Botany krishna series 2nd semester Only Mcq type questions
Newton's Laws of motion Lec2
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
2.
3. Review
Last 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
4. 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
6. 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
7. The Displacement vector
and
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
8. Speed and velocity
Speed is measured using distance and time:
Velocity is measured using displacement and time:
t
x
tt
xx
vav
∆
∆
=
−
−
=
12
12
t
d
tt
dd
∆
∆
=
−
−
=
12
12
speed
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.
9. 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
11. t
x
tt
xx
av
∆
∆
=
−
−
=
12
12
v
x2=1680 m
t2 = 85 sec
sm
s
m
t
x
av /23
73
1675
v ≈=
∆
∆
=
Example- Find the average velocity.
x1=5 m
t1= 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
13. 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
15. Example 2.1 Average and instantaneous velocity
( )mtx 2
520
ischeetahofPosition
+=
16. 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.
b) Find the average velocity during this time interval.
( )mtx 2
520 +=
17. Example 2.1Average and Instantaneous velocity
(cont)
sm
s
m
t
x
So
mxst
mxst
st
av /5.10
1.0
05.1
v
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.
sm
s
m
t
x
So
mxt
mxst
statst
av /05.10
01.0
1005.0
v
1005.25;s01.1
25;1
101.0Using
22
11
==
∆
∆
=
==
==
==∆
( ) 1−
= nn
ntt
dt
d
( )mtx 2
520 +=
18. 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
20. 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
ttt
aav
∆
∆
=
−
−
=
vvv
onaccelaratiAverage
12
12
2
2
0
instant
vv
dt
xd
dt
dx
dt
d
dt
d
t
a
t
=
=
=
∆
∆
=
→∆
lim
22. 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.
b) Find the average acceleration in this time interval
( )smt /50.060v 2
+=
23. 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
/05.1
1.0
5.60605.60
v
1.0using1timeAt
sm
s
s
m
s
m
t
a
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
/005.1
01.0
5.6051005.60
v
01.0using1timeAt
sm
s
s
m
s
m
t
a
stst
av =
−
=
∆
∆
=
=∆=
( ) 1−
= nn
ntt
dt
d
( )smt /50.060v 2
+=
27. 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.
simpleist.com
28. 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
30. Remember…
• Average velocity=Δx/ Δt
• Instantaneous Velocity = dx/dt as Δt→0….
… the derivative of x w.r.t t
• Average acceleration = Δv/ Δt
• Instantaneous acceleration = dv/dt as Δt→0…. …
the derivative of v w.r.t t
starwars.com
31. Lecture 2-What you need to
know and where to find it
• Average, and instantaneous velocity
• Average, and instantaneous acceleration
All found in University Physics, 2.2-2.4
33. Displacement Time Average Velocity
• (2.1) To study the dragster’s motion, you need:
1) a coordinate system x-axis to lie on the dragster’s straight line path.
2) With the origin O at the starting line.
3) Front end of the dragster is where the entire dragster is represented
by that point.
To describe the motion of the particle you say :the change in the
particles coordinate x over a time interval.
The front of the dragster is at point P1
