Chapter 2
Describing Motion: Kinematics
in One Dimension
#______HOMEWORK – CHAPTER 2 :
KINEMATICS 20 %
P 57-115 ALL ODD NUMBERS ONLY
1 WEEK TO DO THE HOMEWORK
#1. NOTES CHAPTER 2 KINEMATICS
P 32-34 ( Read , Summarize, Big Group)
1. Mechanics
2. Kinematics
3. Dynamics
4. Distance
5. Motion
6. Speed
7. Scalar Quantity
8. Average Speed
Equation
Units
2-1 Reference Frames and Displacement
Any measurement of position, distance, or speed must be made with respect to a
reference frame.
For example, if you are sitting on a train and someone walks
down the aisle, their speed with respect to the train is a few
miles per hour, at most. Their speed with respect to the
ground is much higher.
2-1 Reference Frames and Displacement
We make a distinction between distance and displacement.
Displacement (blue line) is how far the object is from its starting point, regardless of
how it got there.
Distance traveled (dashed line) is measured along the actual path.
2-1 Reference Frames and Displacement
The displacement is written:
Left:
Right:
Displacement is positive.
Displacement is negative.
2-2 Average Velocity
Speed: how far an object travels in a given time interval
(2-1)
Velocity includes directional information:
#1. NOTES CHAPTER 2 KINEMATICS
9. Instantaneous Speed
10. Example 2.1
P 34 Copy the problem
11. Follow Up Exercise
P 34 Copy the Problem
#1. NOTES CHAPTER 2 KINEMATICS
2.2 One- Dimensional Displacement and Velocity : vector Quantities
P35-36
12. Displacement –
Examples :
Fig 2.4 p 35
13. Average Velocity
Equation
Units
2-3 Instantaneous Velocity
The instantaneous velocity is the average velocity, in the limit as the time interval
becomes infinitesimally short.
(2-3)
These graphs show (a) constant velocity and
(b) varying velocity.
#1. NOTES CHAPTER 2 KINEMATICS
14. Ex 2.2 p 37
Copy the problem
15.Follow up Exercise
Copy the problem p37
#2. CW THINK, SOLVE, SMALL GROUP, BIG GROUP
AP Physics B Practice Workbook
#1
#1. NOTES CHAPTER 2 KINEMATICS
16. Graphical Analysis use Graph Paper
Copy Figure 2.6 a Data Table
Copy or Draw Graph b
Calculate slope
17. Graphical Analysis use Graph Paper
Copy and Draw Figure 2.7
Calculate slope
#3 CW THINK, SOLVE, SMALL GROUP, BIG
GROUP
AP Physics Workbook
#2. Copy the question
#1. Notes Acceleration
Acceleration is the rate of change of velocity.
2-4 Acceleration
Acceleration is a vector, although in one-dimensional motion we only need the sign.
The previous image shows positive acceleration; here is negative acceleration:
#1. NOTES CHAPTER 2 KINEMATICS –ACCELERATION
18. Example 2.3 Slowing it down
P 41 Copy the problem
#4. CW THINK, EXPLAIN, SMALL G, BIG G
2.3 Follow Up Exercise p 42
2-4 Acceleration
There is a difference between negative acceleration and
deceleration:
19. Negative acceleration is acceleration in the negative
direction as defined by the coordinate system.
20. Deceleration occurs when the acceleration is opposite in
direction to the velocity. The magnitude of the velocity is
decreasing.
# 1 Notes 2-4 Acceleration
21. The instantaneous acceleration is the average acceleration, in the limit as the
time interval becomes infinitesimally short.
(2-5)
CW Think, Explain, Small G, Big G
22. A car accelerates along a straight road from rest to 75 km/h in 5.0 s. What is the
magnitude of its average acceleration?
#6CW Think, Explain, Small G, Big G
If the velocity of an object is zero, does it mean that the acceleration is zero?
If the acceleration is zero, does it mean that the velocity is zero?
a=?
a=?
#1. Notes 2-5 Motion at Constant Acceleration
23. The average velocity of an object during a time interval t is
The acceleration, assumed constant, is
2-5 Motion at Constant Acceleration
24. In addition, as the velocity is increasing at a constant rate, we know that
Combining these last three equations, we find:
(2-8)
(2-9)
2-5 Motion at Constant Acceleration
We can also combine these equations so as to eliminate t:
We now have all the equations we need to solve constant-acceleration problems.
(2-10)
(2-11a)
(2-11b)
(2-11c)
(2-11d)
#1. NOTES CHAPTER 1 KINEMATICS
CONSTANT ACCELERATION
#25. Example 2.4 p 42 A drag racer starting form rest accelerates in a straight line at
a constant rate of 5.5 m/s2 for 6 sec.
a. What is the racer’s velocity at the period of time?
b. If a parachute deployed at this time causes the racer to slow down uniformly at a
rate of 2.4 m/s2, how long will it take the racer to come to stop?
#8. CW THINK, EXPLAIN, SMALL G, BIG G
P 43 Follow Up Exercise
What is the racer’s instantaneous velocity 10 sec after the parachute is deployed?
Figure 2.10 p 43
Draw graphs of the following
1. V=Vo + at
2. V=Vo –at
3. -V=-Vo –at
4. -V= Vo –at
#9. CW THINK, SOLVE, SMALL G, BIG G
AP PHYSICS B WORKBOOK –
#3-7, 10
#1 NOTES CHAPTER 2 KINEMATICS
26. Example 2.6 p 46.
Two riders on dune buggies sit 10 m apart on a long straight track , facing in opposite
directions . Starting at the same time , both riders accelerate at a constant rate of
2m/s2
a. Make a problem sketch
b. How far apart will the dune buggies be at the end of 3 sec?
# 1 Notes 2-7 Falling Objects
27. Near the surface of the Earth, all objects experience approximately the same
acceleration due to gravity.
This is one of the most common examples of
motion with constant acceleration.
2-7 Falling Objects
28. In the absence of air resistance, all
objects fall with the same acceleration,
although this may be hard to tell by
testing in an environment where there is
air resistance.
2-7 Falling Objects
28. The acceleration due to gravity at the Earth’s
surface is approximately 9.80 m/s2.
FIGURE 2-22
TOSSING AN OBJECT UP
FREE FALL
•
Really, just a special case of one dimensional kinematics.
•
Uses the same equations.
– But, since displacement is vertical, it is written as y (instead of x).
– Acceleration is no longer a variable, it is a fixed value 'g'.
Free Fall
Free Fall: The only acceleration or force is due
to a constant gravity.
29. All objects moving under the influence of
“only gravity” are said to be in free fall.
• 30. Free fall acceleration does not depend on
the object’s original motion, or on the mass.
• 31. Acceleration is the one thing constant to
all falling objects. It is constant for the entire
time they are in the air.
Free Fall
32. Free Fall: The only acceleration or force is
due to a constant gravity.
•This means that the significant figures of force
due to other forces like Air Resistance are
InSignificant.
Freely Falling Objects
• A freely falling object is any object moving under the
influence of gravity alone.
• Acceleration does not depend upon the initial motion of
the object!!!!!!!
33. Examples of Freefall with different initial motions.
– Dropped –released from rest
– Thrown downward
– Thrown upward
– Throwing a ball Up or DOWN AFTER the ball leaves the
hands.
Examples of NonFreefall or NOT FREEFALL.
– Throwing a ball Up or DOWN BEFORE the ball leaves the
hands, while the hands are still pushing on it is NOT
freefall.
– A rocket engine is still burning and pushing up on a
rocket.
Freely Falling Objects
• A freely falling object is any object moving under the
influence of gravity alone.
• Acceleration does not depend upon the initial motion of
the object!!!!!!!
Examples of Freefall with different initial motions.
– Dropped –released from rest
– Thrown downward
– Thrown upward
– Throwing a ball Up or DOWN AFTER the ball leaves the
hands.
So what does initial motion affect then???
It changes the vi in vf = vi + at.
It does not change the a.
Acceleration of Freely Falling Object
• The acceleration of an object in free fall is directed
downward, regardless of the initial motion
34.The magnitude of free fall acceleration is g= 9.80 m/s2
which is the number used in your textbooks, so use 9.80
so that you match the black and white textbooks.
• Only valid on the Earth’s surface at OUR ALTITUDE.
• Not to be confused with g for grams
• British textbooks use 9.81 m/s2 instead of 9.80 m/s2 like
American textbooks. Why?
STRENGTH OF GRAVITY
•
The value of 'g' is affected by altitude.
•
At the top of Mt. Everest, g = -9.77
(.028 % less)
m/s2
Acceleration of Free Fall, cont.
• 35. We will neglect air resistance
• 36. Free fall motion is constantly accelerated motion
in ONLY one dimension, which is y.
DO NOT USE ax = -9.80 m/s2
• Let upward be positive
• 37. Use the kinematic equations with ay= -g= -9.80
m/s2
38.
Acceleration of Free Fall, cont.
• Use the kinematic equations with a= g= -9.80 m/s2
REMEMBER: + means up, - means down
• Write these equations on page:
Kinematics Equation Page (Choose 1 side or other.)
Y Axis
Y Axis
2
2
y
=
y
+
v
t
+
½
at
y = yi + vi t - ½ gt
i
i
v f = vi + a t
v f = vi - g t
2 = v 2 + 2a Δy
2
2
v
vf = vi -2g Δy
f
i
g= -9.80 m/s2
g= 9.80 m/s2
(Because g = Neg)
(Because a = g)
#1 NOTES 1
39. P 52 A boy on the bridge throws a stone vertically downward with an initial
velocity of 14.7 m/s toward the river below. If the stone hits the water 2.0 sec
later , what is the height of the bridge above the water?
# 9 CW Think, Solve , Small G, Big G
Follow up Exercise p 53
#10 CW : THINK, SOLVE, SMALL GROUP, BIG
GROUP
AP PHYSICS WORKBOOK B –
# 8-9
Free Fall –an object dropped
• If I drop a soccer ball off the roof of a
very, very tall building, what is the
velocity 8 seconds later?
• What is the displacement?
Solve these in your powerpoint notes.
Free Fall –an object dropped
If I drop a soccer ball off the roof of a very, very tall
building, what is the velocity 8 seconds later?
v f = vi + a t
Vf = vi - g t
2)(8 s)
v
=
0
+
(-9.80
m/s
8
V8 = 0 - (9.80 m/s2)(8 s)
v8 = -78.4 m
V8 = -78.4 m
What is the displacement?
What is the displacement?
2
y
=
y
+
v
t
+
½
at
8
i
i
y8 = yi + vi t - ½ gt2
2)(8 s)2
y
=
0
+
0
(8)+½
(-9.80
m/s
y8 = 0 + 0 (8) - ½ (9.80 m/s2)(8 s)2 8
y8 = -313 m
y = -313 m
8
Free Fall of an object thrown upward
• Initial velocity is upward, so
positive
• The instantaneous velocity
at the maximum height is
zero
• Vy=max = 0
• ay= -g = -9.80 m/s2
everywhere in the motion:
at the start, top, and end
And everywhere inbetween.
At Top
vy = 0
ay = -g
vi≠0
ay = g
Free Fall -object thrown upward
A soccer ball is kicked directly
upwards with a velocity of 8 m/s.
At Top
vy = 0
ay = -g
At what time will it’s velocity be
zero?
At what time will its height peak?
What’s the highest it will go?
How much time will it spend in the
air before it hits the ground?
With what velocity will it strike the
ground?
Vi > 0
ay = -g
Free Fall -object thrown upward
A soccer ball is kicked directly upwards
with a velocity of 8 m/s.
At Top
At what time will it’s velocity be zero?
Ans: vf = vi + a t  t=.816s = .8s
vy = 0
ay = -g
At what time will its height peak?
Ans: Same Question. Notice vy = 0
At the top.
What’s the highest it will go?
Ans: yf = yi + vi t + ½ at2  3.27m =
3m
Or use vf2 = vi2 + 2a Δy
How much time will it spend in the air
before it hits the ground? Double 1st
ans
t = 1.6s
With what velocity will it strike the
Vi > 0
ay = -g
thrown upwards, free fall only
• The motion may be symmetrical IF it lands at
the same height as thrown.
• Then tup= tdown
• Then vf = -vi
• Break the motion into two parts
• up and down
Free Fall –an object thrown
downward
• ay= -g= -9.80 m/s2
• Initial velocity ≠0
• Since up is positive,
initial velocity will be
negative.
Vo< 0
a = -g
Free Fall --object thrown Downward
Same Equations, but Vo will be
negative
A person fires a handgun
straight downwards from a
helicopter. The muzzle
velocity of the gun is 300
m/s. The helicopter is 250 m
above the ground.
With what velocity will the
bullet hit the ground?
Free Fall --object thrown Downward
Same Equations, but Vo will be negative
A person fires a handgun straight
downwards from a helicopter. The
muzzle velocity of the gun is 300 m/s.
The helicopter is 250 m above the
ground.
With what velocity will the bullet hit the
ground?
ANS: vf2 = vi2 + 2a Δy = -308 m/s
All vi ,a, Δy = negative
vf = choose neg sq.root.
Problem Solving tips
• Think about and understand the situation
• GIVENS: List the given information. REMEMBER g = -9.80 m/s2
and at the top of the arc, vy = 0 (y direction only.)
• DIAGRAM: Make a quick drawing of the situation
• FIND: List the unknown you are solving for.
• EQUATIONS: List the equations used to solve.
• CALCULATIONS: Show your work. Think about units
• SOLUTION: List your answer.
• Focus on the expected result
• Think about what a reasonable answer should be
Example
We throw a bowling ball upwards
from the edge of the roof.
What happens at points B, C and
D?
Example
• Initial velocity at A is upward (+)
and acceleration is -g(-9.80 m/s2)
• At B, the velocity is 0 and the
acceleration is -g(-9.80 m/s2)
• At C, the velocity has the same
magnitude as at A, but is in the
opposite direction
• The displacement is –50.0 m (it
ends up 50.0 m below its starting
point)
Problem Solving –
Question: If a flea can jump .440 m, what is the
initial velocity as it leaves the ground?
Problem Solving –
Question: If a flea can jump .440 m, what is the
initial velocity as it leaves the ground?
y = yi +vi t + ½ at2
v = vi + a t
vf2 = vi2 +2ay
Problem Solving –
Question: If a flea can jump .440 m, what is the
initial velocity as it leaves the ground?
Just look at flea on the way up.
y is .440 m
vi is what we’re looking for
a is -9.8 m/s2 vf is zero, since we’re looking at
the top of his jump.
Problem Solving –
Question: If a flea can jump .440 m, what is the
initial velocity as it leaves the ground?
vf2 = vi2 +2ay
Just look at flea on the way up.
y is .440 m
vi is what we’re looking for
a is -9.8 m/s2 vf is zero, since we’re looking at
the top of his jump.
Problem Solving –
Question: If a flea can jump .440 m, what is the
initial velocity as it leaves the ground?
Vf2 = vi2 +2ay
0 = vi 2 + 2 (-9.80m/s2)(.440m)
Problem Solving –
Question: If a flea can jump .440 m, what is the
initial velocity as it leaves the ground?
Vf2 = vi2 + 2
a
y
0 = vi 2 + 2 (-9.80m/s2)(.440m)
= vi 2 - 8.62 m2/s2
so
vi = + / - 2.94 m/s
Which makes sense, + or -?
Problem Solving –
Question: If a flea can jump .440 m, how much
time does he spend going up?
yf = yi +vi t + ½ at2
v f = vi + a t
2
2
vf = vi +2ay
Problem Solving –
Question: If a flea can jump .440 m, how much
time does he spend going up?
v f = vi + a t
a is -9.8 m/s2
vi = 2.94 m/s
-9.8 m/s2 = (0 - 2.94 m/s) / t
t = .3sec
vf = 0
Problem Solving –
What is the velocity when he lands?
vf2 = vi2 +ay
Just look at the problem on the way down.
vf2 = 0 + (-9.80m/s2)(-.440m)
= 0 + 8.62 m2/s2
vf = +/- 2.94 m/s
Which makes sense, + or - ?
PROBLEM SOLVING
•
How much time does he spend in the air?
yf = yi +vi t + ½ at2
v f = vi + a t
2
2
vf = vi +2ay
PROBLEM SOLVING
•
How much time does he spend in the air?
•
f
-2.94m/s = 2.94m/s + (-9.81m/s2)*t
•
-5.82m/s = -9.81m/s2*t
•
t = .6 sec
•
Twice the time it took to go up
v = vi + a t
BENCHMARK :KINEMATICS
PRINCETON REVIEW P 31 1-10
FREE RESPONSE 1-2