Physics 231 Topic 1b: One dimensional motion Wade Fisher August 31 2012

Physics 231
Topic 1b: One dimensional motion
Wade Fisher
August 31 2012
MSU Physics 231 Fall 2012
1
Nuances of Lon-CAPA
Reporting sig figs and units is nuanced.
Lon-CAPA is not always 100% consistent.
Example:
You calculate your answer to be 590.212 cm2
The problem gives you 3 sig figs (eg, area of a
circle is 74.4 cm2)
You should report 590 cm2 (3 sig figs)
Lon-CAPA may not accept 590 cm2. But it accepts
5.90E2 cm2, 5.90E-2 m2, and 590.2 cm2 (??)
My advice: Think carefully about sig figs and try a few
different ways to report answers
MSU Physics 231 Fall 2012
2
Recall from Last Lecture:
Syllabus
SI Units
 Mass: kilograms (kg)
 Length: meters (m)
 Time: seconds (s)
Units and conversions
Dimensional Analysis
Scientific Notation (e.g., 4.3x1012 and 2.3E4)
Significant Figures
MSU Physics 231 Fall 2012
3
Clicker Quiz:
MSU Physics 231 Fall 2012
4
Key Concepts: 1D Motion
Position and coordinate systems
Displacement vs. Distance
Velocity
 Distinguish from speed
 Average vs instantaneous velocity
Acceleration
 Relationship to velocity
 Impact on position in 1D motion
Constant velocity and constant acceleration equations
 How to set up & solve problems
 Understanding graphical representations of distance, velocity, acceleration
Covers chapter 2 in Rex & Wolfson
MSU Physics 231 Fall 2012
5
Galileo (1564 – 1642)
What mathematical forces
govern accelerated motion?
A wooden and stone sphere are
dropped from the tower of pisa.
Which one reaches the earth first?
Answer: they hit the ground at the same time!
MSU Physics 231 Fall 2012
6
Coordinate Systems
We use coordinate systems as a
graphical representation of the
relationships of measurements
 Distance vs time
 Espressos sold per shift
 Cell phone minutes used per day
MSU Physics 231 Fall 2012
7
Coordinate Systems
MSU Physics 231 Fall 2012
8
Position & Displacement
Displacement is a vector and thus has direction (sign)
Displacement is not equal to Distance
MSU Physics 231 Fall 2012
9
Coordinate Systems
MSU Physics 231 Fall 2012
10
Position vs. Time
Equally-spaced photographs of a runner
reveal his position as a function of time
MSU Physics 231 Fall 2012
11
Position vs Time Graph
These measurements can be
translated to a 2D “Position vs
Time” coordinate system.
MSU Physics 231 Fall 2012
12
Average Velocity & Speed
c
Average velocity=
Displacement
Time interval
x = 13.6m
x = 20.4m
t = 2.0s
t = 2.0s
x x f  xi
v

t t f  ti
MSU Physics 231 Fall 2012
13
Vectors vs Scalars
X=-30
t=1
X=0
t=0
X=30
x x f  xi
=-30
v

t
t f  ti
speed 
x
t

x f  xi
t f  ti
X=45
Average velocity: vector
=+30
Average speed: scalar
MSU Physics 231 Fall 2012
14
Speed vs Velocity
A driver commutes from home in the suburbs to work in the city
(each way = 60 km). The morning trip takes 60 minutes (1 hour).
The drive home takes 90 minutes (1.5 hours).
What was his average speed and average velocity over
The whole trip?
a)
b)
c)
d)
e)
Average speed: 60 km/h
Average speed: 48 km/h
Average speed: 60 km/h
Average speed: 0 km/h
No way to tell.
Average velocity: 48 km/h
Average velocity: 0 km/h
Average velocity: 0 km/h
Average velocity: 60 km/h
MSU Physics 231 Fall 2012
15
Instantaneous Velocity
“But officer, my
average speed was
only 25 miles per
hour...”
Sometimes we want to know the speed at one particular point in time.
MSU Physics 231 Fall 2012
16
Instantaneous Velocity
The velocity at a single point in time is
given by the slope of the line tangent to
the x-t curve at that time.
Conceptualize this condition as a
distance measurement in a very
tiny time interval
MSU Physics 231 Fall 2012
x
v  lim
t 0 t
17
x
Acceleration
A change in slope in the
x-t graph…
t
v
…means a change in velocity:
Acceleration
MSU Physics 231 Fall 2012
t
18
Acceleration
vf
Average acceleration:
average change in velocity
v
vi
ti
tf
t
MSU Physics 231 Fall 2012
19
Instantaneous Acceleration
acceleration at one point in time:
v
a  lim
t 0 t
v
Slope of the tangent to the
v-t curve at that point in time
t
MSU Physics 231 Fall 2012
20
Question
v
t=0
v = 20
t=1
v=5
t
What is the average acceleration between t=0 and t=1
A) 15
B) -15
C) 0
MSU Physics 231 Fall 2012
D) infinity
21
Question
GALILEO!
Which v-t diagram matches which a-t diagram?
MSU Physics 231 Fall 2012
22
Clicker Question
Which v-t diagram matches the
x-t diagram on the right?
MSU Physics 231 Fall 2012
23
Overview up to this point
VECTORS
• Displacement
• Average Velocity
• Instantaneous velocity
• Average acceleration
• Instantaneous acceleration
SCALARS
• Distance
•Average speed
•Instantaneous speed
MOTION DIAGRAMS
MSU Physics 231 Fall 2012
24
Constant Acceleration
Velocity at time
t equals…
velocity v (m/s)
v(t )  v0  at
v(t)
v0
t0
Velocity at t=0…
t time (s)
Plus the gain in velocity per second multiplied by
the time elapsed
(every second, the velocity increases with a m/s)
MSU Physics 231 Fall 2012
25
Constant Acceleration II
x(t )  xo  v t  xo 
( v0  vt )
2
t  xo  vot  at
1
2
2
Start position plus average
velocity multiplied by time
Substitute
v 0  vt
v
2
Substitute
MSU Physics 231 Fall 2012
v(t )  v0  at
26
Constant motion
x(m)
x(t)=x0
t
v
m/s
v(t)=0
t
v(t)=v0+at
t
a
m/s2
a(t)=0
t
v
m/s
v(t)=v0
t
a
m/s2
Constant acceleration
2
x(t)=x
+v
t
x(t)=x
+v
t+½at
0
0
0
0
x(m)
x(m)
t
v
m/s
Constant velocity
t
a
m/s2
a(t)=0
t
MSU Physics 231 Fall 2012
a(t)=a0
t
27
v
(m/s)
t=1,v=2
2
0
v
(m/s)
0
T (s)
t=1,v=2
2
0
0
1)What is the distance covered in 1 second?
2)What is the area indicated by
?
MSU Physics 231 Fall 2012
T (s)
Q 1.
2.
a)
b)
c)
d)
1.
2.
1.
2.
1.
1.
2.
2.
28
Gravitational Free Fall
5 kg
A
B
1 kg
100 m
x(t )  xo  vo t  at
v(t )  v0  at
1
2
2
a is the acceleration felt due to gravitation
(it is called ‘g’ and its value is 9.81 m/s2)
g does not depend on the mass of falling object
MSU Physics 231 Fall 2012
29
Clicker Question
A person throws a ball straight up in the air.
Which of the following is true about the acceleration and
velocity at the highest point the ball reaches?
a)
b)
c)
d)
Velocity is 0,
acceleration is not 0.
Velocity is not 0, acceleration is 0.
Both velocity and acceleration are 0.
Neither velocity nor acceleration are 0.
MSU Physics 231 Fall 2012
30
Galileo’s Free Fall Experiment
MSU Physics 231 Fall 2012
31
Galileo’s Free Fall Experiment
Galileo throws a canon ball from the tower of Pisa. Ignoring
friction what is the distance covered between t=1 and t=3
seconds? The initial velocity was 0 m/s.
MSU Physics 231 Fall 2012
32
For Next Week
Rex & Wolfson Chapter 3:
Motion in 2 Dimensions
Homework Set 1: Covers Chapters 1 & 2
Due Friday 9/7 @ 11PM
MSU Physics 231 Fall 2012
33
Point Particle Treatment
The skier can be represented by a
simple, point-like object

MSU Physics 231 Fall 2012
x
34
velocity
acceleration
motion
+
0
traveling with constant speed in the positive direction
+
+
traveling with increasing speed in the positive direction
+
-
traveling with decreasing speed in the positive direction
0
0
0
+
standing still
0
-
accelerating towards the negative direction
-
0
traveling with constant speed in the negative direction
-
+
traveling with decreasing speed in the negative direction
-
-
traveling with increasing speed in the negative direction
accelerating towards the positive direction
PHY 231
MSU Physics 231 Fall 2012
35
35
V vs T Diagram Example
The figure shows velocity versus time.
a) What is the displacement after 4 s?
b) What is the average velocity?
a) The distance covered is the area under the v-t diagram.
x= ½x2x4 + 1x4 + (2+½x1x2) = 11 m
b) Average velocity = x/t = 11m/4s = 2.75 m/s
MSU Physics 231 Fall 2012
36
Instantaneous Velocity
Example
x x f  xi
v

t
t f  ti
The velocity at one point in time is
the slope of the tangent to the x-t
curve at that time
x
t
What is the velocity at point C?
v = x/t = (105-85)/(3-1)
= 20/2 = 10 m/s
x
t
What is the velocity at point E?
v = x/t = (35-75)/(4.5-3.5)
= -40/1 = -40 m/s
Be careful with the sign: velocity is a vector and can be negative!
MSU Physics 231 Fall 2012
37
Ball Toss Problem
Dave throws a ball straight up with a velocity of 5.0 m/s.
1) How high does it go relative to the height at which it was
released?
2) After how much time does it reach that height?
x(t) = x(0) +v(0)t + ½at2
v(t) = v(0) + at
x(t) = 0 + 5t + ½(-9.8)t2
v(tmax) = 0 = 5 - 9.8tmax
tmax = 5/9.8 = 0.51 s
it decelerates
at it highest point, the velocity is zero
time to reach highest point
xmax = x(t=0.51s) = 50.51 - ½9.8(0.51) 2 = 1.3 m
MSU Physics 231 Fall 2012
38
A vs T Diagram Example
a (m/s2)
Given this a-t diagram, calculate:
a) Velocity after 4 s
b) Distance traveled in 4 s
The object was originally at rest.
3
2
1
0
1
2
3
4
t (s)
a) v(t) = v(0)+at
After 2 seconds: v(2) = 0 + 1x2 = 2 m/s
After 4 seconds: v(4) = v(2) + 2x2 = 2+4 =6 m/s
b) x(t) = x(0) + v(0)t + ½at2
After 2 seconds: x(2) = 0+0 + ½x1x22 = 2 m
After 4 seconds: x(4) = x(2) + v(2)t + ½at2 = 2 + 2x2 + ½x2x22 = 10 m
MSU Physics 231 Fall 2012
39
Area of triangle: ½ base x height = ½x4ux5u = 10u2
(u: unit of length in drawing above)
10u2 = 68.4cm2

1 u2 = 6.84cm2
Full square: 9x9u2 = 81u2 corresponds with 81x6.84 = 554 cm2
MSU Physics 231 Fall 2012
40