Download Unit * 2:

Unit – 2:
KINEMATICS AND VECTORS
Kinematics –
Analyzing motion under the condition of constant acceleration
Kinematic Symbols
x,y
t
vo
v
a
g
Displacement
Time
Initial Velocity
Final Velocity
Acceleration
Acceleration due to
gravity
Kinematic #1
v  vo
v
a

t
t
v  vo  at
v  vo  at
Kinematic #1
Example: A boat moves slowly out of a marina (so as to not leave a
wake) with a speed of 1.50 m/s. As soon as it passes the breakwater,
leaving the marina, it throttles up and accelerates at 2.40 m/s/s.
a) How fast is the boat moving after accelerating for 5 seconds?
What do I
know?
vo= 1.50 m/s
a = 2.40 m/s/s
t=5s
What do I
want?
v=?
v  vo  at
v  (1.50)  (2.40)(5)
v
13. 5 m/s
Kinematic #2
2
1
x  voxt  at
2
b) How far did the boat travel during that time?
2
1
x  voxt 
at
2
2
1
x  (1.5)(5)  (2.40)(5 )
2
x  37.5 m
Does all this make sense?
A  bh  A  (5)(1.5)
A  7.50 m
1
1
A  bh  (5)(12)
2
2
A  30 m
Total displacement = 7.50 + 30 = 37.5 m = Total AREA under the line.
Kinematic #3
v  v  2ax
2
2
o
Example: You are driving through town at 12 m/s when suddenly a ball rolls
out in front of your car. You apply the brakes and begin decelerating at
3.5 m/s/s.
How far do you travel before coming to a complete stop?
What do I
know?
vo= 12 m/s
a = -3.5 m/s/s
V = 0 m/s
What do I
want?
x=?
v  v  2ax
2
2
o
0  12 2  2( 3.5) x
 144  7 x
x  20.57 m
Common Problems Students Have
I don’t know which equation to choose!!!
Equation
Missing Variable
v  vo  at
x
x  voxt  1 at
2
2
v  v  2ax
2
2
o
v
t
Kinematics for the VERTICAL Direction
All 3 kinematics can be used to analyze one dimensional
motion in either the X direction OR the y direction.
v  vo  at  v y  voy  gt
2
2
1
1
x  voxt  at  y  voyt 
gt
2
2
2
2
2
2
v  vox  2ax  v y  voy  2 gy
Examples
A pitcher throws a fastball with a velocity of 43.5 m/s. It is determined that
during the windup and delivery the ball covers a displacement of 2.5 meters.
This is from the point behind the body to the point of release. Calculate the
acceleration during his throwing motion.
What do I
know?
vo= 0 m/s
What do I
want?
a=?
Which variable is NOT given and
NOT asked for? TIME
v  v  2ax
2
2
o
x = 2.5 m
V = 43.5 m/s
43.52  02  2a(2.5)
a  378.45 m / s
2
Examples
How long does it take a car at rest to cross a 35.0 m intersection after
the light turns green, if the acceleration of the car is a constant 2.00
m/s/s?
What do I
know?
vo= 0 m/s
x = 35 m
a = 2.00 m/s/s
What do I
want?
t=?
Which variable is NOT given and
NOT asked for?
x  voxt  1 at 2
2
35  (0)  1 (2)t 2
2
t  5.92 s
Examples
A car accelerates from 12.5 m/s to 25 m/s in 6.0 seconds.
What was the acceleration?
What do I
know?
vo= 12.5 m/s
v = 25 m/s
t = 6s
What do I
want?
a=?
Which variable is NOT given and
NOT asked for?
v  vo  at
25  12.5  a(6)
a  2.08 m / s 2
Examples
A stone is dropped from the top of a cliff. It is observed to hit
the ground 5.78 s later. How high is the cliff?
What do I
know?
v = 0 m/s
oy
g = -9.8 m/s2
t = 5.78 s
What do I
want?
y=?
Which variable is NOT given and
NOT asked for?
y  voyt  1 gt 2
2
y  (0)(5.78)  4.9(5.78) 2
y  163.7 m
h  163.7 m
Vectors and Scalars
Scalar
A SCALAR is ANY quantity in
physics that has
MAGNITUDE, but NOT a
direction associated with it.
Scalar
Example
Magnitude
Speed
20 m/s
Magnitude – A numerical
value with units.
Distance
10 m
Age
15 years
Heat
1000
calories
Vector
A VECTOR is ANY quantity in
physics that has BOTH
MAGNITUDE and DIRECTION.
Vector
Velocity
Magnitude
& Direction
20 m/s, N
Acceleration 10 m/s/s, E
   
v , x, a, F
Force
5 N, West
Vectors are typically illustrated by drawing
an ARROW above the symbol. The arrow is
used to convey direction and magnitude.
Applications of Vectors
VECTOR ADDITION – If 2 similar vectors point in the SAME direction, add
them.
Example: A man walks 54.5 meters east, then another 30 meters east.
Calculate his displacement relative to where he started?
54.5 m, E
+
84.5 m, E
30 m, E
Notice that the SIZE of
the arrow conveys
MAGNITUDE and the way
it was drawn conveys
DIRECTION.
Applications of Vectors
VECTOR SUBTRACTION - If 2 vectors are going in opposite
directions, you SUBTRACT.
Example: A man walks 54.5 meters east, then 30 meters west.
Calculate his displacement relative to where he started?
54.5 m, E
30 m, W
24.5 m, E
-
Non-Collinear Vectors
When 2 vectors are perpendicular, you must use the
Pythagorean theorem.
The hypotenuse in Physics is
called the RESULTANT.
A man walks 95 km, East then 55 km,
north. Calculate his RESULTANT
DISPLACEMENT.
Finish
c2  a2  b2  c  a2  b2
55 km, N
Horizontal Component
Vertical
Component
c  Resultant  952  552
c  12050  109.8 km
95 km,E
Start
The LEGS of the triangle are called the COMPONENTS
BUT……what about the direction?
In the previous example, DISPLACEMENT was asked for and since
it is a VECTOR we should include a DIRECTION on our final
answer.
N
W of N
E of N
N of E
N of W
N of E
E
W
NOTE: When drawing a right triangle that
conveys some type of motion, you MUST draw
your components HEAD TO TOE.
S of W
S of E
W of S
E of S
S
BUT…..what about the VALUE of the angle???
Just putting North of East on the answer is NOT specific enough for the
direction. We MUST find the VALUE of the angle.
To find the value of the angle
we use a Trig function called
TANGENT.
109.8 km
55 km, N
 N of E
95 km,E
Tan 
opposite side 55

 0.5789
adjacent side 95
  Tan 1 (0.5789)  30
So the COMPLETE final answer is : 109.8 km, 30 degrees North of East
What if you are missing a component?
Suppose a person walked 65 m, 25 degrees East of North. What were his
horizontal and vertical components?
H.C. = ?
The goal: ALWAYS MAKE A RIGHT TRIANGLE!
To solve for components, we often use the trig
functions sine and cosine.
V.C = ?
25
65 m
adjacent side
hypotenuse
adj  hyp cos 
cosine  
opposite side
hypotenuse
opp  hyp sin 
sine  
adj  V .C.  65 cos 25  58.91m, N
opp  H .C.  65 sin 25  27.47m, E
Example
A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders
another 12 meters west then 6 meters south. Calculate the bear's displacement.
-
12 m, W
-
=
6 m, S
20 m, N
35 m, E
=
23 m, E
14 m, N
R  14 2  232  26.93m
14 m, N
R

14
 .6087
23
  Tan 1 (0.6087)  31.3
Tan 
23 m, E
The Final Answer: 26.93 m, 31.3 degrees NORTH or EAST
Example
A boat moves with a velocity of 15 m/s, N in a river which flows
with a velocity of 8.0 m/s, west. Calculate the boat's resultant
velocity with respect to due north.
Rv  82  152  17 m / s
8.0 m/s, W
15 m/s, N
Rv

8
Tan 
 0.5333
15
1

  Tan (0.5333)  28.1
The Final Answer : 17 m/s, @ 28.1 degrees West of North
Example
A plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate the plane's
horizontal and vertical velocity components.
adjacent side
cosine  
hypotenuse
adj  hyp cos 
H.C. =?
32
63.5 m/s
opposite side
sine  
hypotenuse
opp  hyp sin 
V.C. = ?
adj  H .C.  63.5 cos 32  53.85 m / s, E
opp  V .C.  63.5 sin 32  33.64 m / s, S
Example
A storm system moves 5000 km due east, then shifts course at 40 degrees
North of East for 1500 km. Calculate the storm's resultant displacement.
1500 km
adjacent side
hypotenuse
V.C.
adj  hyp cos 
cosine  
opposite side
hypotenuse
opp  hyp sin 
sine  
40
5000 km, E
H.C.
adj  H .C.  1500 cos 40  1149.1 km, E
opp  V .C.  1500 sin 40  964.2 km, N
1500 km + 1149.1 km = 2649.1 km
R  2649.12  964.2 2  2819.1 km
964.2
 0.364
2649.1
  Tan 1 (0.364)  20.0
Tan 
R
964.2 km

2649.1 km
The Final Answer: 2819.1 km @ 20 degrees,
East of North