Chapter 3: Motion in a Plane
Chapter 3
Motion in a
plane
Vectors
A
•Vector Addition
A vector is a quantity that has both a magnitude and a
direction. Position is an example of a vector quantity.
•Velocity
•Acceleration
•Projectile motion
•Relative Velocity
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Notation:
A scalar is a quantity with no direction. The mass of an
object is an example of a scalar quantity.
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If we lived in 1-Dimension, we would not need vectors, but…
Would you like to be always standing in line, and the person in
front of you only changed if he (or she) disappeared?
Vector Addition and Subtraction
F
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2-D Rules!
r
Vector: F or F
Vectors let us keep track of our way in any number of dimensions
by carrying all the numbers-components-we need in one symbol
v
The magnitude of a vector: F or F or F .
The direction of vector might be “65° north of east”; “65°
above the +x-axis”; or….
• Here is a vector that shows us
going at an angle θ with the x
axis
• It is useful to use rectangular
components to manipulate
vectors
– These are the projections of
the vector along the x- and y-
Vectors may be moved any way you please (to place them
tip to tail) provided that you do not change their length nor
rotate them.
Scalar: m (not bold face; no arrow)
axes Ax and Ay
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5
Ax and Ay are scalars
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Vector Subtraction and Scalar Multiplication
Graphically Adding Vectors
Vector
Components
• When you add vectors, just put the
tail of one on the head on the
next…
ur
• The resultant R is drawn from the
origin of the first vector to the end of
the last vector
• Vectors obey the
commutative law of addition
The order in which the vectors are
addedrdoesn’t
r affect
r the
r result
• The x-component of a vector
is the projection along the x-axis
uur
ur
A x = A cos θ x
Ax = A cos θ
The y-component of a vector is the
projection along the y-axis
uur
ur
A y = A sin θ y
Ayr = Ar sin θr
Then, A = A x + A y
These equations are valid only if θ is
measured with respect to the x-axis
Adding Vectors Algebraically
• Choose a coordinate system and sketch the vectors
• Find the x- and y-components of all the vectors
Add all the x-components
Add all the y-components
2
x
R= R +R
Use inverse tangent function
To find the direction of R:
θ = tan −1
( )
r r
r
r
A − B = A + −B
• Continue with standard
vector addition procedure
A +B =B+ A
The result of the multiplication or division of a vector by a scalar is a
vector-the magnitude of the vector is multiplied or divided by the scalar
Note that vectors are unchanged
by being moved as long as their
direction or magnitude is not
changed
•If the scalar is positive, the direction of the result is the same as of the
original vector
•If the scalar is negative, the direction of the result is opposite that of the
original vector
Example: Vector A has a length of 5.00 meters and points
along the x-axis. Vector B has a length of 3.00 meters and
points 120° from the +x-axis. Compute A+B (=C).
y
B
By
B
60°
Bx
C
120°
2
y
A
120°
A
x
Bx = Bcos120° = (3.00m )cos120° = −1.50 m
x
By = B sin 120° = (3.00m ) sin 120° = 2.60 m
Ry
Rx
Example continued:
y
R x = ∑ Ax
R y = ∑ Ay
Use Pythagorean theorem find
The magnitude of the resultant:
• Special case of vector
addition--add the negative of
the subtracted vector
and Ax = 5.00 m and Ay = 0.00 m
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12
2
Example continued:
The components of C:
C x = Ax + Bx = 5.00 m + (− 1.50 m ) = 3.50 m
C y = Ay + B y = 0.00 m + 2.60 m = 2.60 m
Example: Margaret walks to the store using the following
path: 0.500 miles west, 0.200 miles north, 0.300 miles east.
What is her total displacement? Give the magnitude and
direction.
Example continued:
The displacement is ∆r = rf − ri. The initial position is the
origin; what is rf?
y
The length of C is:
C = C = Cx + C y
C
2
Cy = 2.60 m
=
θ
Cx = 3.50 m
The final position will be rf = r1 + r2 + r3. The components
are rfx = −r1 + r3 = −0.2 miles and rfy = +r2 = +0.2 miles.
r3
(3.50 m)2 + (2.60 m)2
∆r
r2
= 4.36 m
x
Take north to be in
the +y direction and
east to be along +x.
y
2
y
x
r1
∆ry
The direction of C is: tan θ =
Cy
Cx
=
2.60 m
= 0.7429
3.50 m
θ = tan −1 (0.7429) = 36.6°
Using the figure, the magnitude and
direction of the displacement are
∆r
θ
∆rx
∆r = ∆rx2 + ∆ry2 = 0.283 miles
x
tan θ =
From the +x-axis
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∆ry
∆rx
= 1 and θ = 45°
N of W.
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A particle moves along the blue path as shown. At time t1 its
position is ri and at time t2 its position is rf.
Motion, Velocity, Acceleration
y
vi
∆r
vf
The instantaneous
velocity points
tangent to the path.
Average velocity = v av =
∆r
∆t
∆x 

 The x - component would be : vav , x =

∆t 

Instantaneous velocity = v = lim
∆t → 0
ri
rf
∆r
∆t
This is represented by the slope of a line tangent to the curve
on the graph of an object’s position versus time.
x
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v av =
∆r
∆t
Points in the direction of ∆r
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3
Example: Consider Margaret’s walk to the store in the
example on previous slides. If the first leg of her walk takes
10 minutes, the second takes 8 minutes, and the third 7
minutes, compute her average velocity and average speed
during each leg and for the overall trip.
Example: Margaret walks to the store using the following
path: 0.500 miles west, 0.200 miles north, 0.300 miles east.
What is her total displacement? Give the magnitude and
direction.
Take north to be in
the +y direction and
east to be along +x.
y
Use the definitions:
Average speed =
∆r
∆t
∆r
r2
x
r1
distance traveled
time of trip
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A particle moves along the blue path as shown. At time t1 its
position is r0 and at time t2 its position is rf.
y
a av =
∆v
∆t
∆t
(hours)
vav
(miles/hour)
Average
speed
(miles/hour)
1
0.167
3.00 (west)
3.00
2
0.133
1.50 (north)
1.50
3
0.117
2.56 (east)
2.56
Total
trip
0.417
0.679
2.40
∆v
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A nonzero acceleration changes an object’s state of motion.
Example (text problem 3.42): At the beginning of a 3 hour
plane trip you are traveling due north at 192 km/hour. At the
end, you are traveling 240 km/hour at 45° west of north.
(a) Draw the initial and final velocity vectors.
Average acceleration = a av =
∆v
∆t
∆v
Instantaneous acceleration = a = lim
∆t → 0 ∆ t
rf
These have
interpretations
similar to vav
and v.
y (north)
vf
x
The instantaneous acceleration
can point in any direction.
(45° N of W)
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Points in the
direction of ∆v.
vf
ri
Leg
r3
Average velocity = v av =
vi
Example continued:
vi
x (east)
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Example continued:
Example continued:
(b) Find ∆v.
(c) What is aav during the trip?
The components are
a av =
∆v x = v fx − vix = −v f sin 45° − 0 = −170 km/hr
a y ,av
∆v y = v fy − viy = +v f cos 45° − vi = −22.3 km/hr
The magnitude and direction are:
The magnitude and direction are:
∆v = ∆v x + ∆v y = 171 km/hr
2
tan ϕ =
∆v y
∆v x
2
= 0.1312 ⇒ φ = tan −1 (0.1312) = 7.5°
∆vx − 170 km/hr
=
= −56.7 km/hr 2
∆t
3 hr
∆v y − 22.3 km/hr
=
=
= −7.43 km/hr 2
∆t
3 hr
ax ,av =
∆v
∆t
a av = a x2,av + a y2,av = 57.2 km/hr 2
tan φ =
South of
west
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a y ,av
a x ,av
= 0.1310 ⇒ φ = tan −1 (0.1310) = 7.5°
South of
west
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