.6)
= 65º
23
Conceptual Question 2
REASONING AND SOLUTION An object thrown upward at
an angle  will follow the trajectory shown below. Its
acceleration is that due to gravity, and, therefore, always
points downward. The acceleration is denoted by ay in the
figure. In general, the velocity of the object has two
components, vx and vy. Since ax = 0, vx always equals its initial
value. The y component of the velocity, vy, decreases as the
object rises, drops to zero when the object is at its highest
point, and then increases in magnitude as the object falls
downward.
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a.) Since vy = 0 when the object is at its highest point, the velocity
of the object points only in the x direction. As suggested in the
figure below, the acceleration will be perpendicular to the
velocity when the object is at its highest point and vy = 0.
ay
ay
v0 y
v0 x
ay

v0 x
v0 x

–v0 y
vf
b.) In order for the velocity and acceleration to be parallel, the
x component of the velocity would have to drop to zero.
However, vx always remains equal to its initial value; therefore,
the velocity and the acceleration can never be parallel.
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Conceptual Question 4
REASONING AND SOLUTION If a baseball were
pitched on the moon, it would still fall downwards as
it travels toward the batter. However the acceleration
due to gravity on the moon is roughly 6 times less
than that on earth. Thus, in the time it takes to reach
the batter, the ball will not fall as far vertically on the
moon as it does on earth. Therefore, the pitcher's
mound on the moon would be at a lower height than
it is on earth.
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Conceptual Questions 13
REASONING AND SOLUTION Since the plastic
bottle moves with the current, the passenger is
estimating the velocity of the boat relative to the
water. Therefore, the passenger cannot conclude
that the boat is moving at 5 m/s with respect to the
shore.
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Conceptual Questions 16
REASONING AND SOLUTION The time required for
any given swimmer to cross the river is equal to the
width of the river divided by the magnitude of the
component of the velocity that is parallel to the width
of the river. All three swimmers can swim equally fast
relative to the water; however, all three swim at
different angles relative to the current. Since swimmer
A heads straight across the width of the river, swimmer
A will have the largest velocity component parallel to
the width of the river; therefore, swimmer A crosses the
river in the least time.
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Problem 4
REASONING AND SOLUTION The increase in
altitude represents vy = 6.80 m/s. The movement
of the shadow represents vx = 15.5 m/s. The
magnitude of the glider's velocity is therefore
v
vx2
 v2y
 15.5 m/s    6.80 m/s   16.9 m/s
2
2
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Problem 7
REASONING Trigonometry indicates that the x and
y components of the dolphin’s velocity are related to
the launch angle  according to tan  = vy /vx.
SOLUTION Using trigonometry, we find that the y
component of the dolphin’s velocity is
vy  vx tan  vx tan35  (7.7m / s ) tan35  5.4m / s
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Problem 14
55o
15 m/s
vx=?
REASONING
The vertical component of the ball’s
velocity v0 changes as the ball approaches the opposing
player. It changes due to the acceleration of gravity.
However, the horizontal component does not change,
assuming that air resistance can be neglected.
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Hence, the horizontal component of the ball’s velocity
when the opposing player fields the ball is the same as
it was initially.
SOLUTION Using trigonometry, we find that the
horizontal component is
vx  v0 cos  (15m / s ) cos55  8.6m / s
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Problem 16
v0y=0
B
v0=11m/s
A
h=?
65O
REASONING AND SOLUTION Using vy = 0 and
voy = vo sin  = (11 m/s) sin 65 = 1.0  101 m/s
and vy2 = voy2 + 2ayy, we have
y
v02y
2a y

 1.0 10 m/s 
1
2
2  9.80 m/s 2 
 5.1 m
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Problem 27
34
REASONING AND SOLUTION The time of flight of the
motorcycle is given by
y=0,
y=v0
v0y=33.5sin18, t=?
t+(1/2)at2
t=0, or
t
1
0  v0 sin  0t   9.8t 2
2
2v0 sin 0
g
2  33.5 m/s  sin18.0

 2.11 s
2
9.80 m/s
The horizontal distance traveled by the motorcycle is then
x = vo cos o t = (33.5 m/s)(cos18.0°)(2.11 s) = 67.2 m
The daredevil can jump over (67.2 m)/(2.74 m/bus) = 24.5
buses. In even numbers, this means 24 buses
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Problem 40
36
REASONING Using the data given in the problem, we can
find the maximum flight time t of the ball using Equation 3.5b
y  v0 yt 
1
a yt 2
2
Once the flight time is known, we can use the definition of
average velocity to find the minimum speed required to
cover the distance x in that time.
SOLUTION Equation 3.5b is quadratic in t and can be
solved for t using the quadratic formula. According to
Equation 3.5b, the maximum flight time is (with upward
taken as the positive direction)
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1
 v0 y  v  4 a y  y 
2

v

v
0y
0 y  2a y y
2
t

ay
1
2 a y
2
2
0y

 15.0m / s sin 50.0 
15.0m / s sin 50.02  2 9.8m / s 2 2.1m
 9.8m / s 2
 0.200s and 2.145s
where the first root corresponds to the time required for the
ball to reach a vertical displacement of y  2.10 m as it travels
upward, and the second root corresponds to the time required
for the ball to have a vertical displacement of y  2.10 m as
the ball travels upward and then downward. The desired
flight time t is 2.145 s.
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During the 2.145 s, the horizontal distance traveled by the ball is
x  v x t  (v0 cos ) t  (15.0m / s) cos50.0 (2.145s)  20.68m
Thus, the opponent must move 20.68 m –10.0 m  10.68 m
in 2.145 s – 0.30 s  1.845 s . The opponent must,
therefore, move with a minimum average speed of
v min
10.68 m

 5.79 m / s
1.845 s
39
Problem 41
40
REASONING AND SOLUTION In the absence of air
resistance, the bullet exhibits projectile motion. The x
component of the motion has zero acceleration while the y
component of the motion is subject to the acceleration due
to gravity. The horizontal distance traveled by the bullet is
given by Equation 3.5a (with a  0 ):
x
x  v0 xt  (v0 cos )t
with t equal to the time required for the bullet to
reach the target. The time t can be found by
considering the vertical motion. From Equation 3.3b
v y  v0 y  a yt
41
When the bullet reaches the target, v y  v0 y . Assuming
that up and to the right are the positive directions, we
have
 2v sin  
2v0 y 2v0 sin 
0

t

and
x  (v0 cos  ) 
ay


ay


ay
Using the fact that 2sin  cos   sin 2 , we have
x
2v02 cos  sin 
ay

v02 sin 2
ay
Thus, we find that
sin 2  
x ay
v02
2 2
(91.4
m)
(–9.80
m/s

91.4m 9.8m / s  )
3 3
 

4.91

10

4
.
91

10
22
(427mm/s)
(427
/ s)
42
and
2  0.281 or 2  180.000  0.281  179.719
Therefore
  0.141 and 89.860
43
Problem 48
REASONING Since car A is moving faster, it will
eventually catch up with car B. Each car is traveling
at a constant velocity, so the time t it takes for A to
catch up with B is equal to the displacement between
the two cars (x = +186 m) divided by the velocity vAB
of A relative to B. (If the relative velocity were zero, A
would never catch up with B). We can find the
velocity of A relative to B by using the subscripting
technique developed in Section 3.4 of the text.
24.4 m/s
18.6 m/s
B
A
186 m
44
vAB = velocity of car A relative to car B
vAG = velocity of car A relative to the Ground = +24.4 m/s
vBG = velocity of car B relative to the Ground = +18.6 m/s
We have chosen the positive direction for the
displacement and velocities to be the direction in
which the cars are moving. The velocities are related
by
vAB = vAG + vGB
45
SOLUTION The velocity of car A relative to car B is
vAB = vAG + vGB = +24.4 m/s + (18.6 m/s) = +5.8 m/s,
where we have used the fact that vGB =  vBG = 18.6
m/s. The time it takes for car A to catch car B is
x
186 m
t

 32.1 s
v AB
+5.8 m/s
46
Problem 53
REASONING Let vHB represent the velocity of the
hawk relative to the balloon and vBG represent the
velocity of the balloon relative to the ground. Then, as
indicated by Equation 3.7, the velocity of the hawk
relative to the ground is vHG  vHB  vBG. Since the
vectors vHB and vBG are at right angles to each other,
the vector addition can be carried out using the
Pythagorean theorem. (B: Balloon, H: Hawk, G: Ground)
SOLUTION Using the drawing at the right, we have
from the Pythagorean theorem
47
B: Balloon
H: Hawk
G: Ground
NO RTH
vHG

vB G=6.0
vHB=2.0
EAST
2
2
v HG  v HB
 v BG
 (2.0 m / s) 2  (6.0 m / s) 2  6.3 m / s
The angle  is
vHB
1 2.0m / s
  tan
 tan
 18, north of east
vBG
6.0m / s
1
48
Problem 56
49
REASONING AND SOLUTION
The velocity of the
raindrops relative to the train is given by vRT = vRG + vGT
where vRG is the velocity of the raindrops relative to the ground
and vGT is the velocity of the ground relative to the train
Since the train moves horizontally, and the rain
falls vertically, the velocity vectors are related
as shown in the figure at the right. Then
vRT
25°
vRG
vGT
vGT = vRG tan  = (5.0 m/s) (tan 25°) = 2.3 m/s
The train is moving at a speed of
2.3 m/s
50