elegy for the arctic week 3, tuesday oct 10th by ludovico...
TRANSCRIPT
Instructor:HeatherHillEmail:[email protected]
Office:MH111
GS104Week3,Tuesday
Oct10thElegyfortheArcticbyLudovicoEinaudi
Which position-versus-time graph goes with this velocity-versus-time graph on the left? The particle’s position at ti = 0 s is xi = –10 m .
1. A2. B3. C4. D
Which position-versus-time graph goes with this velocity-versus-time graph on the left? The particle’s position at ti = 0 s is xi = –10 m .
1. A2. B3. C4. D
Theslopeatapointonaposition-versus-timegraphofanobjectis
A. Theobject’sspeedatthatpoint.B. Theobject’svelocityatthatpoint.C. Theobject’saccelerationatthatpoint.D. Thedistancetravelledbytheobjectto
thatpoint.
Theslopeatapointonaposition-versus-timegraphofanobjectis
A. Theobject’sspeedatthatpoint.B. Theobject’svelocityatthatpoint.C. Theobject’saccelerationatthatpoint.D. Thedistancetravelledbytheobjectto
thatpoint.
Theareaunderavelocity-versus-timegraphofanobjectis
A. Theobject’sspeedatthatpoint.B. Theobject’saccelerationatthatpoint.C. Thedistancetravelledbytheobjecttothat
point.D. Theobject’svelocityatthatpoint
Theareaunderavelocity-versus-timegraphofanobjectis
A. Theobject’sspeedatthatpoint.B. Theobject’saccelerationatthatpoint.C. Thedistancetravelledbytheobjecttothat
point.D. Theobject’svelocityatthatpoint
Theslopeatapointonavelocity-versus-timegraphofanobjectis
A. Theobject’sspeedatthatpoint.B. Theobject’saccelerationatthatpoint.C. Thedistancetravelledbytheobject.D. Theobject’svelocityatthatpoint.
Theslopeatapointonavelocity-versus-timegraphofanobjectis
A. Theobject’sspeedatthatpoint.B. Theobject’saccelerationatthatpoint.C. Thedistancetravelledbytheobject.D. Theobject’svelocityatthatpoint.
slope slope
Position Velocity Acceleration
area under curve area under curve
x v a
Graphical Representation
Position Velocity Acceleration
x v a
a = g = 9.8 m/s2
You drop a ball from rest off the second story of MH, the ball starts 5 meters off the ground. How many seconds does it take to reach the ground?
First thing that you do in problem solving is
always: Draw a PictureAnd, write down what you know:
v0 = 0 m/s g = 9.8 m/s2
x0 = 5 m xf = 5m
And, what you don’t know and want to know:t = ?
First thing that you do in problem solving is
always: Draw a PictureAnd, write down what you know:
v0 = 0 m/s g = 9.8 m/s2
x0 = 5 m xf = 0 m
And, what you don’t know and want to know:t = ?
First thing that you do in problem solving is
always: Draw a PictureAnd, write down what you know:
v0 = 0 m/s g = 9.8 m/s2
x0 = 5 m xf = 0 m
And, what you don’t know and want to know:t = ?
A quick note on direction
• There is no universal coordinate system in physics.
• Each problem will you require to define a new coordinate system
• Choose a coordinate system that makes your life easy
If I drop a lead brick and a wood brick off the second floor of MH, which is going to hit the
ground first?
1. The lead brick will hit the ground first.
2. The wood block will hit first.
3. They will both hit at the same time.
4. Neither will ever hit the ground.
If I drop a lead brick and a wood brick off the second floor of MH, which is going to hit the
ground first?
1. The lead brick will hit the ground first.
2. The wood block will hit first.
3. They will both hit at the same time.
4. Neither will ever hit the ground.
Kinematic Equations
Answer: “–
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
Kinematic Equations
Answer: “–
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
Kinematic Equations
Answer: “–
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
Kinematic Equations
Answer: “–
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
Kinematic Equations
Answer: “–
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
Kinematic Equations
Answer: “–
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
Kinematic Equations
Answer: “–
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
Caution:Onlyusewhenaccelerationisconstant
Kinematic Equations
Answer: “–
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
GS 104, Homework 1 Due Sept 28th
1 Physics Subtopic
⌃
~
F
net
= m~a (1)
Distance
Time
(2)
Velocity
Time
(3)
v
f
= v
0
+ at (4)
x
f
= x
0
+ v
0
t +
1
2
at
2
(5)
�x = v
0
t +
1
2
at
2
, �x = x
f
� x
0
(6)
�x =
✓v
f
� v
0
2
◆t (7)
1
Caution:Onlyusewhenaccelerationisconstant
Disclaimer:Onlyuseforonedirectionatatime
Review question:
Which of the following are vectors: a.Positionb.Velocityc. Accelerationd.All of the abovee.B and C only
Review question:
Which of the following are vectors: a.Positionb.Velocityc. Accelerationd.All of the abovee.B and C only