Transcript Vectors and 2D Motion - Ohio Wesleyan University

Vectors
• Vector: quantity that has magnitude and direction
• Scalar: quantity that has magnitude only
– Example: 20 mi north (vector) vs. 20 mi (scalar)
Vector
Scalar
Position
Distance
Velocity
Speed
Weight
Mass
• Arrows represent vectors graphically

A (or A)

Magnitude = A (or A)
(arrow length proportional to vector magnitude)
Vectors
• Two vectors A and B are equal if |A| = |B| AND A  B:
A
B
• For vector C antiparallel to A and B, with same
length as A and B  A = B = – C
A
B
C
• Vector addition: Must take direction into account!
– Use tail-to-tip rule: place tail of 2nd vector to tip of 1st
vector
– Works for any number of vectors
Vector Addition and Subtraction
• For A + B = C:
B
A
B
A
C
– Vector addition is commutative (order doesn’t
matter) – try it!
• For multiple vectors A + B + D = E:
A
B
D
E
Vector Addition and
Subtraction
Interactive
• Rules for vector subtraction are the same –
–B
just consider adding a negative
vector:
A – B = A + (– B)
A–B
A
Component Vectors
• Imagine the hassle of measuring a triangle
each time you want to add vectors!
– Would need protractor, ruler, etc.
• Use of vector components provides simple
yet accurate method for adding vectors
y
Ax and Ay are determined from
A=A +A
x
Ax
Ay
q
Ax
y
A
Ay
A = (Ax2 + Ay2)1/2
trigonometry:
Ax = Acosq and Ay = Asinq
x
Ax (Ay) could be positive or negative
depending on whether or not it points in
the +x or –x (+y or –y) direction
Component Vectors
• It may be helpful to envision components as
“shadows” that are projected along each axis:
x
y
A
Ay
q
Ax
y
x
– The length of the “shadow” depends on the
orientation of A
Component Vectors
• Any vector can be represented by its x and y
component vectors
• Sometimes it is convenient to determine the
components of vectors in a tilted coordinate
system:

CQ1: Consider vector Q shown below. Which of
the following could not be a pair of component
vectors for vector Q ?

Q
A)
B)
C)
D)
Component Vectors
• Using components to add vectors A + B = C:
y
B Ax + Bx = Cx
C
By
Ay + By = Cy
Ay
Cy
A
Ax
Cx

C   Ax  Bx iˆ  Ay  By  ˆj
x
Bx
Unit vector
In terms of components:

A  Axiˆ  Ay ˆj

B  Bxiˆ  By ˆj
Magnitude
Direction
iˆ
1
+x
ˆj
1
+y
kˆ
1
+z
CQ2: A man entered a cave and walked 100 m
north. He then made a sharp turn 150° to the
west and walked 87 m straight ahead. How far is
the man from where he entered the cave?
(Note: sin30° = 0.50; cos30° = 0.87.)
A)
B)
C)
D)
25 m
50 m
100 m
150 m
Example Problem #3.20
The helicopter view in the figure at right
shows two people pulling on a stubborn
mule. Find
(a) the single force that is equivalent to the
two forces shown, and
(b) the force that a third person would have to
exert on the mule to make the net force
equal to zero.
The forces are measured in units of Newtons
(N).
Solution (details given in class):
(a) 185 N, q = 77.8° from x axis
(b) 185 N, q = 258° from x axis
2–D and 3–D Motion
• Strong similarity between 1–D motion and 2–D (or
3–D) motion
• Same kinematic quantities (displacement, velocity,
acceleration) are used
• Only difference is vector quantities can have up to
three non-zero components

r2
• First, we describe position:

r1  x1iˆ  y1 ˆj  z1kˆ

r2  x2iˆ  y2 ˆj  z2kˆ
z
y

r1
x
• Displacement (change in position):
 

r2  r1  r  x2  x1 iˆ   y2  y1  ˆj  z2  z1 kˆ
2–D and 3–D
Velocity



r2  r1 r

• Average velocity: vave 
t 2  t1 t

x ˆ y ˆ z ˆ
i
j
k
– Or: vave 
t
t
t
• Instantaneous velocity:


r
v  lim
 v x iˆ  v y ˆj  v z kˆ
t 0 t

• Instantaneous speed: v  vx2  v y2  vz2
– Think Pythagorean theorem!
• Just like in 1–D motion, average velocity only depends
on beginning and ending position

• Instantaneous velocity is tangent to the curve of r
versus time
2–D and 3–D Acceleration
 

• Average acceleration defined as: aave  v2  v1  v
(same as 1–D)
t2  t1
t
 
– v2 , v1 are the instantaneous velocities at t2 and t1,
respectively
 

– aave points in the same direction as v2  v1

• Instantaneous acceleration: 
v
a  lim
t 0 t
(same as 1–D)

– a points toward the inside of any turn particle is making
• Remember that an object can have a non-zero
acceleration even though its speed remains
unchanged
• Example: uniform circular motion
– Speed remains the same
– Velocity keeps changing due to changing direction
CQ3: A weather balloon travels upward for 6 km
while the wind blows it 10 km north and 8 km
east. Approximately what is its final displacement
from its initial position?
A)
B)
C)
D)
7 km
10 km
14 km
20 km
Projectile Motion
• Have you ever wondered how to predict how far a
batted baseball will travel, or how far a skier will travel
after jumping from a ramp?
• These are examples of projectile motion
– Projectile (baseball, skier) follows a trajectory (path in space)
determined by its initial velocity and effects of gravity
y

a
v0
• Start with simple model
(Usually put start of motion
at origin of coordinates)
x
– Projectile represented by single particle
– Neglect air resistance, curvature & rotation of the earth
Projectile Motion
• Notice that acceleration is constant (= g) and is in
vertical direction (due to gravity) pointing toward
earth’s surface
• This is motion with constant acceleration in 2–D
• Particle is given an initial velocity v0:
y
v0x
v0y
q
v0x
ax = 0  vx = v0x = v0cosq
(constant in time)

v0
ay = –g  vy varies with time
v0y
x
• We can analyze projectile motion as:
– Horizontal motion with constant velocity (ax = 0)
– Vertical motion with constant acceleration (due to gravity)
CQ4: Two skydivers are playing catch with a ball
while they are falling through the air. Ignoring air
resistance, in which direction should one skydiver
throw the ball relative to the other if the one wants
the other to catch it?
A)
B)
C)
D)
above the other since the ball will fall faster
above the other since the ball will fall more slowly
below the other since the ball will fall more slowly
directly at the other since there is no air resistance
Projectile Motion
Velocity Components
Interactive
Position and Time
Interactive
• The equations describing the motion follow directly from our
discussion of 1–D motion with constant acceleration:
x–direction (ax = 0):
y–direction (ay = –g):
vy   gt  v0 y
vx  constant v0 x
1 2
y   gt  v0 y t  y0
x  v0 xt  x0
2
• Notice that we treat the x– and y–components separately
Projectile Motion
• You can answer lots of questions concerning the
path the projectile takes by using these equations
• Many times you need to restate the question to
determine what the question is looking for
y
y = ymax
x
x = xmax
What is the maximum height reached? (or At what value of y is vy = 0?)
What is the maximum range? (or What is the value of x when y = 0
[other than at the origin]?)
Example Problem #3.23
A student stands at the edge of a cliff and throws a
stone horizontally over the edge with a speed
of 18.0 m/s. The cliff is 50.0 m above a flat,
horizontal beach, as shown in the figure at
right. (a) What are the coordinates of the initial
position of the stone? (b) What are the
components of the initial velocity? (c) Write the
equations for the x- and y-components of the
velocity of the stone with time. (d) Write the
equations for the position of the stone with
time. (e) How long after being released does
the stone strike the beach below the cliff?
(f) With what speed and angle of impact does
the stone land?
Partial solution (details given in class):
(e) 3.19 s
(f) 36.1 m/s, q = 60.1° below +x axis
Example Problem #3.31
A car is parked on a cliff overlooking the ocean on an
incline that makes an angle of 24.0° below the
horizontal. The negligent driver leaves the car in
neutral, and the emergency brakes are defective.
The car rolls from rest down the incline with a
constant acceleration of 4.00 m/s2 for a distance of
50.0 m to the edge of the cliff, which is 30.0 m above
the ocean. Find (a) the car’s position relative to the
base of the cliff when the car lands in the ocean, and
(b) the length of time the car is in the air.
Solution (details given in class):
(b) 1.78 s
(a) 32.5 m
CQ5: Interactive Problem:
Target Practice
The airplane cargo needs to be dropped at a
horizontal position of:
A)
B)
C)
D)
E)
0m
0.775 m
1.35 m
4.65 m
6.00 m
ActivPhysics Problem #3.7, Pearson/Addison Wesley (1995–2007)