Relative motion
ƒ Recall, that observers
stationary with respect to
each other measure the same
displacement of a moving
particle
ƒ Therefore, since the time
interval is measured by them is
also the same (GalileoNewton), all of these
observers will measure the
same velocity of this particle.
ƒ What if one of the observers
is moving with respect to
others?
D=D’+D0 , D’=D-D0
D
A
O1
A
D
B
O2
B
D’
O1
O2 Do
O’2
Relative velocity
ƒ Thus, it turns out that if one observer is moving with
respect to another, they will not measure the same
displacement of a particle.
ƒ Consequently, they will measure different velocities of
the particle
r r r
r r r
D = D'+ Do ,
D' = D − Do
r r
r r
r
r
D D'+ Do
D' D − Do
,
=
=
∆t
∆t
∆t
∆t
v = v'+vo ,
v' = v − vo
Example: Moving sidewalk
ƒ Observer 1 standing on the ground
ƒ Observer 2 standing on the sidewalk
Bill’s velocity
wrt sidewalk
Annie’s
velocity wrt
sidewalk
v’A
v’B
Velocity of moving
sidewalk
vO
Find velocities of Bill and Annie with respect to the
stationary observer on the ground
v = v'+ vo
Bill moves faster!
Annie moves slower!
How to jump from the moving train ?
ƒ Moving train – “primed frame” – moving observer –
the question “how” means we want to find v’
ƒ Trains velocity is obviously vO
ƒ What do we want to achieve? Perhaps minimize the
velocity of the jumper with respect to the ground
v = v'+vo
ƒ Therefore, for v to be smaller than vO, v’ has to be
negative (opposite to vO). Ideally, v’=-vO, so that v=0
It works the same way in two dimensions
A bird is flying in the wind.
Which way is it moving with
respect to the ground ?
v = v'+vo
Acceleration
ƒ If a material point is not moving uniformly, the velocity
must be changing, it is accelerating.
ƒ Acceleration is always associated with a change of the
velocity: its magnitude (speed), direction, or both
ƒ The average acceleration is introduced similarly to the
average velocity:
r
r ∆v
a=
∆t
The average acceleration = the change in velocity
divided by the time over which the change occurs.
Example
The dealer tells you that the Porsche he is selling can
accelerate from 0 to 30 m/s (about 67mph) in 3
seconds. What’s the average acceleration of the car?
m
30
m
∆v
s
=
= 10 2
a=
s
∆t
3s
Units for acceleration:
Velocity = displacement / time length/time
m/s
Acceleration = velocity/time = (length/time)/time
= length/(time x time)= length/(time2)
m/s2
Acceleration, some features
ƒ Can introduce the instantaneous acceleration similarly to
the introduction of instantaneous velocity
ƒ Acceleration is a vector quantity directed in the direction
of velocity change, not velocity itself.
ƒ The magnitude of acceleration does not have a specific
name
ƒ If the acceleration changes the magnitude of the velocity
so that it decreases (in some frame of reference), the
acceleration is sometimes called a deceleration
ƒ It’s easy to prove that if the second observer moves with
respect to the first at a constant velocity, they will measure
the same acceleration of the moving object
v = v'+ vo
⇒
∆v = ∆v'+ ∆vo ,
but ∆vo = 0
⇒
∆v = ∆v'
⇒
a = a'
Simple Types of Motion
1. Motion with constant (zero) velocity – uniform motion
2. Motion with constant acceleration (in a straight line)
3. Circular motion at a constant speed
1. Motion with constant (zero) velocity:
∆D
v=
⇒ ∆D = v ⋅ ∆t , or D(t) = D0 + v ⋅ t
∆t
v
d
t
v>0
v=0
t
v<0
2. Motion with constant acceleration (in a straight
line)
v
Zero initial
velocity, a>0
v
v0 > 0, a > 0
v0 > 0, a < 0
t
ƒ Acceleration constant – velocity changes linearly with
time – directly proportional to time – the graph is a
straight line on a v vs. t diagram
∆v
a=
∆t
⇒
∆v = a ⋅ ∆t or v(t ) = vo + a ⋅ t
How the displacement depends on time?
t
v
∆d = vaverage ⋅ ∆t
d = vaverage ⋅ t
But what is the
average velocity?
t
Recall: The average velocity of any motion is a velocity of
such uniform motion that brings the object from the point
of departure to the destination for the same time interval
1
1
1
vaverage = (v0 + v f ) = (v0 + (v0 + at )) = v0 + at
2
2
2
1 2
d = vaveraget = v0t + at
Equation of motion
2
FREE FALL
Galileo: Any object allowed to fall equally fast in the
absence of air friction.
Acceleration due to gravity is the same for all
objects, regardless of mass.
Close to Earth’s surface it is a
= g = 9.8 m/s2
(directed downward)
NOTE: When air friction is present, the object
will reach terminal speed which depends on
mass and shape of the object.
O
For an object with low air resistance (e.g. stone)
dropped from a building with zero initial velocity
Velocity:
v = a t = g t = (9.8 m/s2) × t
(positive – downward)
Displacement:
d = ½ a t2 = ½ g t2 = (4.9 m/s2) t2
v
O
y
Slope=g
t
O
Slope increases
as the velocity
does - parabola
t
v=gt
d = ½ g t2
Example:
g = 9.8 m/s2
A stone is released at the top of a high cliff and
falls for 2.3 seconds before hitting the water.
(a) How fast is the stone moving at impact?
(b) How high is the cliff ?
Solution:
t = 2.3 s
v = g × t = (9.8 m/s2) × (2.3 s) = 22.54 m/s
(a) d = ½ g t2 = ½ (9.8 m/s2)×(2.3s)2 = 25.92 m
O
A stone is thrown upward at 10 m/s. What
altitude will it reach?
1. When it reaches the maximum height, its velocity is
ZERO.
∆v = -10 m/s, g = -9.8 m/s2, ∆t = ∆v / g = 1.02 s
2. The time up = the time down, the height up = the height
down
height = ½ g ∆t2 = 5.1 m
v
10 m/s
0 m/s (maximum height)
2.04 s
v = v0-g t
d = v0t-½ g t2
g = -9.8 m/s2
0s
1.02 s
t
- 10 m/s
Direction of motion – direction of the velocity
Which of the following statements are true ?
1. Acceleration always directed in the direction of
motion
2. If the acceleration is opposite of the direction of
motion the object will slow down.
3. If the direction of acceleration is not in line with
the motion of the object, it will change direction
of motion.
4. Positive acceleration always refers to an object
that is speeding up.
5. (Non zero) acceleration always changes the speed
6. (Non zero) acceleration always changes the
velocity
Acceleration can act in any direction. The effect it
will have on the motion depends on the current
direction of the motion.
Example:
Centripetal acceleration: Acceleration directed
towards a central point, perpendicular wit respect to
the velocity at any instant.
What happens?
r
r ∆v
a=
∆t
r r
∆v = a × ∆t
r
New v
r
Old v
Centripetal acceleration
To keep an object on a circular path, a centripetal
acceleration (an acceleration perpendicular to the velocity
and directed towards the center of this circle) is needed.
The magnitude of the centripetal
acceleration needed to keep an
object moving on a circular path of
radius r at speed v is given by:
a = v2/r
A little sloppy derivation: during one revolution v makes a full circle, hence
∆v=2πv, the time of one revolution is ∆t=2πr / v
ac = ∆ v /∆ t = v2 / r
and the acceleration