This document discusses motion in a straight line and key concepts related to velocity and acceleration, including: - The differences between average and instantaneous velocity and acceleration. Average values are calculated over a time interval, while instantaneous values are the limit as the time interval approaches zero. - Velocity is a vector quantity that involves displacement over time, while speed is a scalar quantity involving distance over time. - Acceleration is the rate of change of velocity with respect to time. It can be calculated as the change in velocity over a time interval for average acceleration or the derivative of velocity with respect to time for instantaneous acceleration. - Examples are provided to demonstrate calculating average and instantaneous values and interpreting velocity-

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

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

5. 011
22
7.48
1.5
8.5
tantan
7.7)8.5()1.5(magnitudeResultant
≈





=





=
≈+=
−−
m
m
R
R
Angle
mmm
x
y
θ
5.8m
5.1m
Example- Adding vectors (cont)
5.8m
5.1m
X X
Y Y
θ

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

10. Average velocity
t
x
tt
xx
av
∆
∆
=
−
−
=
12
12
v
Average velocity is the displacement divided by the time
interval

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

12. Average velocity from graph

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

14. Instantaneous velocity
t
x
tt
xx
av
∆
∆
=
−
−
=
12
12
vvelocityAverage
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
=
∆
∆
=
→∆
lim0
instantv
howstuffworks.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

19. Average acceleration
t
x
tt
xx
av
∆
∆
=
−
−
=
12
12
v
Average velocity is the displacement divided by
the time interval
Average acceleration is the change in velocity
divided by the time interval
ttt
aav
∆
∆
=
−
−
=
vvv
12
12
Units of acceleration are m/s2

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

21. Example2.3- Instantaneous acceleration
( )smt /50.060v
bygiveniscartheofvelocityThe
2
+=

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
+=

24. Instantaneous acceleration from a graph

25. Review: What does the displacement vs
time graph look like?

26. Review: What does the velocity vs time
graph look like?

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

29. legacy.shadowlordinc.com
However around the surface of the Earth
the acceleration is fairly constant so a
velocity time graph is a straight line.
v
m/s
t sec

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