### Power Point

```SCALARS
35 km long
35 mm wide
3 hours 35
minutes
35.8 kg
2000 kcal
20% efficiency
Vectors
5 km south
35 m/s south-west
45 N upward
9.8 m/s2 downward
Symbol for vectors:
Typed: v, a, F, x
Written:
v, a, F, x
Multiplying vectors by scalars
Dx = v × Dt
Vector x scalar = vector
Multiplying vectors by scalars
a
2a
-2a
1
a
2
Comparing vectors
Which vectors have the same magnitude?
Which vectors have the same direction?
Which arrows, if any, represent the same vector?
a
b
a+b
a+b
b+a
b+a
Subtracting vectors (+ negative)
a
a-b
a-b
b
b-a
b-a
Subtracting vectors (“fork”)
a
a-b
b
a-b
b-a
b-a
Construct and label a diagram that shows the vector sum
2A + B. Construct and label a second diagram that shows
B + 2A.
Construct and label a diagram that shows the vector sum A – B/2.
Construct and label a second diagram that shows B/2 - A.
algebraically
that are
perpendicular
Example
A person walks 5 km east and 3 km
south. What is the person’s total
displacement?
Hint: displacement is a vector
magnitude (km) and direction
(degrees)
1km
Dx = Dx1 + Dx2
Dx = x + x
2
1
2
2
Dx = 25 + 9
æ 3ö
q = tan ç ÷
è 5ø
-1
Relative Motion
• A person is sitting in a bus that is moving
north 2 m/s.
a)What is the passenger’s velocity relative to a
person standing on the sidewalk nearby?
b) What is the passenger’s velocity relative to
the person sitting next to him?
c) What is his speed relative to a car moving at 4
m/s in the opposite late towards the bus?
Relative motion
• A swimmer moves at 2 m/s in a pool.
a)What will be her speed relative to a person on
the river bank if she swims downstream and the
current is 5 m/s?
b) What will her speed be if she tries to swim
against the current?
• A boat is crossing a river moving south with a
speed of 4.0m/s relative to the current. The
current moves due east at a speed of 2.0 m/s
relative to the land.
a) How far across the river will the boat be in 2
seconds?
b) How far down the river the boat be in 2
seconds?
c) What will the boat’s total displacement be
(from the place where it started).
Relative motion
• A motorboat is crossing a 350-m wide river
moving 10 m/s relative to water. If current is 3
m/s east, how far down the stream will the
boat end up when it reaches the opposite
bank? What will the boat’s total displacement
be? What will the boat’s velocity be relative to
water?
• (more than one way to solve!)
are NOT
perpendicular
RESOLVING VECTORS
b
a
yacomponent
y
q
x component
a
x
ax = a × cosq
ay = a ×sinq
ax = a ×sin b
ay = a × cos b
NE
EN
How fast must a truck travel to stay
beneath an airplane that is moving 105
km/h at an angle of 25° to the ground?
vt = v p cos 25
vt = 105×cos25 = 95(km / h)
Practice finding components
•
•
•
•
35 km/h 20o NE
120 m 35o WN
3.2 m/s2 40o SW
65 km 70o ES
X: 33 km/h Y: 12 km/h
X: - 69 m
Y: 99 m
X: - 2.5 m/s2 Y: - 2.1 m/s2
X: 61 km
Y: - 22km
40
20
70
35
In groups, on big graph paper, illustrate
the following
• One of the holes on a golf course lies due east of
the tee. A novice golfer flubs his tee shot so that
the ball lands only 64 m directly northeast of the
tee. He then slices the ball 30° south of east so
that the ball lands in a sand trap 127 m away.
Frustrated, the golfer then blasts the ball out of
the sand trap, and the ball lands at a point 73 m
away at an angle 27° north of east. At this point,
the ball is on the putting green and 14.89 m due
north of the hole. To his amazement, the golfer
then sinks the ball with a single shot.
• Use algebraic formulas to find the x and y
components of each displacement vector.
Shot 1 x component _____________
y component _____________
Shot 2 x component _____________
y component _____________
Shot 3 x component _____________
y component _____________
Shot 4 x component _____________
y component _____________
Calculate
• Find the total displacement (to the nearest
meter) the golf ball traveled from the tee to
the hole. Assume the golf course is flat. (Hint:
Which component of each displacement
vector contributes to the total displacement of
the ball between the tee and the hole?)
R= R +R
2
x
2
y
Rx = ax + bx
Ry = ay + by
æ Rx ö
q = tan ç ÷ or
è Ry ø
-1
æ Ry ö
q = tan ç ÷
è Rx ø
-1
Example, p. 94, #1
• A football player runs directly down the field
for 35 m before turning to the right at an
angle of 25° from his original direction and
running an additional 15 m before getting
tackled. What is the magnitude and direction
of the runner’s total displacement?
ax = 0m; ay = 35m
bx = 15sin 25 = 6.34m
b = 15m
by = 15 cos 25 = 13.6m
Rx = 0 + 6.34 = 6.3m
Ry = 35 +13.6 = 49m
a = 35m
R = R2x + R2y = 6.32 + 49 2 = 49(m)
æ 6.3 ö
q = tan ç ÷ = 7.3
è 49 ø
-1
Example 2, p 94, #4
• An airplane flying parallel to the ground
undergoes two consecutive displacements.
The first is 75 km 30.0° west of north, and the
second is 155 km 60.0° east of north. What is
the total displacement of the airplane?
Answer: 171 km at 34° east of north
Table hockey lab
Projectile motion
Free-falling while moving horizontally
Moving in two directions at the same time
Projectile motion
Parabolic trajectory
Vertical motion: with acceleration of -9.8 m/s2 and
initial velocity vy and vertical displacement Δy
Horizontal motion: constant vx and horizontal
displacement Δx
Projectile motion
g 
Vertical
Horizontal
Accelerated
Constant speed
v
t
 9.8 m / s
2
v
v f  vi  g  t
 y  vi  t 
1
t
vi  v f  v
g (t )
2
x  vt
2
v f  vi  2 g  y
2
x
2
v  vertical com ponen t
t 
x
v
v  horizontal com ponent
An object launched horizontally
Find the initial speed of the bike
Given
Dy = -50.0m
g = -9.81m / s 2
vyi = 0
Dx = 90.0m
vx = ?
Equations / Substitution
Horizontal
Vertical
Dx
vx =
Dt
1
Dy = viy Dt + g(Dt)2
2
1
Dy = g(Dt)2
2
2Dy
Dt =
g
Solving
vx =
90
= 3.2(s)
3.2
2 ×(-50)
-9.81
Dt = 3.19(s)
90.0
vix =
3.19
Dt =
vix = 28.2m / s
Lab – Launching a Projectile
Horizontally
• Work in pairs
• Each person should have his / her set of data
• Each student turns in a report with his / her
name on it.
Projectile Launched at an angle
Projectile has initial velocity in both directions
vxi = vcos 30
vx = const
vyi = vsin 30
vyf = vyi + gDt
Example
• A football is kicked 22 m/s at 40o to the
horizontal. Calculate the maximum height and
horizontal range the football will reach before
it falls down (providing no one catches it).
1st step:
v ® vx and v y
v = 22m/s at q =40
v y = 22 ×sin 40
vx = 22 ×cos 40
H  y
v  v  2 g  y ( v f  0)
2
f
2
i
y 
v
2
i
g 
t
t 
x  vt
t  ?
2g
v
R  x
0  14
9.8
 t  2  1.44  2.9( s )
R  17  2.8  49( m )
Example 2 (p.110, #35)
• A place kicker must kick a football from a point
36.0 m from the goal. As a result of the kick, the
ball must clear the crossbar, which is 3.05 m
high. When kicked, the ball leaves the ground
with a speed of 20.0 m/s at an angle of 53° to
the horizontal.
• a. By how much does the ball clear or fall short
of clearing the crossbar?
• b. Does the ball approach the crossbar while still
rising or while falling?
Given
vi = 20m / s
Units
Equations
vx = 20×cos53
v y = 20×sin53
q = 53
hc = hg - Dy
hg = 3.05m
Dy = ?
vy ¹ 0
hc = ?
1
Dy = viy Dt + g(Dt)2
2
- or ¯
Dt =
Dx
vx
- or ¯
Dt > or < Dtup
0 - v yi
Dtup =
g
Substitute/Solve
Dt =
36
= 2.99(s)
1.204
1
Dy = 16 × 2.99 - 9.8×(2.99)2
2
Dy = 4.03m
hc = 4.03- 3.05 = .98(m)
Dtup =
0 -16
= 1.63(s)
-9.8
Dt > Dtup Þ ¯
```