Motion in One Dimension Physics 2053 Lecture Notes dx dt x t Motion in One Dimension (2053)

Motion in One Dimension

Physics 2053Lecture Notes

dx

dt

x

t

Motion in One Dimension (2053)

Motion in 1 Dimension

v

In the study of kinematics, we consider a moving object as a particle.

A particle is a point-like mass having infinitesimal size and a finite mass.


0 2 4 6246

x

Displacement


The displacement of a particle is defined as its change in position.

(m)

x = x xo

= 6 m 2 m

= 4 m

Note: Motion to the right is positiveMotion in One Dimension (2053)

0 2 4 6246

x

Displacement



(m)

x = x xo

= 6 m 6 m

= 12 m

Note: Motion to the left is negativeMotion in One Dimension (2053)

0 2 4 6246

x

Displacement



(m)

x = x xo

= ( m) (6 m)

= 8 m

Note: Motion to the right is positiveMotion in One Dimension (2053)


Average velocity

The average velocity of a particle is defined as

x

t

x1

x2

t1 t2

x

tVelocity is represented

by the slope on a displacement-time graph



Average speed

The average speed of a particle is defined as



Average acceleration

The average acceleration of a particle is defined

as the change in velocity vx divided by the time

interval t during which that change occurred.

v

t

v1

v2

t1 t2

v

tAcceleration is

represented by the slope on a

velocity-time graph



Instantaneous acceleration

The instantaneous acceleration equals the derivative of the velocity with respect to time

v

t

v

t

Instantaneous acceleration is represented by

the slope of a velocity-time graph



t

x

t

v

t

a

Displacement, velocityand acceleration graphs

The slope of a velocity-timegraph represents acceleration

The slope of a displacement-timegraph represents velocity



t

x

t

v

t

a

t

Displacement, velocityand acceleration graphs

The area under an acceleration-timegraph represents change in velocity.

v

The area under a velocity-timegraph represents displacement.

x



Definitions of velocity and acceleration

Average velocity




For constant acceleration

An object moving with an initial velocity vo undergoes

a constant acceleration a for a time t. Find the final velocity.vo

time = 0 time = t

Solution:

Eq 1

a ?


What are we calculating?

0 t

a

V


Motion in 1 DimensionFor constant acceleration

An object moving with a velocity vo is passing position xo when it undergoes a constant acceleration a for a time t. Find the object’s final position.

Solution:

time = 0 time = t

xo?

avo


Eq 2

What are we calculating?

at


0 t

vi

v


Eq 1 Eq 2

Solve Eq 1 for a and sub into Eq 2:

Solve Eq 1 for t and sub into Eq 2:

Eq 3

Eq 4



More Graphs


0 1 2 31 5 64-1-2-3-4-5-6


0 1 2 31 5 64-1-2-3-4-5-6


0 1 2 31 5 64-1-2-3-4-5-6


01

23

15

64

-1-2

-3-4

-5-6

2 4 6 8 10 12


01

23

15

64

-1-2

-3-4

-5-6

2 4 6 8 10 12


01

23

15

64

-1-2

-3-4

-5-6

2 4 6 8 10 12


2 4 6 8 10 12

01

23

15

64

-1-2

-3-4

-5-6

m

s


2 4 6 8 10 12

02

64

-2-4

-6

m

s

v

(m/s)t (s)4 8 12

2

1

0

-1

-2

-3Motion in One Dimension (2053)

2 4 6 8 10 12

02

64

-2-4

-6

m

s

v

(m/s)t (s)12

2

1

0

-1

-2

-3

+4 m

-12 m

+8 m

4 8


16 2412840 20 28

(s)

x

4

8

12

16

20

24

28

(m)

1 2 3 4t

5


(s)

x

48

1216202428

(m)

1 2 3 4 t5

1 2 3 4 5 t (s)

2

4

6

8

10

v

(m/s)Displacement

25 m


Review:

Average velocity


Kinematics with Constant Acceleration


Definitions

t

x

t

v

t

a

t

v

x


t

x

t

v

t

a

Review:

Problem Solving Skills

1. Read the problem carefully

2. Sketch the problem

3. Visualize the physical situation

4. Strategize

5. Identify appropriate equations

6. Solve the equations

7. Check your answers


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Motion in One Dimension Physics 2053 Lecture Notes dx dt x t Motion in One Dimension (2053)