Transcript Slide 1

Chapter 3
Kinematics in Two Dimensions; Vectors
Trigonometry Review
opposite side
sin  
hypotenuse
adjacent side
cos  
hypotenuse
opposite side
tan  
adjacent side
sin
More Trigonometry
• Pythagorean Theorem
• To find an angle, you need the inverse trig
function
– for example,
3-2 Addition of Vectors – Graphical Methods
For vectors in one
dimension, simple
addition and subtraction
are all that is needed.
You do need to be careful
about the signs, as the
figure indicates.
3-2 Addition of Vectors – Graphical Methods
If the motion is in two dimensions, the situation is
somewhat more complicated.
Here, the actual travel paths are at right angles to
one another; we can find the displacement by
using the Pythagorean Theorem.
3-2 Addition of Vectors – Graphical Methods
Adding the vectors in the opposite order gives the
same result:
3-2 Addition of Vectors – Graphical Methods
Even if the vectors are not at right
angles, they can be added graphically by
using the “tail-to-tip” method.
3-2 Addition of Vectors – Graphical Methods
The parallelogram method may also be used;
here again the vectors must be “tail-to-tip.”
3-3 Subtraction of Vectors, and
Multiplication of a Vector by a Scalar
In order to subtract vectors, we
define the negative of a vector, which
has the same magnitude but points
in the opposite direction.
Then we add the negative vector:
3-4 Adding Vectors by Components
Any vector can be expressed as the sum
of two other vectors, which are called its
components. Usually the other vectors are
chosen so that they are perpendicular to
each other.
3-4 Adding Vectors by Components
If the components are
perpendicular, they can be found
using trigonometric functions.
3-4 Adding Vectors by Components
The components are effectively one-dimensional,
so they can be added arithmetically:
3-4 Adding Vectors by Components
Adding vectors:
1. Draw a diagram; add the vectors graphically.
2. Choose x and y axes.
3. Resolve each vector into x and y components.
4. Calculate each component using sines and cosines.
5. Add the components in each direction.
6. To find the length and direction of the vector, use:
3-5 Projectile Motion
A projectile is an object
moving in two
dimensions under the
influence of Earth's
gravity; its path is a
parabola.
3-5 Projectile Motion
It can be understood by
analyzing the horizontal and
vertical motions separately.
3-5 Projectile Motion
The speed in the x-direction
is constant; in the ydirection the object moves
with constant acceleration g.
This photograph shows two balls
that start to fall at the same time.
The one on the right has an initial
speed in the x-direction. It can be
seen that vertical positions of the
two balls are identical at identical
times, while the horizontal
position of the yellow ball
increases linearly.
3-5 Projectile Motion
If an object is launched at an initial angle of θ0
with the horizontal, the analysis is similar except
that the initial velocity has a vertical component.
3-7 Projectile Motion Is Parabolic
In order to demonstrate that
projectile motion is parabolic,
we need to write y as a function
of x. When we do, we find that it
has the form:
This is
indeed the
equation for
a parabola.
3-6 Solving Problems Involving
Projectile Motion
Projectile motion is motion with constant
acceleration in two dimensions, where the
acceleration is g and is down.
3.3 Projectile Motion
Example 3 A Falling Care Package
The airplane is moving horizontally with a constant velocity of
+115 m/s at an altitude of 1050m. Determine the time required
for the care package to hit the ground.
3.3 Projectile Motion
y
ay
-1050 m -9.80 m/s2
vfy
viy
t
0 m/s
?
3.3 Projectile Motion
y
ay
vfy
-1050 m -9.80 m/s2
y  viyt  ayt
1
2
t
2y

ay
2
viy
t
0 m/s
?
y  ayt
1
2
2
2 1050 m 

14
.
6
s
2
 9.80 m s
3.3 Projectile Motion
Example 4 The Velocity of the Care Package
What are the magnitude and direction of the final velocity of
the care package?
3.3 Projectile Motion
y
ay
-1050 m -9.80 m/s2
vfy
viy
t
?
0 m/s
14.6 s
3.3 Projectile Motion
y
ay
vfy
viy
t
?
0 m/s
14.6 s
-1050 m -9.80 m/s2


v fy  viy  a y t  0   9.80 m s 14.6 s 
 143m s
2
3.3 Projectile Motion
Conceptual Example 5
I Shot a Bullet into the Air...
Suppose you are driving a convertible with the top down.
The car is moving to the right at constant velocity. You point
a rifle straight up into the air and fire it. In the absence of air
resistance, where would the bullet land – behind you, ahead
of you, or in the barrel of the rifle?
3.3 Projectile Motion
Example 6 The Height of a Kickoff
A placekicker kicks a football at and angle of 40.0 degrees and
the initial speed of the ball is 22 m/s. Ignoring air resistance,
determine the maximum height that the ball attains.
3.3 Projectile Motion
vi

viy
vix
viy  vi sin   22m ssin 40  14m s

vix  vi cos  22m scos40  17m s

3.3 Projectile Motion
y
ay
vfy
viy
?
-9.80 m/s2
0
14 m/s
t
3.3 Projectile Motion
y
ay
vfy
viy
?
-9.80 m/s2
0
14 m/s
v  v  2ay y
2
fy
2
iy
y
t
v v
2
fy
2a y
0  14 m s 
y


10
m
2
2  9.8 m s
2


2
iy
3.3 Projectile Motion
Example 7 The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
3.3 Projectile Motion
y
ay
0
-9.80 m/s2
vfy
viy
t
14 m/s
?
3.3 Projectile Motion
y
ay
vfy
0
-9.80 m/s2
viy
t
14 m/s
?
y  viyt  ayt
1
2

2

0  14m st   9.80m s t
1
2

2
2

0  214m s   9.80m s t
t  2.9 s
2
3.3 Projectile Motion
Example 8 The Range of a Kickoff
Calculate the range R of the projectile.
x  vixt  a xt  vixt
 17 m s 2.9 s   49 m
1
2
2
Projectile Motion Example
A canon is fired with a muzzle velocity of 1000 m/s at an
angle of 30°. The projectile fired from the canon lands in the water
40 m below the canon.
a) What is the range of the projectile.
b) What id the velocity of the projectile in x and y when it lands
c) What is the landing angle of the projectile
Known
vi = 1000m/s
θ = 30°
Δy = -40m
x  vox t  a x t  vox t
1
2
vi
Unknown
voy
X, vfx, vfy, θ, vf
2
vox  (1000m / s) cos30  866m / s
voy  (1000m / s) sin 30  500m / s
30°
vix
Projectile Motion Example
Use y info to find t
y  voyt  ayt
1
2
2
-40m = (500m/s)t + ½(-9.8m/s2)t2
4.9t2 – 500t – 40 = 0
a
b
c
Solve for t using quadratic equation
t = -(-500) ± (-500)2 – 4(4.9)(-40)
2(-40)
Take the positive root:
t = 102.1s
Projectile Motion Example
Calculate x:
x  vox t
x  (866m / s)102.1s  88418.6m
Calculate vy
v  v  2a y y
2
y
2
oy
v y  v 2 oy  2a y y
v y  (500m / s) 2  2(9.8m / s 2 )(40m)
v y  500.8m / s  500.8m / s
Projectile Motion Example
Find landing angle and velocity
vx = 866m/s
θ
vy = -500.8m/s
vf
vf =  (866m/s)2 + (-500.8m/s)2 = 1000.4m/s
tan θ = -500.8m/s = -0.578
866m/s
θ = tan-1(-0.578)
θ = -30.04°
Projectile Motion Example
Find landing angle and velocity
vx = 866m/s
θ
vy = -500.8m/s
vf
vf =  (866m/s)2 + (-500.8m/s)2 = 1000.4m/s
tan θ = -500.8m/s = -0.578
866m/s
θ = tan-1(-0.578)
θ = -30.04°
1.7 The Components of a Vector
Example
A displacement vector has a magnitude of 175 m and points at
an angle of 50.0 degrees relative to the x axis. Find the x and y
components of this vector.
sin   y r




y  r sin   175m sin 50.0  134m
cos  x r
x  r cos  175m cos50.0  112m

r  112mxˆ  134myˆ
A hiker walks 100m north, 130m northeast, and
120m south. Find the displacement vector and the
angle measured from the positive x axis.
N
B
C
W
A
S
D
θ
E
Example: Adding Vectors
In an experiment, an object moves along the path described by
vectors A, B, C. The resultant of the three vectors is R. The length
of the three vectors is as follows: A=10m, B=12m, and C=15m.
Find the magnitude and direction of R
+y
Rx
θ
Ry R
A
30°
+x
B
C
X
Y
A
10mcos 30° =
8.66m
10msin 30° =
5m
B
0
-12m
C
-15m
0
R
-6.34m
-7m
R = (-6.34m)2 + (-7m)2 = 9.44m
tan θ = -7m
-6.34m
 θ= 47.8° below the
negative x axis