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0( )v t v at
Recap: Constant Acceleration
0 0( )
tx x v t dt
Area under the functionv(t).
2
Recap: Constant Acceleration
ad
t
v
d
0v av t
210 0 2x x v t at
2 20 02 ( )v xav x
3
Recap: Acceleration due to Gravity (Free Fall)
In the absence of air resistance all objects fall with the same constant acceleration of about
g = 9.8 m/s2
near the Earth’s surface.
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Recap: Example
A ball is thrown upwards at 5 m/s, relative to the ground, from a height of 2 m.
We need to choose a coordinate system.
2 m
5 m/s
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Recap: Example
Let’s measuretime from whenthe ball is launched.
This defines t = 0.
Let’s choose y = 0 to be ground leveland up to be the positive y direction.
y0 = 2 m
v0 = 5 m/s
y
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Recap: Example
1. How high above the ground will the ball reach?
with a = –gand v = 0.
2 20 02 ( )v yav y use
y0 = 2 m
v0 = 5 m/s
y
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Recap: Example
210 0 2y v t ay t Use
2. How long does it take the ball to reach the ground?
y0 = 2 m
v0 = 5 m/s
y
with a = –gand y = 0.
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Recap: Example
2 20 02 ( )yav v y Use
3. At what speed does the ball hit the ground?
y0 = 2 m
v0 = 5 m/s
y
with a = –gand y = 0.
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Demonstrate an understanding that two dimensional motion is analyzed by resolving vectors into separate components.
Topic: Motion in two or three dimension
Suppose a particle follows a path in the xy plane. At time t1, the particle is at point P; and at time t2, it is
at point P2.
The vector r1is the position vector of the particle at time t1 (it represents the displacement of the particle from the origin of the coordinate system); and r2 is the position vector at the time t2.
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In one dimension we defined displacement as the change in position of the particle.
In two or three dimensions, the displacement vector is defined as the vector representing the change in position of the particle, it is the vector r.
r = r2 – r1
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Unit VectorsIt is convenient to define unit vectorsparallel to the x, y and zaxes, respectively.
ˆˆ ˆx y zA A i A j A k
ˆˆ ˆ, ,i j k
Then, we can write avector A as follows:
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Position vector in three dimensions
-vector that goes from the origin of the coordinate system to a certain point in space.
written as:
.ˆˆˆ
,ˆˆˆ
kzjyixr
kzjyixr
r = (x2-x1)i + (y2-y1)j + (z2-z1)k
Average and Instantaneous Velocity
kt
zzj
t
yyi
t
xxv
)()(ˆ)( 121212
_
Average velocity is the displacement of an object over some time interval.
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Instantaneous velocity
The components of are :
td
rdv
v
td
zdvand,
td
ydv,
td
xdv zyx
Is the limit of v as t approaches zero and can also be written as the derivative
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kdt
dzj
dt
dyi
dt
dxv
Which can be expanded to:
and then rewritten or simplified to
v = vxi + vyj + vzk.
Particle’s Path vs Velocity
Displacement: The velocity vector
The direction of the instantaneous velocity of a particle is always tangent to the particle’s path at the particle’s position.
Average acceleration
t
v
tt
vvaav
12
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The Acceleration vector
dt
vd
t
va
t
0lim
Instantaneous acceleration is equal to the instantaneous rate of change of velocity with time
• Parallel and Perpendicular Components of Acceleration
vvv
12
vvv
||
dt
vd
dt
vdt
v
t
v
t
va
ttt
||
0
||
00limlimlim
Review kdt
dzj
dt
dyi
dt
dx
dt
rdv ˆˆˆ
kdt
zdj
dt
ydi
dt
xd
kdt
dvj
dt
dvi
dt
dv
dt
vda zyx
ˆˆˆ
ˆˆˆ
2
2
2
2
2
2
By differentiation
When the acceleration vector, a, is parallel to the velocity vector,v, its effect is to change the speed (magnitude of the velocity) of the object but NOT the direction.
In most cases a has a component that is parallel to the velocity vector and another component that is perpendicular to the velocity vector and these statements still apply to the individual components.
When the acceleration vector, a, is perpendicular to the velocity vector,v, its effect is to change the direction but NOT the speed (magnitude of velocity).
Speed up or slow down
If the velocity and acceleration components along a given axis have the same sign then they are in the same direction. In this case, the object will speed up.
If the acceleration and velocity components have opposite signs, then they are in opposite directions. Under these conditions, the object will slow down.
How to solve two-dimensional motion problem?
One ball is released from rest at the same instant that another ball is shot horizontally to the right
The horizontal and vertical motions (at right angles to each other) are independent, and the path of such a motion can be found by combining its horizontal and vertical position components.
By Galileo
Projectile Motion
A particle moves in a vertical plane with some initial velocity but its acceleration is always the free-fall acceleration g, which is downward. Such a particle is called a projectile and its motion is called projectile motion.
Properties of Projectile Motion
The Horizontal Motion:•no acceleration •velocity vx remains unchanged from its initial value throughout the motion•The horizontal range R is maximum for a launch angle of 45°
The vertical Motion:•Constant acceleration g •velocity vy=0 at the highest point.
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In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. A projectile with an initial velocity can be written as
The horizontal motion has zero acceleration, and the vertical motion has a constant downward acceleration of - g.
jvivv y0x00
0v
00000x0 sinvvandcosvv y
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The range R is the horizontal distance the projectile has traveled when it returns to its launch height.
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Examples of projectile motion
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0r
r
Projectile Motion under Constant Acceleration
Coordinate system: î points to the right, ĵ points upwards
ji
30
0r
) ˆ0 (i ja g
210 2v t at
r
22
0
100 vr r
v v
t
t
t
a
a
R = Range
Impact point
Projectile Motion under Constant Acceleration
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Projectile Motion under Constant Acceleration
0 0
0 0
cos
sinx
y
v v
v v
Strategy: split motion into x and y components.
210 0 2
0 0
y
x
y y v t gt
x x v t
R = Range R = x - x0
h = y - y0
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Horizontal motion
No acceleration
tvxx x00
t)cosv(xx 000
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Vertical motion (Equations of Motion ):-
)yy(g2)sinv(v
tgsinvv
tg2
1t)sinv(
tg2
1tvyy
02
002y
00y
,200
2y00
1 )
2 )
3 )
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The equation of the path
In this equation, x0 = 0 and y0 = 0. The path, or trajectory, is a parabola. The angle is between and the + direction.
)trajectory()cosv(2
xgx)(tany
200
2
0
0 0v
x
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The horizontal range
This equation for R is only good if the final height equals the launch height. We have used the relationsin 2 = 2 sin cos .
The range is a maximum when = 45o
t)cosv(R 00
To find t = time of flight, y - y0 = 0 means that :
0
20
00
20
200
2sing
vR
cossing
v2R
tg2
1t)sinv(0
0 0 0
0
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Special case: y = y0, i.e., h = 0
0 0
20
2
sin 2
x yv vR
g
v
g
R
y0 y(t)
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Sample Problem
In Fig. 4-15, a rescue plane flies at 198 km/h (= 55.0 m/s) and a constant elevation of 500 m toward a point directly over a boating accident victim struggling in the water. The pilot wants to release a rescue capsule so that it hits the water very close to the victim.
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(a) What should be the angle of the pilot's line of sight to the victim when the release is made?
Solving for t, we find t = ± 10.1 s (take the positive root).
Solutionh
xtan 1
t)cosv(xx 000
2000 tg
2
1t)sinv(yy
)s1.10()0(cos)s/m0.55(0x m5.555x
48m500
m5.555tan 1
22 t)s/m8.9(2
1t)0(sin)s/m0.55(m500
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(b) As the capsule reaches the water, what is its velocity in unit-vector notation and as a magnitude and an angle?
When the capsule reaches the water,
s/m0.55)0(cos)s/m0.55(cosvv 00x
s/m0.99
)s1.10()s/m8.9()0(sin)s/m0.55(
tgsinvv2
00y
j)s/m0.99(i)s/m0.55(v
61ands/m113v
v
40
Sample Problem 2Figure 4-16 shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v0 = 82 m/s.
(a) At what angle from the horizontal must a ball be fired to hit the ship?
0
41
SOLUTION:
Which gives
There are two solutions
0
20 2sing
vR
816.0sin
)s/m82(
)m560)(s/m8.9(sin
v
Rgsin2
1
2
21
20
10
63)3.125(2
1
27)7.54(2
1
0
0
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(b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannonballs?
SOLUTION: Maximum range is :-
The maximum range is 690m. Beyond that distance, the ship is safe from the cannon.
.m690m686
)45x2(sins/m8.9
)s/m82(2sin
g
vR
2
0
20
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Sample Problem 3Figure 4-17 illustrates the flight of Emanuel Zacchini over three Ferris wheels, located as shown and each 18 m high. Zacchini is launched with speed v0 = 26.5
m/s, at an angle = 53° up from the horizontal and with an initial height of 3.0 m above the ground. The net in which he is to land is at the same height.
(a) Does he clear the first Ferris wheel?
0
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SOLUTION
The equation of trajectory when x0 = 0 and y0 = 0 is given by :
Solving for y when x = 23m gives
Since he begins 3m off the ground, he clears the Ferris wheel by (23.3 – 18) = 5.3 m
200
2
0 )cosv(2
xgx)tan(y
m3.20
)53(cos)s/m5.26(2
)m23()s/m8.9()m23()53tan(
22
22
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(b) If he reaches his maximum height when he is over the middle Ferris wheel, what is his clearance above it?
SOLUTION:
At maximum height, vy is 0. Therefore,
and he clears the middle Ferris wheel by
(22.9 + 3.0 -18) m =7.9 m
0gy2)sinv(v 200
2y
m9.22)s/m8.9()2(
)53(sin)s/m5.26(
g2
)sinv(y
2
22200
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(c) How far from the cannon should the center of the net be positioned?
SOLUTION:
m69
)53(2sins/m8.9
)s/m5.26(2sin
g
vR
2
2
0
20
47
Uniform Circular Motion A particle is in uniform circular motion if it travels
around a circle at uniform speed. Although the speed is uniform, the particle is accelerating.
The acceleration is called a centripetal (center seeking) acceleration.
T is called the period of revolution.
).period(v
r2T
),onacceleratilcentripeta(r
va
2
48
sinvtd
xd,cosv
td
yd pp
.jtd
xd
r
vi
td
yd
r
v
td
vda
.jr
xvi
r
yvv
.j)cosv(i)sinv(jvivv
pp
pp
yx
.jsinr
vicos
r
va
22
49
Thus, , which means that is directed along the radius r, pointing towards the circle’s center.
tancosr
v
sinrv
a
atan
,r
v)(sin)(cos
r
vaaa
2
2
x
y
222
22
y2
x
a
50
Sample Problem 4-9
“Top gun” pilots have long worried about taking a turn too tightly. As a pilot's body undergoes centripetal acceleration, with the head toward the center of curvature, the blood pressure in the brain decreases, leading to loss of brain function.
There are several warning signs to signal a pilot to ease up: when the centripetal acceleration is 2g or 3g, the pilot feels heavy. At about 4g, the pilot's vision switches to black and white and narrows to “tunnel vision.” If that acceleration is sustained or increased, vision ceases and, soon after, the pilot is unconscious—a condition known as g-LOC for “g-induced loss of consciousness.”
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What is the centripetal acceleration, in g units, of a pilot flying an F-22 at speed v = 2500 km/h (694 m/s) through a circular arc with radius of curvature r = 5.80 km?
SOLUTION:
If a pilot caught in a dogfight puts the aircraft into such a tight turn, the pilot goes into g-LOC almost immediately, with no warning signs to signal the danger.
g5.8s/m0.83m5800
)s/m694(
r
va 2
22
52
Relative Motion in One Dimension
The coordinate xPA of P as measured by A is equal to the coordinate xPB of P as measured by B plus the coordinate xBA of B as measured by A. Note that x is a vector in one dimension.
BAPBPA xxx
53
The velocity vPA of P as measured by A is
equal to the velocity vPB of P as measured by
B plus the velocity vBA of B as measured by A.
Note that v is a one dimensional vector. We have deleted the arrow on top.
BAPBPA
BAPBPA
vvv
),x(td
d)x(
td
d)x(
td
d
54
Observers on different frames of reference (that move at constant velocity relative to each other) will measure the same acceleration for a moving particle. Note that the acceleration is a one dimensional vector.
Because VBA is constant, the last term is zero.
)v(td
d)v(
td
d)v(
td
dBAPBPA
PBPA aa
55
Sample Problem For the situation of Fig. 4-20, Barbara's velocity relative to Alex is a constant vBA = 52 km/h and car P is
moving in the negative direction of the x axis.
(a) If Alex measures a constant velocity vPA = -78 km/h
for car P, what velocity vPB will Barbara measure?
SOLUTION:
h/km130v
h/km52vh/km78
vvv
PB
PB
BAPBPA
56
(b) If car P brakes to a stop relative to Alex (and thus the ground) in time t = 10 s at constant acceleration, what is its acceleration aPA relative to Alex?
2
0PA
s/m2.2
h/km6.3
s/m1
s10
)h/km78(0
t
vva
57
(c) What is the acceleration aPB of car P relative to
Barbara during the braking?
SOLUTION: To calculate the acceleration of car P relative to
Barbara, we must use the car's velocities relative to Barbara. The initial velocity of P relative to Barbara is vPB = -130 km/h. The final velocity of P relative to Barbara is -52 km/h (this is the velocity of the stopped car relative to the moving Barbara).
This result is reasonable because Alex and Barbara have a constant relative velocity, they must measure the same acceleration.
2
0PB
s/m2.2
h/km6.3
s/m1
s10
)h/km130(h/km52
t
vva
58
Relative Motion in Two Dimension
PBPA
BAPBPA
BAPBPA
aa
vvv
rrr
