Download Lecture 10 - Purdue Physics

PHYSICS 149: Lecture 10
• Chapter 4
– 4.1 Motion along a Line due to a Constant Net Force
– 4.2 Visualizing Motion along a Line with Constant
Acceleration
Lecture 10
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Exam Schedule
• Midterm Exam 1
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Wednesday, February 23, 18:30 – 19:30
Place: PHY 333
Chapters 1 - 4
Things to bring:
• Purdue ID Card, Crib Sheet, #2 Pencil, Calculator
• See syllabus for detailed information
• Midterm Exam 2
– Wednesday, April 6, 18:30 – 19:30
– Place: PHYS 333
• Final Exam
– Thursday, May 5, 15:20 – 17:20
– Place: PHY 333
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Midterm Exam 1
• The exam is closed book.
• The exam is a multiple-choice test.
• There will be 15 multiple-choice problems.
– Each problem is worth 10 points.
– Maximum possible score will be 150 points.
• Note that total possible score for the course is 1,000 points
(see the course syllabus)
• The difficulty level is about the same as the level of
textbook problems.
• You may make a single crib sheet
– You may write on both sides of an 8.5” × 11.0” sheet
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Midterm Exam 1
• Do not forget to bring
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–
–
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Purdue ID Card,
Crib Sheet,
#2 Pencil (with an eraser),
Calculator
• Adaptive learners should contact Prof. Neumeister ASAP.
• Academic Dishonesty:
– Do not cheat! Cheaters will be given an F grade in the course and
will be reported to the Dean of Students.
• Review session:
– During next recitation session
– Come and see your TA in the help center (private mini-reviews)
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ILQ 1
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)
B)
C)
D)
Both average acceleration and average velocity are zero.
Average acceleration is zero but average velocity is not zero.
Average velocity is zero but average acceleration is not zero.
Neither average acceleration nor average velocity are zero.
C correct
Vave = Δy/Δt = (yf – yi) / (tf – ti) = 0
aave = Δv/Δt = (vf – vi) / (tf – ti)
Not 0 since Vf and Vi are
not the same !
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ILQ 2
How could you determine acceleration from a
graph of velocity versus time?
A) From the slope of the line
B) Impossible – you need a plot of velocity vs
acceleration
C) By reading it off the horizontal axis
D) Impossible – you need a plot of acceleration versus
time
E) By reading it off the vertical axis
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ILQ 3
A)
B)
C)
Which x vs t plot shows positive acceleration?
“The amount of distance traveled increases with each second that passes, so the
slope should get steeper and steeper as long as the acceleration is positive.”
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Review: Notions in Chapter 3
• Position Vector
• Displacement vs. Distance
• Average Velocity vs. Average Speed
• Instantaneous Velocity (often called velocity)
• Average Acceleration
• Instantaneous Acceleration (often called acceleration)
(They are vectors expect for distance and average speed)
• Newton’s 2nd Law of Motion:
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Determination of vx on Graph of x vs. t
• The instantaneous velocity vx is the slope of the
line tangent to the graph of x vs. t (or of x(t)) at
the chosen time. Recall
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Positive, Zero, and Negative Velocity
vx: zero
vx: positive
vx: negative
positive slope Î vx: positive & moving in +x-direction
steeper slope Î faster moving
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ILQ
•
The graph for position vs. time is given below for a car.
What can you say about the velocity of the car over
time?
A)
B)
C)
D)
E)
It speeds up all the time.
It slows down all the time.
It moves at constant velocity.
Sometimes it speeds up and sometimes it slows.
Not really sure.
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Determination of ax on Graph of vx vs. t
• The instantaneous acceleration ax is the slope of
the line tangent to the graph of vx vs. t (or of vx(t))
at the chosen time. Recall
lower
positive
acceleration
higher
positive
acceleration
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Determination of ∆x on Graph of vx vs. t
• The displacement of ∆x during any time interval
equals the area under the graph of vx(t).
Recall:
– Note: If vx is negative, the displacement is also
negative. So, we count the area as negative.
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ILQ
•
The graph at right represents the velocity of a car over
time. The displacement of the car can be found by
A)
B)
C)
D)
adding the slopes of each section of the graph.
adding the area 1 to area 2.
subtracting area 2 from area 1.
subtracting area 1 from area 2.
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Summary for Interpreting Graphs
• On a graph of x(t), the slope at any point is vx
• On a graph of vx(t), the slope at any point is ax
• On a graph of vx(t), the area under the graph
during any time interval is the displacement ∆x
during that time interval.
– Note: If vx is negative, the displacement is also
negative. So, we count the area as negative.
• On a graph of ax(t), the area under the curve is
∆vx(t), the change in vx during that time interval.
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Problem Solving Strategy
• Decide what objects will have Newton’s second law
applied to them.
• Identify all the external forces acting on that object.
• Draw a free-body diagram to show all the forces acting on
the object.
• Choose a coordinate system. If the direction of the net
force is known, choose axes so that the net force is along
one of the axes.
• Find the net force by adding the forces as vectors.
• Use Newton’s second law to relate the net force to the
acceleration.
• Relate the acceleration to the change in the velocity
vector during a time interval.
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Newton’s Second Law: F = ma
A tractor T is pulling a trailer M with a constant acceleration. If the forward
acceleration is 1.5 m/s2, calculate the force on the trailer (m=400 kg) due to
the tractor (m=500 kg).
x–direction
y
N
T
x
W
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1-D Motion with Constant Acceleration
• If the net force on an object is constant, the
acceleration of the object is also constant,
both in magnitude and direction (recall
Newton’s 2nd Law).
• If the acceleration a is constant,
xf = xi + vi⋅(tf–ti) + ½⋅a⋅(tf–ti)2
(xf – xi) = ½⋅(vf+vi)⋅(tf–ti)
vf = vi + a⋅(tf–ti)
vf2 – vi2 = 2⋅a⋅(xf–xi)
where xi and vi are the initial position and
velocity at an initial time ti.
And, xf and vf are the final position and
velocity at a final time tf.
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1-D Motion with Constant Acceleration
• vf = vi + a⋅(tf–ti)
(if a is constant)
– This is simply the definition of
average acceleration. In this
case, aav = a (= const).
• (xf – xi) = ½⋅(vf+vi)⋅(tf–ti)
(if a is constant)
– vav = ½⋅(vf+vi) (if a is constant)
– From the definition of average
velocity,
(xf – xi) = vav⋅(tf–ti)
= ½⋅(vf+vi)⋅(tf–ti)
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1-D Motion with Constant Acceleration
• xf = xi + vi⋅(tf–ti) + ½⋅a⋅(tf–ti)2
(if a is constant)
– (xf – xi) = ½⋅(vf+vi)⋅(tf–ti)
= ½ ⋅ {[vi+a⋅(tf–ti)]+vi} ⋅ (tf–ti)
= vi⋅(tf–ti) + ½⋅a⋅(tf–ti)2
• vf2 – vi2 = 2⋅a⋅(xf–xi)
(if a is constant)
– (xf – xi) = ½⋅(vf+vi)⋅(tf–ti)
= ½⋅(vf+vi) ⋅ (vf–vi)/a
= (vf2–vi2) / (2⋅a)
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Constant Acceleration
x
• x = x0 + v0t + 1/2 at2
• v = v0 + at
• v2 = v02 + 2a(x-x0)
t
v
at2
Δx = v0t + 1/2
Δv = at
v2 = v02 + 2a Δx
t
a
t
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Example
A tractor T (m=500 kg) is pulling a trailer M (m=400 kg). It starts from rest
and pulls with constant force such that after 10 seconds it has moved 30
y
meters to the right. Calculate the horizontal force on the tractor.
x-direction: Tractor
ΣF = ma
Fw – T = mtractora
Fw = T + mtractora
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N
T
W
x-direction: Trailer
ΣF = ma
T = mtrailera
x
T
N
Fw
W
Combine:
Fw = mtrailera + mtractora
Fw = (mtrailer + mtractor ) a
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Example
A tractor T (m=500 kg) is pulling a trailer M (m=400 kg). It starts from rest
and pulls with constant force such that after 10 seconds it has moved 30
y
meters to the right. Calculate the horizontal force on the tractor.
x
Combine:
Fw = mtrailera + mtractora
Fw = (mtrailer + mtractor ) a
Acceleration:
Δx = v0t +0.5 a
t2
a = 2 Δx / t2 = 0.6 m/s2
Lecture 10
N
W
T
T
N
Fw
W
FW = 900kg×0.6m/s2
FW = 540 Newtons
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Kinematics Example
A car is traveling 30 m/s and applies its breaks to stop after
a distance of 150 m. How fast is the car going after it has
traveled ½ the distance (75 meters) ?
A) v < 15 m/s
B) v = 15 m/s
C) v > 15 m/s
This tells us v2 proportional to Δx
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ILQ: Acceleration
A car accelerates uniformly from rest. If it travels a
distance D in time t then how far will it travel in a time 2t?
A) D/4
B) D/2
C) D
D) 2D
E) 4D
Correct x=1/2 at2
Follow up question: If the car has speed v at time t
then what is the speed at time 2t?
A) v/4
B) v/2
C) v
D) 2v
E) 4v
Lecture 10
Correct v=at
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