physics 2111 unit 1 · 2021. 1. 31. · physics 2111 unit 1 things for today: displacement,...
TRANSCRIPT
Physics 2111
Unit 1
Things for Today:▪ Displacement, Velocity, Acceleration – graphically
▪ Displacement, velocity, acceleration - numerically
▪ 1-D Kinematics with constant acceleration
▪ Free-fall
Mechanics Lecture 1, Slide 1
What we asked about
Mechanics Lecture 1, Slide 2
• It honestly took me a while to do the second checkpoint exercise because I get a bit
confused as to when to use certain equations. Are there key signals in word
problems that make it easier on where to start?
• I found constant acceleration to be the most difficult for me.
• for now, some of the parts seem to be easy and the problems are also easy to
solve.
• In the pre-lecture some of the integral equations were slightly confusing, but it might
just be because the videos themselves were condensed.
• Im not sure why the ball rolling down the ramp had a velocity of 3 feet/sec instead of
2 feet/sec or even 4 feet/sec.
• I remember most of the topics since when I took physics in high school I feel that I
just need to see some more examples and do practice to understand it fully.
• The rolling down the ramp question was kinda confusing. I am not sure how the
answer was 3. Is there a specfic equation to use or is it more common sense?
• I would like to discuss the relationships between the position/velocity/acceleration
graphs and how they affect each other. If there is time to review kinematic
equations that would be great as well.
• I think the constant acceleration equations were the hardest to understand.
• I found it more confusing that the equation is r = r0 + v0t + at^2/2 in the formula
sheet and not x = x0 + v0t + at^2/2 since I tend to think of r as radius and x as
distance/displacement.
• Reading graphs
Mechanics Lecture 1, Slide 3
Think of what the position graph would look like for someone who started 10 meters from the origin and walked away at 2 meters per second.
po
siti
on
Displacement and Velocity in One Dimension
time
Mechanics Lecture 1, Slide 4
Displacement (rise)
Time taken (run)
Displacement and Velocity in One Dimension
Rise
= Slope
Run
Speed = |v(t)| when moving in one direction
The v(t) vs. t plot is just theslope of the x(t) vs. t plot
Displacement and Velocity in One Dimension
Mechanics Lecture 1, Slide 5
Definition:
• Speed = path / time
• Average Velocity = displacement / time
Displacement and Velocity in One Dimension
Mechanics Lecture 1, Slide 6
A) YESB) NO
Are the plots shown at the left correctly related
The velocity vs. time plot of some object is shown to the right.
Which diagram below could be the Displacement vs. time plot for the same object?
A B C
QUESTION
Mechanics Lecture 1, Slide 7
QUESTION
Mechanics Lecture 1, Slide 8
The graph represents the position as a function of time for an object in one-dimensional motion. Identify which statement below correctly identifies the times at which the velocity of the object is greater than zero.
va) The velocity is greater than 0 at t = 0 s and t = 4 s.
b) The velocity is greater than 0 at t = 2 s.
c) The velocity is greater than 0 at t = 0 s and t = 1 s.
d) The velocity is greater than 0 at t = 3 s and t = 4 s.
e) The velocity is greater than 0 at t = 0, 1, 3, and 4 s.
Mechanics Lecture 1, Slide 9
Acceleration
Question
At what point on the velocity graph is the acceleration zero?
A) A
B) B
C) C
D) both A and C
Mechanics Lecture 1, Slide 10
A
B
C
Mechanics Lecture 1, Slide 11
NOTES
Mechanics Lecture 1, Slide 12
For the Displacement and Velocity curves shown on the left, which is the correct plot of acceleration vs. time?
A
B
Checkpoint 1
QUESTION
Mechanics Lecture 1, Slide 13
The graph represents the velocity as a function of time for an object in one-dimensional motion. Identify which statement below correctly describes the motion at t = 3 s
va) The object is slowing down and the acceleration is negative.
b) The object is speeding up and the acceleration is negative.
c) The object is slowing down and the acceleration is positive.
d) The object is speeding up and the acceleration is positive.
QUESTION
Mechanics Lecture 1, Slide 14
Which pair of these graphs could represent the same motion?
v
x v
v a
(1) (2)
(3) (4)
a) (1) and (3)
b) (2) and (3)
c) (3) and (4)
d) (1) and (2)
e) None of them
QUESTION
Mechanics Lecture 1, Slide 15
In which of the above graphs could the object reverse its direction?
v
x v
v a
(1) (2)
(3) (4)
a) (1)
b) (2)
c) (3)
d) (4)
e) (1) and (2)
As shown to the right, a physics student is moving to the right with the positive direction to the right. She originally is jogging at 3m/sec, but in a period of 4 seconds, she slows to a walk of 1m/sec. What is her acceleration?
QUESTION
Mechanics Lecture 1, Slide 16
+
a) 4.0 m/sec/secb) -4.0 m/sec/secc) 0.5m/sec/secd) -0.5 m/sec/sece) none of the above
QUESTION
Mechanics Lecture 1, Slide 17
As shown above, a physics student is moving to the right with the positive direction to the right. She originally is walking at 1m/sec, but in a period of 2 seconds, she speeds up to a jog of 3m/sec. What is her average acceleration?
+
a) 1.0m/sec/secb) -1.0 m/sec/secc) 4.0 m/sec/secd) -4.0 m/sec/sec e) none of the above
As shown above, a physics student is moving to the left with the positive direction to the right. She originally is walking at 1m/sec, but in a period of 2 seconds, she speeds up to a jog of 5m/sec. What is her average acceleration?
QUESTION
Mechanics Lecture 1, Slide 18
+
a) 2.0m/sec/secb) -2.0 m/sec/secc) 4.0 m/sec/secd) -4.0 m/sec/sec e) none of the above
QUESTION
Mechanics Lecture 1, Slide 19
In the above sketch, to the right is positive. A student is originally walking at 2m/sec to the right, but during a period of 2 seconds, she has reversed direction and is walking to the left at 2m/sec. What is her average acceleration during these 2 seconds?
+
a) 0.0 m/sec/secb) 2.0m/sec/secc) -2.0 m/sec/secd) 4.0 m/sec/sece) -4.0 m/sec/sec
Mechanics Lecture 1, Slide 20
In physics, what is the difference between “accelerating” and “deaccelerating”?
QUESTION
a) One is speeding up, the other is slowingdown.
b) One is moving to the left, the other ismoving to the right.
c) We don’t say “deaccelerating”in physics
Does “acceleration” always mean “speeding up”?
(A) Yes (B) NO
Does “positive acceleration” always mean “speeding up”?
(A) Yes (B) NO
Mechanics Lecture 1, Slide 21
Example Problem 1.1
The position in meters of an object moving along a straight line is given by the equation
x = 6m + 9.0m/sec2*t2 -2.0m/sec*t -6.0m/sec3*t3 .
a) What is its average velocity between 2 seconds and 4 seconds?
b) What is its instantaneous velocity at t=3sec?
Mechanics Lecture 1, Slide 22
To the right is the position vs time graph for two objects. When is the velocity of the objects the same?
QUESTION
xa) (A)
b) (B)
c) (C)
d) (D)
e) (A) and (C)
timeA B DC
Mechanics Lecture 1, Slide 23
Constant Acceleration
constant
a(t) = a
Mechanics Lecture 1, Slide 24
Example Problem 1.2
You are cruising along at 15m/sec, when you pass a police car stopped along the side of the road. 10 meters after you pass the car, you begin to speed up at 2.0m/sec2. How far are you from the police car 5 seconds after you begin to accelerate?
1) Draw a sketch2) Draw a coordinate
system3) What are you looking
for?
Example 1.10 (When does clock start?)
Objects don’t always start to move at
t = 0.
Mechanics Lecture 1, Slide 25
Example:
3 sec after t=0, a cart initially at
rest begins to accelerate at
a=2m/sec2.
What is its position at t=6sec?
Question
Mechanics Lecture 1, Slide 26
3 sec after t=0, a cart initially at rest
begins to accelerate at a=2m/sec2.
What is its position at t=6sec?
a) x = ½*(2)*t2
b) x = ½*(2)*t2 -3
c) x = 3 + ½*(2)*t2
d) x = ½*(2)*(t-3)2
e) x = ½*(2)*(t+3)2
Example 1.11 (2 Carts, 2 Times)
Mechanics Lecture 1, Slide 27
At t=0, blue toy cart starts with a velocity of
10m/sec and an acceleration of -2m/sec2.
4 sec later, a red toy cart starts from the
same position and moves with a constant
velocity of 6m/sec in the same direction.
How long before the blue cart briefly
comes to rest?
How long before the carts collide?
QUESTION
When I ask how long before the carts collide,
algebraically, I’m asking what is t when
Mechanics Lecture 1, Slide 28
a) |xf-blue| is max
b) |xf-red| is max
c) xf-red = xf-blue
d) |xf-red + xf-blue| is max
Mechanics Lecture 1, Slide 29
At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1 foot betweent = 0 sec and t = 1 sec.
How far does it move between t = 1 sec and t = 2 sec?
A) 1 foot B) 2 feet C) 3 feet D) 4 feet E) 6 feet
Checkpoint 2
Mechanics Lecture 1, Slide 30
the ball is accelerating (gaining speed) while it is rolling down. so that means at t = 1 the ball will be traveling faster than it was at t = 0 therefore it will travel a greater distance in the same amount of time for those two intervals..
B (2 feet)
I used the equation for displacement from motion with constant Acceleration to solve for acceleration using the information from the first second. I found that the acceleration was 2 ft/sec^2. I then put that into the displacement equation using the time and velocity for the interval from 1 to 2 seconds and got 3 feet.
C (3 feet)
The ball is moving 1ft every second, thus it is moving at a velocity of 1ft/sec. The elapsed time between 1 and 2 seconds is 1 second which means that it has only moved 1ft that second
A (1 foot)
Checkpoint 2 Responses
I chose this answer because acceleration is meter*second*second, meaning that the acceleration of the ball is exponential throughout its trip down the ramp with respect to time. From this logic, I found that the ball would have covered 4 feet over the time interval
D (4feet)
Mechanics Lecture 1, Slide 31
1ft4ft
9ft
?
16ft
At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1 foot betweent = 0 sec and t = 1 sec.
How far does it move between t = 1 sec and t = 2 sec?
A) 1 foot B) 2 feet C) 3 feet D) 4 feet E) 6 feet
QUESTION
Mechanics Lecture 1, Slide 32
1ft4ft
9ft
?
16ft
Graphical View
4m - 1m = 3m
Mechanics Lecture 1, Slide 33
At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1 foot betweent = 0 sec and t = 1 sec.
How far does it move between t = 3 sec and t = 4 sec?
A) 3 feet B) 4 feet C) 7 feet D) 9 feet E) 16 feet
Follow up
Mechanics Lecture 1, Slide 34
1ft4ft
9ft
?
16ft
ACT 5
16m - 9m = 7m
Free fall
Straight forward example of constant
acceleration is “free fall”
If air resistance is negligible, all
objects fall with the same rate of
acceleration
g = 9.81m/sec2
Mechanics Lecture 1, Slide 35
Example Problem 1.4
You drop a small rock from the top of
a 300meter tall building.
What is its velocity after 5
seconds?
What is its position after 5
seconds?
Mechanics Lecture 1, Slide 36
(Note: This is a pretty tall building. The Sears Tower is 527 meters tall.)
Example Problem 1.5
You drop a small rock from the top of
a 300meter tall building.
How long does it take to
hit the ground?
What is its velocity just
before it hits?
Mechanics Lecture 1, Slide 37
Example Problem 1.6
You throw a small rock downwards
with a velocity of 20m/sec from the
top of a 300meter tall building.
How long does it take to
hit the ground?
What is its velocity just
before it hits?
Mechanics Lecture 1, Slide 38
Example Problem 1.7
You throw a small rock upwards with
a velocity of 20m/sec from the top of
a 300meter tall building.
What is its velocity just before it hits?
Mechanics Lecture 1, Slide 39
Example Problem 1.7
You throw a small rock upwards with
a velocity of 20m/sec from the top of
a 300meter tall building.
What is its velocity just before it hits?
Mechanics Lecture 1, Slide 40
How long does it take to hit the
ground?
Example Problem 1.8 (Ball Toss)
Mechanics Lecture 1, Slide 41
A person throws a ball vertically upward into the air with an initial velocity of 15.0 m/s.
➢ How high does the ball go?
➢ How long is the ball in the air before it comes back to the persons hand?
Example 1.9 (Non-constant acceleration)
Can use kinematic concepts with non-
constant acceleration.
Mechanics Lecture 1, Slide 42
Example:
a(t) = 8.0m/sec2 – 0.4m/sec3*t
At t = 0sec, x=0m, v=20m/sec.
What is xf at t=5sec?
Example 1.9 (Non-constant acceleration)
Can use kinematic concepts with non-
constant acceleration.
Mechanics Lecture 1, Slide 43
Example:
a(t) = 8.0m/sec2 – 0.4m/sec3*t
At t = 0sec, x=0m, v=20m/sec.
What is xf at t=5sec?
Motion Diagrams
Mechanics Lecture 1, Slide 44
Imagine you took multiple exposures
every second of an object as it moved.
QUESTION
Mechanics Lecture 1, Slide 45
1 2 3 4 5
1 2 3 4 5
Both cars in the above
motion diagram are
moving in the positive
direction. Which position
graph might represent
the red Car A?
+
A
B
tt
x
(A) (B)
(C) (D)
t
x
t
x
t
x
You are driving your fire-engine red Toyota Prius (with the great CD player) along at 50km/hr(14m/s) when a dog runs out into the road in front of you. It takes you 0.5 second to react and slam on the brakes.
The plot to the right shows your velocity as function of time.
How far will you travel before coming to a stop?
Example 1.12 (Position from velocity)
Mechanics Lecture 1, Slide 46
Method I: Algebraically
Breaking Distance of a Car
Mechanics Lecture 1, Slide 47
A B
Method II: Graphically
Breaking Distance of a Car
Mechanics Lecture 1, Slide 48
Find the area under
the velocity vs time
curve
Integrate!
Question
A car has a negative acceleration of
-4.2m/sec2. We can say magnitude of
the car’s velocity is
A. constant
B. increasing
C. decreasing
D. We can’t really say any of the above
Mechanics Lecture 1, Slide 49