Newton's Laws of Motion L2.1
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Transcript of Newton's Laws of Motion L2.1
Motion in a straight line
• Average, and instantaneous velocity
• Average, and instantaneous acceleration
All found in University Physics, 2.2-2.4
Lecture 2
Today… weekendnotes.com
ReviewLast lecture we looked at :• The difference between a vector and a scalar quantity.
Vector notation.
• The distinction between distance and displacement. Distance is a scalar quantity. Displacement is a vector
• Addition and subtraction of vectors
• Components of vectors
• Unit vectors. The scalar (dot) product
Example- Adding vectors• Find A + B using the component method if A has
magnitude 5.0m at elevation of 600 and B - magnitude 3.0m at elevation of 300 as shown below.
mji
jmm
imm
jBAiBABAR yyxx
ˆ8.5ˆ1.5
ˆ)30sin0.360sin0.5(
ˆ)30cos0.360cos0.5(
ˆ)(ˆ)(
00
00
A
B
300
300
5m
3m
5.8m
5.1m
X
Y
011
22
7.481.5
8.5tantan
7.7)8.5()1.5(magnitudeResultant
m
m
R
RAngle
mmm
x
y
5.8m
5.1m
Example- Adding vectors (cont)
5.8m
5.1m
X X
Y Y
θ
Example- Calculation of components
5.8m
5.1m
X
Y
θ=120°
A
0
0
120sin
120cos
AA
AA
y
x
θ1=30° 0
0
30cos
30sin
AA
AA
y
x
θ1=30°
000000
000000
30sin90cos30cos90sin3090sin
30sin90sin30cos90cos3090cos
AAA
AAA
y
x
The Displacement vectorand
Distance scalar
If P2 is at P1 then displacement is zero but distance travelled still has a value
Average velocity =
displacement/time
Average speed =
distance/time
Speed and velocity
Speed is measured using distance and time:
Velocity is measured using displacement and time:
t
x
tt
xxvav
12
12
t
d
tt
dd
12
12speed
So speed is a scalar, velocity is a vector.For example 15 km/hour is a scalar. 58.4 m/s in a northerly direction is a velocity and is a vector.
Example - Average velocity vs speed Ian Thorpe’s personal best for 100 m freestyle in a 50 m pool in 2001 was 48.81 seconds.
His average SPEED would have been 2.05 m/s
His average VELOCITY however is zero m/s!webswimming.tripod.com
Average velocity
t
x
tt
xxav
12
12v
Average velocity is the displacement divided by the time interval
t
x
tt
xxav
12
12vx2=1680 mt2 = 85 sec
sms
m
t
xav /23
73
1675v
Example- Find the average velocity.
x1=5 mt1= 12 sec
stop
Only includes initial and final conditions, the average does not have details about the path or events. Such as stopping or drive over the mountain.
old.risk.ru
Example - Average velocity
It takes Penny 37.0 seconds to drive along a 1.2 km airport runway, from a standing start at one end to the other end. What is her average velocity in completing this displacement.
To become a velocity this speed needs to include a direction.Average velocity v=32.4m/s in a due North direction.
hunternissan.com
Instantaneous velocity
t
x
tt
xxav
12
12vvelocityAverage
Instantaneous velocity is the limit of the average velocity as the time interval approaches zero. It is the instantaneous rate of change of position with time
dt
dx
t
x
t
lim
0instantv
howstuffworks.com
Example 2.1Average and Instantaneous velocity (cont)
a) Find the displacement of the cheetah during the interval between t = 1.0 sec and t = 2.0 sec.
vav=Δx/Δt=(40 m– 25m) /(2s – 1s)=15 m/s in the positive x direction
b) Find the average velocity during this time interval.
mtx 2520
Example 2.1Average and Instantaneous velocity (cont)
sms
m
t
xSo
mxst
mxst
st
av /5.101.0
05.1v
05.26;1.1
;25;1
1.0
22
11
c) Find the instantaneous velocity at time t = 1.0 sec, by taking Δt = 0.1 sec, then Δt = 0.01 sec.
d) Derive a general expression for the instantaneous velocity as a function of time, and from it find v, at t = 1.0 sec and t = 2.0 sec.
sms
m
t
xSo
mxt
mxst
statst
av /05.1001.0
1005.0v
1005.25;s 01.1
25 ;1
1 01.0 Using
22
11
1 nn nttdt
d
mtx 2520
Instantaneous velocity from a graph
As P2 approaches P1 the slope of the line between P1 and P2 becomes the tangent at point P1 and its slope gives the value for instantaneous velocity at P1 ….. On a graph of position as a function of time for straight-line motion, the instantaneous velocity at any point is equal to the slope of the tangent to the curve at that point
Average acceleration
t
x
tt
xxav
12
12v
Average velocity is the displacement divided by the time interval
Average acceleration is the change in velocity divided by the time interval
tttaav
vvv
12
12
Units of acceleration are m/s2
Instantaneous acceleration
Instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero; it is the instantaneous rate of change of velocity with time
tttaav
vvv
onaccelarati Average12
12
2
2
0instant
vv
dt
xd
dt
dx
dt
d
dt
d
ta
t
lim
Example 2.3 Instantaneous acceleration (cont)
a) Find the change in velocity of the car in the time interval between t =1.0 sec and t = 3.0 sec.
At t = 1 sec..v = 60.5 m/s At t = 3sec..v = 64.5 m/s
b) Find the average acceleration in this time interval
smt /50.060v 2
Example 2.3 Instantaneous acceleration (cont)
c) Find the average acceleration at time t=1.0 s by taking Δt to be first 0.1 s, then 0.01 s.
2/05.11.0
5.60605.60v
1.0 using 1timeAt
sms
s
m
s
m
ta
stst
av
d) Derive an expression for the instantaneous acceleration at any time, and use it to find the acceleration at t = 1.0 s and t = 3.0 s.
2/005.101.0
5.6051005.60v
01.0using 1timeAt
sms
s
m
s
m
ta
stst
av
1 nn nttdt
d
smt /50.060v 2
Example: A dizzy turkey• A turkey moves according to the following
relationship: x = 5 – 2 t + 0.5 t2 (m)• Calculate its instantaneous velocity and
acceleration at a time t = 5 s.
Differentiate: v = -2 + t (m/s); a = 1 m/s2
At t = 5 sv = 3 m/s; acceleration always 1m/s2
simpleist.com
Lets look at an object falling from afar.
An astronaut floats near the space shuttle. His acceleration is a function of his position.
wikipedia.org
legacy.shadowlordinc.com
However around the surface of the Earth the acceleration is fairly constant so a velocity time graph is a straight line.
vm/s
t sec
Remember…
• Average velocity=Δx/ Δt
• Instantaneous Velocity = dx/dt as Δt0….
… the derivative of x w.r.t t
• Average acceleration = Δv/ Δt
• Instantaneous acceleration = dv/dt as Δt0…. …the derivative of v w.r.t t
starwars.com