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Kinematics in One
Dimension
The Cheetah: A cat that is built for speed. Its strength and
agility allow it to sustain a top speed of over 100 km/h. Such
speeds can only be maintained for about ten seconds.
Distance and Displacement
Distance is the length of the actual path
taken by an object. Consider travel from
point A to point B in diagram below:
d = 20 m
A
B
Distance d is a scalar
quantity (no direction):
Contains magnitude only
and consists of a
number and a unit.
Distance and Displacement
Displacement is the straight-line separation
of two points in a specified direction.
Δs = 12 m, 20o B
A
q
A vector quantity:
Contains magnitude
AND direction, a
number, unit & angle.
Distance and Displacement
• For motion along x or y axis, the displacement is
determined by the x or y coordinate of its final
position. Example: Consider a car that travels 8 m, E
then 12 m, W.
Net displacement Δx
is from the origin to
the final position:
Δx = 4 m, W
What is the distance
traveled? d = 20 m !!
Δx
8 m,E
x = -4
x
x = +8
12 m,W
Definition of Speed
• Speed is the distance traveled per unit of
time (a scalar quantity).
d = 20 m
A
Time t = 4 s
B
vs =
d
t
=
20 m
4s
vs = 5 m/s
Not direction dependent!
Definition of Velocity
• Velocity is the displacement per
unit of time. (A vector quantity.)
s = 20 m
A
Δx=12 m
20o
Time t = 4 s
B
= 3 m/s at 200 N of E
Direction required!
Average Speed and
Instantaneous Velocity
 The average speed depends ONLY on the
distance traveled and the time required.
A
s = 20 m
C
Time t = 4 s
B
The instantaneous
velocity is the magnitude and direction of
the velocity at a particular instant. (v at
point C)
Example 1. A runner runs 200 m, east, then
changes direction and runs 300 m, west. If
the entire trip takes 60 s, what is the average
speed and what is the average velocity?
Recall that average
s2 = 300 m
speed is a function
only of total distance
start
and total time:
s1 = 200 m
Total distance: s = 200 m + 300 m = 500 m
total path 500 m
Average speed 

time
60 s
Avg. speed= 8 m/s
Direction does not matter!
Example 1 (Cont.) Now we find the average
velocity, which is the net displacement divided
by time. In this case, the direction matters.
t = 60 s
x = -100 m
xo = 0 m; x = -100 m
x1= +200 m
xo = 0
Direction of final
displacement is to
the left as shown.
Average velocity:
Note: Average velocity is directed to the west.
Definition of Acceleration
 An acceleration is the change in velocity
per unit of time. (A vector quantity.)
 A change in velocity requires the
application of a push or pull (force).
A formal treatment of force and acceleration will
be given later. For now, you should know that:
• The direction of acceleration is same as
direction of force.
• The acceleration is
proportional to the
magnitude of the force.
Acceleration and Force
F
a
2F
2a
Pulling the wagon with twice the force
produces twice the acceleration and
acceleration is in direction of force.
Example 3 (No change in direction): A constant
force changes the speed of a car from 8 m/s to
20 m/s in 4 s. What is average acceleration?
+
Force
t=4s
vo = +8 m/s
Step
Step
Step
Step
1.
2.
3.
4.
v = +20 m/s
Draw a rough sketch.
Choose a positive direction (right).
Label given info with + and - signs.
Indicate direction of force F.
Example 3 (Continued): What is average
acceleration of car?
+
Force
t=4s
vo = +8 m/s
Step 5. Recall definition
of average acceleration.
v = +20 m/s
Graphical Analysis
slope:
velocity:
acceleration:
Position vs time graph (velocity)
9
8
7
p
o
s
i
t
i
o
n
x, (m)
6
No Motion (zero
velocity)
5
Uniform Motion
(constant velocity)
4
3
Accelerated Motion
(Increasing/changing
velocity)
2
1
0
1
2
3
time, t (s)
4
velocity vs time graph (acceleration)
v, (m/s)
Graphical Analysis
x2
Dx
x1
Dt
t1
t2
Instantaneous Velocity:
Dx
vinst 
(Dt  0)
Dt
Displacement, x
Average Velocity:
Dx x2  x1
vavg 

Dt t2  t1
slope
Dx
Dt
Time
Uniform Acceleration
in One Dimension:
• Motion is along a straight line (horizontal,
vertical or slanted).
• Changes in motion result from a CONSTANT
force producing uniform acceleration.
• The velocity of an object is changing by a
constant amount in a given time interval.
• The moving object is treated as though it
were a point particle.
Average velocity for constant a:
setting to = 0
combining both equations:
For constant acceleration:
Formulas based on definitions:
Derived formulas:
For constant acceleration only
Example 6: An airplane flying initially at 400
ft/s lands on a carrier deck and stops in a
distance of 300 ft. What is the acceleration?
vo = 400 ft/s
v=0
+
Step 1. Draw and label sketch.
Step 2. Indicate + direction
Example: (Cont.)
vo = 400 ft/s
v=0
+
Step 3. List given; find information with signs.
Given: vo = 400 ft/s - initial velocity of airplane
v=0
- final velocity after
traveling Δx = +300 ft
Find: a = ?
- acceleration of airplane
Given: vo = +400 ft/s
v=0
Δx = +300 ft
Step 4. Select equation
that contains a and not t.
0
-vo2
-(400 ft/s)2
a=
=
v 2 - vo2 = 2aΔx
2x
2(300 ft)
a = - 300 ft/s2
Why Force
is the isacceleration
negative?
Because
in a negative
direction which
means that the airplane slows down
Example 5: A ball 5.0 m from the bottom of an
incline is traveling initially at 8.0 m/s. Four
seconds (4.0 s) later, it is traveling down the
incline at 2.0 m/s. How far is it from the bottom at
that instant?
+
Δx
-2.0 m/s
5.0 m
8.0 m/s
Given: d = 5.0 m
vo = 8.0 m/s
t = 4.0 s
- distance from initial position of the ball
- initial velocity
v = -2.0 m/s - final velocity after t = 4.0 s
Find: x = ?
- distance from the bottom of the incline
Given: d = 5.0 m
vo = 8.0 m/s
- initial velocity
v = -2.0 m/s - final velocity after t = 4.0 s
Find: x = ?
Solution:
- distance from the bottom of the incline
where
x = 17.0 m
Acceleration Due to Gravity
• Every object on the earth
experiences a common force: the
force due to gravity.
• This force is always directed
toward the center of the earth
(downward).
• The acceleration due to gravity is
relatively constant near the
Earth’s surface.
g
W
Earth
Gravitational Acceleration
• In a vacuum, all objects fall
with same acceleration.
• Equations for constant
acceleration apply as usual.
• Near the Earth’s surface:
a = g = -9.80 m/s2 or -32 ft/s2
Directed downward (usually negative).
Sign Convention:
A Ball Thrown
Vertically Upward
avy==
=-0
+
avy=
==-++
ya
=+-v ==
•
UP = +
Release Point
Displacement is positive
(+) or negative (-) based
on LOCATION.
vya==
=-0-
•
yv=
=a=Negative
Negative
Velocity is positive (+) or
negative (-) based on
direction of motion.
• Acceleration is (+) or (-)
based on direction of force
(weight).
Example 7: A ball is thrown vertically upward with
an initial velocity of 30.0 m/s. What are its
position and velocity after 2.00 s, 4.00 s, and 7.00
s? Find also the maximum height attained
Given: a = -Δ9.8 m/s2
vo = 30.0 m/s
t = 2.00 s; 4.00 s; 7.00 s
Find:
+
a=g
Δy = ? – displacement
v=?
- final velocity
After those three “times”
Δy = ? – maximum height
vo = +30.0 m/s
Given: a = -9.8 m/s2; vo = 30.0 m/s
t = 2.00 s; 4.00 s; 7.00 s
Solutions:
For t = 2.00 s:
For t = 4.00 s:
For t = 7.00 s:
Given: a = -9.8 m/s2; vo = 30.0 m/s
t = 2.00 s; 4.00 s; 7.00 s
Solutions:
For t = 2.00 s:
For t = 4.00 s:
For t = 7.00 s:
Given: a = -9.8 m/s2; vo = 30.0 m/s
t = 2.00 s; 4.00 s; 7.00 s
Solutions:
For maximum height, v = 0 (the ball stops at
maximum height):
Experiment 10
Uniformly Accelerated Motion
(Acceleration due to Gravity) 39
(06A)