# Kinematics

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```					                                      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

 Please  come down and fill in every
available seat – the lecture is full.
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 Remember to set your iclicker to the
right frequency [ push on, hold, B B ]
 Office hours now underway
 Read the course descriptions & FAQ on
the course web site

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
 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

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