# Physics 101: Lecture 1 Notes

Mechanics

Nov 14, 2013 (4 years and 8 months ago)

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Physics 101: Lecture 5, Pg
1

Lecture 5:

Introduction to Physics
PHY101

Chapter 2:

Distance and Displacement, Speed and Velocity
(2.1,2.2)

Acceleration (2.3)

Equations of Kinematics for Constant

Acceleration (2.4)

Physics 101: Lecture 5, Pg
2

Displacement and Distance

Displacement is the vector that points from a body’s
initial position,
x
0
, to its final position,

x
. The length of
the displacement vector is equal to the shortest
distance between the two positions.

x

=
x

x
0

Note:

The length of

x

is (in general) not the same as distance

traveled !

Physics 101: Lecture 5, Pg
3

Average Speed and Velocity

Average speed is a measure of how fast an object
moves on average:

average speed = distance/elapsed time

Average speed does not take into account the direction of

motion from the initial and final position.

Physics 101: Lecture 5, Pg
4

Average Speed and Velocity

Average velocity describes how the displacement of an
object changes over time:

average velocity = displacement/elapsed time

v
av

= (
x
-
x
0
) / (t
-
t
0
) =

x

/

t

Average velocity also takes into account the direction of

motion.

Note:

The magnitude of

v
av
is (in general) not the same as the

average speed !

Physics 101: Lecture 5, Pg
5

Instantaneous Velocity and Speed

Average velocity and speed do not convey any
information about how fast the object moves at a
specific point in time.

The velocity at an instant can be obtained from

the average velocity by considering smaller and smaller
time intervals, i.e.

Instantaneous velocity:

v

= lim

t
-
> 0

x

/

t

Instantaneous speed
is the magnitude of
v
.

Physics 101: Lecture 5, Pg
6

Concept Question

If the average velocity of a car during a trip along a
straight road is positive, is it possible for the
instantaneous velocity at some time during the trip to
be negative?

1
-

Yes

2
-

No

correct

If the driver has to put the car in reverse and back up some time

during the trip, then the car has a negative velocity. However,

since the car travels a distance from home in a certain amount of

time, the average velocity will be positive.

Physics 101: Lecture 5, Pg
7

Acceleration

Average acceleration describes how the velocity

of an object moving from the initial position to

the final position changes on average over time:

a
av

= (
v
-
v
0
) / (t
-
t
0
) =

v

/

t

The acceleration at an instant can be obtained from

the average acceleration by considering smaller and
smaller time intervals, i.e.

Instantaneous acceleration:

a

= lim

t
-
> 0

v

/

t

Physics 101: Lecture 5, Pg
8

Concept Question

If the velocity of some object is not zero, can its
acceleration ever be zero ?

1
-

Yes

2
-

No

correct

If the object is moving at a constant velocity,

then the acceleration is zero.

Physics 101: Lecture 5, Pg
9

Concept Question

Is it possible for an object to have a positive velocity at
the same time as it has a negative acceleration?

1
-

Yes

2

No

correct

An object, like a car, can be moving forward

giving it a positive velocity,

but then brake, causing deccelaration which is

negative.

Physics 101: Lecture 5, Pg
10

Kinematics in One Dimension

Constant Acceleration

Simplifications:

In one dimension all vectors in the previous equations

can be replaced by their scalar component along one axis.

For motion with constant acceleration, average and

instantaneous acceleration are equal.

For motion with constant acceleration, the rate with which

velocity changes is constant, i.e. does not change over time.

The average velocity is then simply given as

v
av

= (v
0
+v)/2

Physics 101: Lecture 5, Pg
11

Kinematics in One Dimension

Constant Acceleration

Consider an object which moves from the initial position x
0
, at time t
0

with velocity v
0
, with constant acceleration along a straight line.

How does displacement and velocity of this object change with time ?

a = (v
-
v
0
) / (t
-
t
0
) =>
v(t) = v
0
+ a (t
-
t
0
)

(1)

v
av

= (x
-
x
0
) / (t
-
t
0
) = (v+v
0
)/2 =>
x(t) = x
0

+ (t
-
t
0
) (v+v
0
)/2
(2)

Use Eq. (1) to replace v in Eq.(2):

x(t) = x
0

+ (t
-
t
0
) v
0
+ a/2 (t
-
t
0
)
2

(3)

Use Eq. (1) to replace (t
-
t
0
) in Eq.(2):

v
2

= v
0
2

+ 2 a (x
-
x
0
)

(4)

Physics 101: Lecture 5, Pg
12

Summary of Concepts

kinematics: A description of motion

position:

displacement:

x
=
change of position

velocity:
rate of change of position

average :

x
/

t

instantaneous:
slope of x vs. t

acceleration:
rate of change of velocity

average:

v
/

t

instantaneous:
slope of v vs. t