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MOTION IN TWO DIMENSIONS
VECTOR ALGEBRA
PROJECTILE MOTION
RELATIVE MOTION
P
P
P
POLAR FORM OF VECTORS:
The polar form of a vector simply lists or states the magnitude ofthe vector together with its direction with respect to a reference.
EXAMPLE: 50m/s at 20o
In print the usual symbolic notation for a vector is a capital letter inbold print. In writing the usual notation for a vector is a capitalletter with a small horizontal arrow above it.
A or A
GRAPHICAL REPRESENTATION OF VECTORS:
A vector quantity may be depicted using an arrow, a straight line witha point. The length of the arrow will be proportional to and representthe vector’s magnitude (which is always positive). The orientation ofthe arrow in space or on a coordinate grid corresponds to the vector’sdirection.
50o
This arrow represents thevelocity vector of 4 m/s at50o .
x
y
GRAPHICAL METHOD OF ADDING VECTORS:[ HEAD - TO - TAIL METHOD ]
Add: 20 at 60o , 30 at 135o , and 40 at 340o
x
y
COMPONENTS OF A VECTOR:
x
= 130o
50
Ax
Ay
as shown is in standard form.
50 130oA at
cos 50cos130 32.14
sin 50sin130 38.30
ox
oy
A A
A A
ADDING VECTORS USING THEIR COMPONENTS:
ADDING VECTORS USING THEIR COMPONENTS:
Add: A = 20 at 40o and B = 30 at 210o C = A + B Vector X Y
20 @ 40o 20cos 40o 20sin 40o
15.32 12.86
30 @ 210o 30cos 210o 30sin 210o
-25.98 -15.00
Cx = -10.66 Cy = -2.14
2 2 2 2
1 1 0 0
( 10.66) ( 2.14) 10.87
2.14tan tan 180 191.4
10.66
x y
y
x
C C C
C
C
When using a calculator to calculate tan-1, add the appropriate correction angle to the result given by the calculator.
USING tan -1:
= tan-1 (Cy/Cx ) + Correction
I
Cx + : C y +
Correction = 0 o
II
Cx - : C y +
Correction = 180 o
III
Cx - : C y -
Correction = 180 o
IV
Cx + : C y -
Correction = 360 o
x
y
IN CLASS EXERCISES....
Plot the following vectors:
20 at 30o
40 at 90o
30 at 160o
70 at 325o
Answers on next slide.
x
y
x
y
x
y
x
y
20 at 30o40 at 90o
30 at 160o70 at 325o
IN CLASS EXERCISES....
Add the following vectors graphically:
20 at 30o
40 at 90o
30 at 160o
70 at 325o
Answer on next slide.
20 at 30o
40 at 90o
30 at 160o
70 at 325o
60 at 25o
Answer: 60 at 25o
IN CLASS EXERCISES....
Find the components of the following vector:
40 at 295o
Answer on next slide.
x-component:
40 cos 295o = 16.9
y-component:
40 sin 295o = -36.3
x
y
ANSWER:
ADD THE FOLLOWING VECTORS USING THEANALYTICAL METHOD:
A = 60 at 75o B = 45 at 225o
Answer on next slide.
ANSWER:
vector x y
60 at 75o 60cos75o 60sin75o
15.5 58.0
45 at 225o 45cos225o 45sin225o
-31.8 -31.8
Cx = -16.3 Cy = 26.2
2 2 2 2
1 1 0 0
( 16.3) (26.2) 30.9
26.2tan tan 180 121.9
16.3
x y
y
x
C C C
C
C
PROJECTILE MOTION:
vo
vxo
vyo
R
parabolic arc
2
c
2 in
os in
s
sxo
yo
o y
o
o
l
o
tota
v vt
v v v v
g g
2
2
2 sincos
2sin c
sin 2
os
xo to
o
oto
a
o
l
vv
g
vR
R v
v
g
g
t
R
ay = -g
ax = 0
Example: In a physics demonstration a golf ball is thrown 70m. Its angle of launch is 45o. What is its initial speed?
ANSWER: In a physics demonstration a golf ball is thrown 70m. Its angle of launch is 45o. What is its initial speed?
2
2
2
o
sin 2, 45
solve for
26
v
9. 70
. 9
8
1
o o
o
mo
s
s
mo
vR
g
vR
g
v gR m
v
Relative Motion:
Consider yourself to be a stationary observer. A friend of yours, Joe, is another observer but he is in motion. Lets say you measure his velocity to be the vector u.
Some object of mutual interest passes by. Both you and Joe measure its velocity. Joe measures its velocity to be v’ and you measure its velocity to be v.
These two velocities, v’ and v, are different vectors because they are measured in different coordinate frames. However, they are related by the following equation:
'v u v
RELATIVE MOTION:
frame S
frame S
x
y x
y
O
O
u vStationary Frame
Moving Framemoving object
u = velocity of moving framev = velocity of object as seen in moving framev = velocity of object as seen in stationary frame
v u v
EXAMPLE: A boat is to cross directly across a river. The boat is able to do 15 knots in still water and the river current is 5 knots. ( A knot is 1 nautical mile per hour, a nautical mile is 6080 ft). (a) At what angle upstream must the boat be pointed? (b) What will the boat’s speed be relative to the ground?
ANSWER: A boat is to cross directly across a river. The boat is able to do 15 knots in still water and the river current is 5 knots. ( A knot is 1 nautical mile per hour, a nautical mile is 6080 ft). (a) At what angle upstream must the boat be pointed? (b) What will the boat’s speed be relative to the ground?
u = vwater
v’=vboat, water v=vboat, gnd
5 kn
15 kn
, ,
'
boat gnd water boat water
v u v
v v v
This vector problem can be solved using the fact that these vectors form a right-triangle.
2 2,
1
14.14(15 ) (5 )
5t 19.4an
14 47
.1
boat gn
o
d kv kn nkn
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