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PHYSICS 220
Lecture 03
Motion with Constant Acceleration
Textbook Sections 3.1
Lecture 3
Purdue University, Physics 220
1
Exercise
x (meters)
100
•
•
•
•
0
-100
position vs. time
-200
-300
0
v (m/s)
5
10
t (seconds)
15
Where is velocity zero?
Where is velocity positive?
Where is velocity negative?
Where is speed largest?
20
20
• Where is acceleration zero?
• Where is acceleration positive?
0
-20
-40
-60
velocity vs. time
-80
-100
0
Lecture 3
5
10
t (seconds)
15
20
If speed is increasing,
v and a are in same direction.
If speed is decreasing,
v and a are in opposite direction.
Purdue University, Physics 220
2
Question
• A skydiver is falling straight down, along the negative y
direction. During the initial part of the fall, her speed
increases from 16 to 28 m/s in 1.5 s. Which of the
following is correct?
A) v>0, a>0
v a
B) v>0, a<0
C) v<0, a>0
D) v<0, a<0
 correct
• During a later part of the fall, after the parachute has
opened, her speed decreases from 48 to 26 m/s in 11 s.
Which of the following is correct?
A)
B)
C)
D)
Lecture 3
v>0, a>0
v>0, a<0
v<0, a>0
v<0, a<0
 correct
v a
Purdue University, Physics 220
4
iClicker
A ball is tossed from the ground up a height of two
meters above the ground and falls back down.
• Draw v vs t
v
3
-2
v
A
3
4 t
t
4
t
E
t
Purdue University, Physics 220
4
-2
v
3
4
C
3
D
-2
Lecture 3
B
-2
v
3
v
-2
4
t
5
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
6
m/s2
Change in v between t=4 and t=1. Dv = +4 m/s
t=1-3: Dv = (3m/s2)(2s) = 6 m/s
24
t (s)
-3
t=3-4: Dv = (-2m/s2)(1s) = -2 m/s
Lecture 2
Purdue University, Physics 220
6
Constant Acceleration
• v = v0 + at
• x = x0 + v0t + 1/2 at2
• v2 = v02 + 2a(x-x0)
Dv = at
Dx = v0t + 1/2 at2
v2 = v02 + 2a Dx
x (meters)
200
150
100
50
0
1 2
x  x0  v0t  at
2
x  x0  v t
v0  v f
2

v0  (v0  Dv)
Lecture 3
2
10
t (seconds)
15
20
5
10
t (seconds)
15
20
5
10
t (seconds)
15
20
20
15
10
5
2
2v0  at
1
 v0  at
2
2
v
5
v (m/s)
0
v0  v f
0
0
2
a (m/s )
2
1.5
1
2v0  at

2
0.5
0
Purdue University, Physics 220
0
7
Tossed Ball
A ball is tossed from the ground up a
height of two meters above the ground
and falls back down.
y
t
v
• Draw v vs t
• Draw y vs t
• Draw a vs t
t
a
t
Lecture 3
Purdue University, Physics 220
8
Question
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 3
Correct v=at
Purdue University, Physics 220
9
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
v 2  vo2  2aDx
B) v = 15 m/s
C) v > 15 m/s
This tells us v2 proportional to Dx
v 2f  vo2
302
2

3
m
/
s
a

2(150) 2(150)
2
v75
 302  2a(75)
2
v75
 302  2(3)(75)
Lecture 3
Purdue University, Physics 220
v75  21m / s
10
Frame of Reference
• Motion is relative.
• Motion can be described only after a frame of
reference is chosen.
• A frame of reference may be in motion with
respect to other frames of reference.
• The description of motion in one frame of
reference may be mathematically transformed
into that in another frame of reference.
• In a frame of reference, different coordinate
systems may be used to describe motion.
Lecture 3
Purdue University, Physics 220
11
Inertial Frame of Reference
Any reference frame in which Newton’s laws are
valid is called an
inertial frame of reference
SF = ma
If no net force acts on a body (FNET = 0) there is no acceleration
acting on the body, i.e. it moves with constant velocity
F3
F2
F1
M
F4
Lecture 3
Purdue University, Physics 220
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Newton’s Third Law
For every action there is an equal an opposite
reaction
Forces in nature come in pairs
N
mg
Lecture 2
Purdue University, Physics 220
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Newton’s Second Law
SF = ma
F
 max
F
 ma y
x
y
Fx
ax 
m
Lecture 3
ay 
Fy
m
Purdue University, Physics 220
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Problem Solving Recipe
• Identify the object(s) of interest.
• Draw the free-body diagram for each object,
showing all the forces acting on it.
• Choose a coordinate system.
• Find the net force along each axis.
• Use Newton’s second law to find the acceleration
along each axis.
• Find the velocity and displacement along each
axis from the acceleration.
Lecture 3
Purdue University, Physics 220
15
Free Body Diagrams
• Isolate the object of interest
• Identify all forces acting on object and represent
them as vectors
• Choose a coordinate system (e.g., x,y,z)
Lecture 3
Purdue University, Physics 220
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Example
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). Neglect friction.
x–direction
Fx  max
T  max

T  400 kg
y
 1.5 m/s 
N
2
T
T  600 Newtons
x
W
Lecture 3
Purdue University, Physics 220
17
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 driving force on the tractor.
x
X direction: Tractor
N
SF = ma
Fw – T = mtractora
Fw = T + mtractora
W
X direction: Trailer
SF = ma
T = mtrailera
Lecture 3
T
T
N
Fw
W
Combine:
Fw = mtrailera + mtractora
Fw = (mtrailer + mtractor ) a
Purdue University, Physics 220
18
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 driving force on the tractor.
x
Combine:
Fw = mtrailera + mtractora
Fw = (mtrailer + mtractor ) a
Acceleration:
Dx = v0t +0.5 a
t2
a = 2 Dx / t2 = 0.6 m/s2
Lecture 3
N
W
T
T
N
Fw
W
FW=900kg0.6m/s2
FW = 540 Newtons
Purdue University, Physics 220
19
iClicker
If an object is acted on by two finite constant
forces, is it possible for the object to move at
constant velocity?
A)
B)
C)
D)
E)
Lecture 3
No, it will accelerate
Yes, the forces must be perpendicular
No, it will follow a curved path
Yes, the forces must be equal and opposite
Yes, the forces must be in the same direction
Purdue University, Physics 220
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