Download T - Purdue Physics - Purdue University

PHYSICS 149: Lecture 9
• Chapter 3
– 3.2 Velocity and Acceleration
– 3.3 Newton’s Second Law of Motion
– 3.4 Applying Newton’s Second Law
– 3.5 Relative Velocity
Lecture 9
Purdue University, Physics 149
1
Velocity (m/s)
• The average velocity is the change in position
(vector) divided by the change in time.
Δr x f − x0
v=
=
Δt t f − t0
x(t)
Δx
Δt
t
• Instantaneous velocity is the limit of average velocity
as ∆t gets small. It is the slope of the x(t) plot.
G
Δr
G
v = lim
Δt →0 Δt
Lecture 9
x(t)
(t)
t
Purdue University, Physics 149
2
ILQ 1
• A go-kart travels around a circular track at a
constant speed. Which of these is a true
statement?
(a) The go-kart has a constant velocity.
(b) The go-kart has zero acceleration.
A) Both (a) and (b) are true
B) Neither (a) nor (b) is true
C) (a)
D) (b)
Lecture 9
Purdue University, Physics 149
3
ILQ 2
If the average velocity of a car during a trip along a straight
road is positive, is it possible for the instantaneous velocity
att some time
ti
during
d i th
the ttrip
i tto b
be negative?
ti ?
A - Yes
correct
The car might have reversed
for a little while along the trip
creating a negative
instantaneous velocity at the
point. If the overall
p
displacement of the car is
positive for that particular time
interval than the average
interval,
velocity is positive as well.
Lecture 9
B - No
If the car is traveling in a
straight
g p
path the velocity
y
will always be positive.
The car needs to travel in
the opposite direction to
get a negative velocity.
Purdue University, Physics 149
4
How to determine ∆x from v(t)
• The total displacement ∆x during and time ∆t is
the area under the graph v(t)
• Assume 1D, and v constant:
Δx
vx = vav , x =
Δt
Δx = v1x Δt
Lecture 9
v
v1x
Purdue University, Physics 149
Δx
t1
t2
t
5
Acceleration (m/s2)
• The average acceleration is the change in velocity
divided by the change in time. v(t)
Δv
a=
Δt
Δv
Δt
t
• Instantaneous acceleration is limit of average
acceleration as ∆t gets small. It is the slope of the
v(t)
()p
plot.
v(t)
G
Δv
G
a = lim
Δt →0 Δt
Lecture 9
Purdue University, Physics 149
t
6
Average Acceleration
• Average acceleration is the change in velocity
during
g the time interval during
g which the velocity
y
changed.
" change"
g in velocity
=
elapsed time
• The x- and y-components of the average velocity
are:
• Typical units for the magnitude of acceleration
are m/s2, km/s2, and so on.
Lecture 9
Purdue University, Physics 149
7
Instantaneous Acceleration
• Instantaneous acceleration is the average
acceleration we measure when the time interval
is infinitesimally short.
• The x- and y-components of the instantaneous
velocity are:
Lecture 9
Purdue University, Physics 149
8
Direction of Acceleration
• If speed is increasing, v and a are in
same direction.
direction
• If speed is decreasing, v and a are in
opposite direction.
Lecture 9
Purdue University, Physics 149
9
ILQ: Acceleration
If the velocity of some object is not zero, can its
acceleration ever be zero?
A) Yes
B) No
If the velocity is constant, the acceleration is
zero ((slope
p of v(t)
( ) is 0)).
v(t)
v(t)
t
Lecture 9
Purdue University, Physics 149
t
10
ILQ: Acceleration
Is it possible for an object to have a positive velocity
at the same time as it has a negative acceleration?
A) Yes
B) No
Yes, when they have different signs, the object is
slowing down.
v(t)
v(t)
Δv
Δt
Lecture 9
t
Purdue University, Physics 149
t
11
Position vs Time Plots
• Gives location at any time
• Displacement is change in position
• Slope gives velocity
x (m)
3
Position at t=3
t=3, x(3) = 1
4
Displacement between t=5 and t=1. Δx = -1.0 m
1.0 m - 2.0 m = -1.0 m
t
-3
Average
g velocity
y between t=5 and t=1. v = -0.25 m/s
-1 m / 4 s = -0.25 m/s
Lecture 9
Purdue University, Physics 149
12
Velocity vs Time Plots
• Gives velocity at any time
• Area gives displacement
• Slope gives acceleration
Velocity at t=2, v(2) = 3 m/s
Displacement between t=0
t 0 and t=3:
t 3: ∆x = 7.5
75m
t=0 to t=1: ½ (3m/s) (1 s) = 1.5 m
v (m/s)
3
1.5
6
4
t
-33
t=1 to t=3: (3m/s) (2 s) = 6 m
Average velocity between t=0 and t=3? v= 7.5 m / 3s = 2.5 m/s
Change in v between t=5 and t=3. ∆v = -2 m/s – 3 m/s = -5 m/s
A
Average
acceleration
l ti b
between
t
tt=5
5 and
d tt=3:
3
a = -5 m/s / (2 s) = -2.5 m/s2
Lecture 9
Purdue University, Physics 149
13
Acceleration vs Time Plots
• Gives acceleration at any time
• Area gives change in velocity
a (m/s2)
3
Acceleration at t=4
t=4, a(4) = -2
6
m/s2
24
Change in v between t=4 and t=1. ∆v = +4 m/s
t=1-3: ∆v = (3m/s2)(2s) = 6 m/s
t
-3
t=3-4: ∆v = (-2m/s2)(1s) = -2 m/s
Lecture 9
Purdue University, Physics 149
14
ILQ: Concepts
• 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
v>0, a<0
C) v<0, a>0
D) v<0, a<0
y
Í 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 9
v>0, a>0
v>0, a<0
v<0,
v<0 a>0
v<0, a<0
Í correct
v a
Purdue University, Physics 149
15
Dropped Ball
A ball is dropped from a height of two
meters above the ground.
• Draw vy vs t
3
v
A
-2
3
4 t
3
B
v
3
4
-2
4
x
v
C
t
3
D
-2
Lecture 9
v
y
v
-2
t
E
t
-2
Purdue University, Physics 149
4
4
t
16
Dropped Ball
x
A ball is dropped from a height of two
ground.
meters above the g
t
v
• Draw v vs t
• Draw x vs t
• Draw a vs t
t
a
t
Lecture 9
Purdue University, Physics 149
17
Tossed Ball
A ball is tossed from the ground up a height of two
meters above the ground and falls back down.
y
• Draw v vs t
3
v
-2
A
3
4 t
3
B
3
4
-2
v
4
v
t
E
t
Purdue University, Physics 149
4
-2
v
x
C
t
3
D
-2
Lecture 9
v
-2
4
t
18
Tossed Ball
A ball is tossed from the ground up a
height of two meters above the ground
and falls back down.
x
t
v
• Draw v vs t
• Draw x vs t
• Draw a vs t
t
a
t
Lecture 9
Purdue University, Physics 149
19
ILQ
•
Which car has a westward acceleration?
A)
B)
C)
D)
E)
a car traveling westward at constant speed
a car traveling
t
li eastward
t
d and
d speeding
di up
a car traveling westward and slowing down
a car traveling
li eastward
d and
d slowing
l i d
down
a car starting from rest and moving toward the
eastt
Lecture 9
Purdue University, Physics 149
20
Newton’s Second Law of Motion
• If there is a net force acting on an object, it experiences
an acceleration in the direction of the net force, and the
magnitude of the acceleration is the net force’s magnitude
divided by the mass of the object.
• In component form,
• Units for force: 1 N = 1 kg⋅
g m/s2
Lecture 9
Purdue University, Physics 149
21
ILQ
•
A single force is acting on an object. Which type
of motion is not physically allowed?
A)
B)
C)
D)
the object speeds up
the object is at rest
the object slows down
the object's motion changes direction
Lecture 9
Purdue University, Physics 149
22
ILQ
If an object is acted on by two finite constant
forces, is it p
possible for the object
j
to move at
constant velocity?
A)
B))
C)
D))
E)
Lecture 9
No, it will accelerate
Yes, the forces must be p
perpendicular
p
No, it will follow a curved path
Yes,, the forces must be equal
q
and opposite
pp
Yes, the forces must be in the same direction
Purdue University, Physics 149
23
ILQ
• A force
f
F acting
i on a mass m1 results
l iin an
acceleration a1.The same force acting on a different
mass m2 results in an acceleration a2 = 2a1. What is
the mass m2?
F
A)
•
•
•
•
m1
2m1
a1
B) m1
m2
F
C)
a2 = 2a1
1/2 m1
F=ma
F= m1a1 = m2a2 = m2(2a1)
Therefore, m2 = m1/2
Or in words…twice the acceleration means half the mass
Lecture 9
Purdue University, Physics 149
24
Mass
• Mass is a measure of an object’s inertia – the amount of
resistance to changes in velocity.
• Mass is an intrinsic property of an object,
object so mass does not
change wherever the object is (unlike weight).
• The units of mass are g, kg, and so on
• Objects with a larger amount of inertia (large mass) are
harder to change the state of motion of compared to
objects with a small amount of inertia (small mass)
For the same net force,
am (=|ΣF/m|) > aM (=|ΣF/M|)
because m < M
•m
Lecture 9
M
Purdue University, Physics 149
25
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 (m=400Kg) due to the
tractor T (m=500Kg).
x – direction
Fx = max
T = max
y
T = (400 kg ) (1.5 m/s
2
)
N
T
T = 600 Newtons
x
W
Lecture 9
Purdue University, Physics 149
26
Example
A tractor
t t T (m=500Kg)
( 500K ) is
i pulling
lli a ttrailer
il M ((m=400Kg).
400K ) It
starts from rest and pulls with constant force such that after
1 seconds its velocityy is 1m/s. Calculate the horizontal force
y
on the tractor due to the ground.
x-direction: Tractor
ΣF = ma
Fw – T = mtractora
Fw = T+ mtractora
x-direction:
di i
T
Trailer
il
ΣF = ma
T = mtrailera
Lecture 9
x
N
W
T
T
N
Fw
W
Combine:
Fw = mtrailera + mtractora
Fw = (m
( trailer+mtractor ) a
Purdue University, Physics 149
27
Example
A ttractor
t T (m=500Kg)
( 500K ) is
i pulling
lli a ttrailer
il M ((m=400Kg).
400K ) It
starts from rest and pulls with constant force such that after
1 seconds its velocityy is 1m/s. Calculate the horizontal force
y
on the tractor due to the ground.
x
N
Combine:
Fw = mtrailera + mtractora
Fw = (mtrailer+mtractor ) a
a=Δv/Δt=(1m/s)/(1s)=1
W
m/s2
T
T
N
Fw
W
Constant
C
t t force
f
=
constant a
Fw=(900kg)×1 m/s2 =900N
Lecture 9
Purdue University, Physics 149
28
ILQ
•
A wrench has a mass of 0.75 kg here on earth.
It is shipped up to be used on the International
Space Station. An astronaut on the Space
Station will measure its mass to be:
A))
B)
C)
D)
0 kg
g
7.5 N
1.7
1 7 lbs
0.75 kg
How about its weight?
Lecture 9
Purdue University, Physics 149
29
Relative Velocity
• W
We often
ft assume that
th t our reference
f
frame
f
is
i attached
tt h d to
t the
th
Earth. What happen when the reference frame is moving at
a constant velocityy with respect
p
to the Earth?
• The motion can be explained by including the relative
velocity of the reference frame in the description of the
motion.
motion
The ground speed of an airplane is
the vector sum of the airspeed and
the wind velocity. Using the air as
the intermediate reference frame,
ground speed is:
Example airplanes
G
G
G
v PG = v PA + v AG
Lecture 9
Purdue University, Physics 149
30
Relative Velocity
• What is the relation between the velocity of an
object as determined in one reference frame and
the velocity of the same object as determined in
another reference frame that is moving
g with
respect to the first reference frame?
vAC = vAB + vBC
where vAC is the velocityy of A relative to C,,
vAB is the velocity of A relative to B, and
vBC is the velocity of B relative to C.
Lecture 9
Purdue University, Physics 149
31
Example
where vWT is the velocity of Wanda relative to Tim,
vTG is the velocity of Tim relative to Greg, and
vWG is the velocity of Wanda relative to Greg.
Lecture 9
Purdue University, Physics 149
32
ILQ
•
As a train travels with a velocity vTG = 15.5 m/s
to right, Wanda walks toward the front of the
train with a velocity vWT = 1.2
1 2 m/s to right.
right What
is her velocity relative to the ground?
A))
B)
C)
D)
0 m/s
16.7 m/s, to right
14.3 m/s, to right
impossible to tell
Lecture 9
Purdue University, Physics 149
33
ILQ: Relative Velocity
You are on a train traveling 40 mph North. If you
walk 5 mph sideways across the car (W)
(W), what is
your speed relative to the ground?
A) < 40 mph
5
40
Lecture 9
B) 40 mph
C) >40 mph
| v |= 40 + 5 = 40.3mph
2
Purdue University, Physics 149
2
34
ILQ: Time to Cross the River
Three swimmers can swim equally
fast relative to the water. Theyy
have a race to see who can swim
across a river in the least time.
Relative to the water
water, Beth (B)
swims perpendicular to the flow,
Ann (A) swims upstream, and
C l (C) swims
Carly
i
d
downstream.
t
Which swimmer wins the race?
A) Ann
B) Beth
C) Carly
Lecture 9
A B C
Beth uses all of her force in the 'y'
direction, the direction of the other bank.
Time to get across = width of river /
y-component of velocity
Purdue University, Physics 149
35