Exam I
Physics 101: Lecture 03
Kinematics
Today’s lecture will cover Textbook
Sections 3.1-3.3
HW 1 & 2 are due on Tuesday Feb 2 at 6am!
Physics 101: Lecture 3, Pg 1
Announcements
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Physics 101: Lecture 3, Pg 2
Force at Angle Example
A person is pushing a 15 kg block across a floor with mk= 0.4 at a
constant speed. If she is pushing down at an angle of 25 degrees, what
is the magnitude of her force on the block?
x- direction: SFx = max Combine:
Fpush cos(q) – Ffriction = 0 Fpush cos(q) / m–mg – FPush sin(q) = 0
Fpush cos(q) – m FNormal = 0 Fpush ( cos(q) / m - sin(q)) = mg
FNormal = Fpush cos(q) / m Fpush = m g / ( cos(q)/m – sin(q))
y- direction: SFy = may Fpush = 80 N
FNormal –Fweight – FPush sin(q) = 0
Normal
FNormal –mg – FPush sin(q) = 0
Pushing y
q x
q Friction
Weight
Physics 101: Lecture 3, Pg 3
Homework 2 Example
Calculate the tension in the left string.
Work through example? y
A: Yes B: No
x
x-direction: SF=ma
-TL+TR cos(q)= 0 TR
TL q
TL = TR cos(q)
y-direction: SF=ma W
TR sin(q) – Mg = 0 Combine:
TR = Mg / sin(q) TL = Mg cos(q)/sin(q)
Physics 101: Lecture 3, Pg 4
Overview
Kinematics: Description of Motion
Position and displacement
velocity
» average
» instantaneous
acceleration
» average
» instantaneous
Physics 101: Lecture 3, Pg 5
Position vs Time Plots
Gives location at any time.
Displacement is change in position.
Slope gives velocity.
x (m)
3
Position at t=3, x(3) = 1
Displacement between t=5 and t=1. Dx = -1.0 m
t
4
1.0 m - 2.0 m = -1.0 m
Average velocity between t=5 and t=1. v = -0.25 m/s -3
-1 m / 4 s = -0.25 m/s
Physics 101: Lecture 3, Pg 6
Velocity vs Time Plots
v (m/s)
Gives velocity at any time.
Area gives displacement 3
Slope gives acceleration. 1.5
6
t
velocity at t=2, v(2) = 3 m/s 4
Displacement between t=0 and t=3: Dx= 7.5 m -3
t=0 to t=1: ½ (3m/s) (1 s) = 1.5 m
t=1 to t=3: (3m/s) (2 s) = 6 m
Average velocity between t=0 and t=3? v= 7.5 m / 3s = 2.5 m/s
Change in v between t=5 and t=3. Dv = -2 m/s – 3 m/s = -5 m/s
Average acceleration between t=5 and t=3: a = -5 m/s / (2 s) = -2.5 m/s2
Physics 101: Lecture 3, Pg 7
Acceleration vs Time Plots
Gives acceleration at any time.
Area gives change in velocity
a (m/s2)
3
Acceleration at t=4, a(4) = -2 m/s2
6
Change in v between t=4 and t=1. Dv = +4 m/s
t
t=1-3: Dv = (3m/s2)(2s) = 6 m/s 24
t=3-4: Dv = (-2m/s2)(1s) = -2 m/s -3
Physics 101: Lecture 3, Pg 8
Acceleration Preflights
Is it possible for an object to have a positive
velocity at the same time as it has a negative
acceleration?
90% 1 - Yes “An object may still be moving forward
10% 2 - No while it is slowing down.”
If the velocity of some object is not zero,
can its acceleration ever be zero ?
92% 1 - Yes “yes because they are two separate
measurements.”
8% 2 - No “the object is still in motion so it will still
have an acceleration”
Physics 101: Lecture 3, Pg 9
Velocity ACT
If the average velocity of a car during a trip along a straight
road is positive, is it possible for the instantaneous velocity at
some time during the trip to be negative?
A - Yes
B - No
Drive north 5 miles, put car in reverse
and drive south 2 miles. Average
velocity is positive.
Physics 101: Lecture 3, Pg 10
Dropped Ball y
•A ball is dropped from a height of x
two meters above the ground.
Draw vy vs t
v v v
9 A 9 B 9 C
0.5 t 0.5 t 0.5 t
-6 -6 -6
v v
9 9 E
D
0.5 t 0.5 t
-6 -6 Physics 101: Lecture 3, Pg 11
Dropped Ball x
A ball is dropped for a height of two
meters above the ground. t
v
Draw v vs t
Draw x vs t t
Draw a vs t
a
t
Physics 101: Lecture 3, Pg 12
Tossed Ball
•A ball is tossed from the ground up a height of two
meters above the ground. And falls back down y
x
Draw v vs t
v v v
9 A 9 B 9 C
1 t 1 t 1 t
-6 -6 -6
v v
9 9
D E
1 t 1 t
-6 -6 Physics 101: Lecture 3, Pg 13
Tossed Ball x
•A ball is tossed from the ground up a
height of two meters above the ground. t
And falls back down v
Draw v vs t
Draw x vs t t
Draw a vs t
a
t
Physics 101: Lecture 3, Pg 14
ACT
A ball is thrown straight up in the air and returns to its
initial position. During the time the ball is in the air,
which of the following statements is true?
A - Both average acceleration and average velocity are zero.
B - Average acceleration is zero but average velocity is not zero.
C - Average velocity is zero but average acceleration is not zero.
D - Neither average acceleration nor average velocity are zero.
Vave = DY/Dt = (Yf – Yi) / (tf – ti) = 0
aave = DV/Dt = (Vf – Vi) / (tf – ti)
Not 0 since Vf and Vi are
not the same ! 101: Lecture 3, Pg 15
Physics
x (meters)
Example
100
0
-100
• Where is velocity zero?
-200
position vs. time • Where is velocity positive?
• Where is velocity negative?
•
-300
0
v (m/s)
5 10 15 20 Where is speed largest?
t (seconds)
20 • Where is acceleration zero?
0 • Where is acceleration positive?
-20
-40
-60 velocity vs. time
-80
-100
0 5 10 15 20
t (seconds) Physics 101: Lecture 3, Pg 16
Summary of Concepts
kinematics: A description of motion
position: your coordinates
displacement: Dx = change of position
velocity: rate of change of position
average : Dx/Dt
instantaneous: slope of x vs. t
acceleration: rate of change of velocity
average: Dv/Dt
instantaneous: slope of v vs. t
Physics 101: Lecture 3, Pg 17