Lesson 8 Kinematics

Describing Motion

There are lots of different ways to describe motion….

1. Words

2. Sketches

3. Time elapsed photographs

4. Physical Expressions (Equations)

Words to know…



Define the following words. Write words on

page 19.



Mechanics



Motion



Scalar



Vector

Let’s Practice

Quantity Category

a. 5 m

b. 30 m/sec, East

c. 5 mi., North

d. 20 degrees Celsius

e. 256 bytes

f. 4000 Calories

Scalar

Vector

Vector

Scalar

Kinematics - (describing

how

things move)

Scalar

(magnitude but

no direction)

Vector

( magnitude w/

direction)

Distance (d)

Displacement (d)

Speed (s) Velocity (v)

Acceleration (a) How far you travel

Change in position

(How far you travel in a given direction)

How fast you travel

How fast you travel (in a given direction)

Kinematics Equations that Make Sense!

Average speed: s_av = d / change in t SI unit: m/s

s_av = d / ∆t = d / t_f - t_i

Average velocity: v_av = ∆d / ∆t SI unit: m/s

v_av = (d_f - d_i) / t

Average acceleration: a_av = ∆v / ∆t SI unit: m/s/s = m/s2

a_av = (v_f - v_i) / t

d_f = d_i + v_av t

v_f = v_i + a_av t

Using Split Times!

Position (m) 0-5 5-10 10-15 15-20 20-25

Split Time (s) 1.6 2.4 3.0 3.5 4.0

Av. Velocity (m/s) _{3.1 2.1 1.7 1.4 1.2(5)}

Determine the average velocity for each distance interval

Determine average velocity of the object over the time recorded

Determine the average acceleration over the time recorded

v_av = d_f - d_i / t = 25m - 0m / 14.5 s = 1.7 m / s

a = v_f - v_i / t = (1.25 m/s - 3.1m/s) / 14.5 s = - 0.13 m / s2

Reference Frames

All measurements are made relative to a frame of reference..

Sitting at your desk right now you don’t seem to be moving but you are infact revolving at 30 km/s (67,000 mi/h) around the sun

The earth also has a rotational speed about its axis. People on earth experience different tangential speeds depending on their latitude. The space shuttle at Kennedy Space Center is already traveling at 410 m/s (917 mi/h) before it even gets off the ground. A geosynchronous satellite that is in a stationary orbit over the earth is traveling at over 3080 m/s (6890 mi/h)

Solving Kinematics Problems

1. Assign a coordinate system – Define which directions are positive and negative.

2. Write down your known variables and show unknowns with a question

mark.

3. Write down the kinematics expression that will allow you to solve for

one variable. All the others in your expression should be known.

Rearrange if necessary.

4. Substitute numbers and units into your physical expression.

5. Solve the equation for your unknown and include the correct units.

6. Check your answer.

Does the magnitude of your answer make sense? Do the units come out right?

Example 1

Distance Run by a Jogger

How far does a jogger run in 1.5 hours (5400 s) if his

average speed is 2.22 m/s?



t = 5400s s

= 2.22 m/s d = ?

s

= d /



t

So…… d = s



t

= (2.22 m/s)(5400s)

d = 12000 m

Example 2

The World’s Fastest Jet-Engine Car

Andy Green in the car ThrustSSC set a world record of 341.1 m/s in

1997. To establish such a record, the driver makes two runs through the course, one in each direction,to nullify wind effects. From the data, determine the averagevelocity for each run.

a) t = 4.740 s x = +1609m v_av = ?

v_av = x / t = (+1609m) / (4.740 s)

v_av = + 339.5 m/s

b) t = 4.695 s x = -1609m v_av = ?

v_av = x / t = (-1609m) / (4.695 s)

Example 3

Acceleration and Increasing Velocity

Determine the average acceleration of the plane.

v

= 0 km/h

v

= 260 km/h t

= 0s t

= 29s

a

= (

v

-

v

) / (t

– t

)

a

= (+260 km/h – 0 km/h) / (29s – 0s)

Graphical Representation of Motion

Kinematics Relationships Through Graphing:

1. The slope of a d-t graph at any time tells you the av. velocity of the object.

2. The slope of a v-t graph at any time tells you the av. acceleration of the object.

3. The area under a v-t graph tells you the displacement of the object during that time.

Constant Motion

0 1 2 3 4 5 6

time (s) p o si ti o n ( m ) 0 2 4 6 8 10 12 14 16 18 20

0 1 2 3 4 5 6

time (s) v e lo ci ty ( m / s) 0 2 4 6 8 10

0 1 2 3 4 5 6

time (s) a c ce le ra ti o n ( m / s )

On the d-t graph at any point in time … v_av = ∆d / ∆t

v_av = (50 - 0)m / (5 - 0)s

v_av = 10 m/s

The slope is constant on this graph so the velocity is constant

On the v-t graph at any point in time… a_av = v_f - v_i / t

a_av = (10 - 10)m/s / (5 - 0)s

a_av = 0 m/s2

Looking at the area between the line and the x-axis….

Area of rectangle = b x h Area = 5s x 10 m/s = 50 m

Which is of course displacement

On the a-t graph the area between the line and the x-axis is….

Area of rectangle = b x h Area = 5s x 0 m/s2 ₌ 0 m/s

The area thus represents….

∆v = a_av ∆t

Changing Motion

0 1 2 3 4 5 6

time (s) p o si ti o n ( m ) -10 0 10 20 30 40 50 60

0 1 2 3 4 5 6

time (s) p o si ti o n ( m ) 0 5 10 15 20 25

0 1 2 3 4 5 6

time (s) v e lo ci ty ( m / s) 0 5 10 15 20 25

0 1 2 3 4 5 6

time (s) v e lo ci ty ( m / s) 0 1 2 3 4 5 6 7 8

0 1 2 3 4 5 6

time (s) a cc e le ra ti o n ( m / s/ s) 0 1 2 3 4 5 6 7 8

0 1 2 3 4 5 6

time (s) a cc e le ra ti o n ( m / s/ s)

On the d-t graph at any point in time … v_av = ∆d / ∆t

The slope is constantly increasing on this graph so the velocity is increasing at a constant rate The slope of a tangent line drawn at a point on the curve will tell you the instantaneous velocity at this position

On the v-t graph at any point in time… a_av = v_f - v_i / t

a_av = (20 - 0)m/s / (5 - 0)s a_av = 4 m/s2

Looking at the area between the line and the x-axis….

Area of triangle = 1/2 (b x h) Area = 1/2 (5s x 20 m/s) = 50 m

Which is of course displacement

On the a-t graph the area between the line and the x-axis is….

Area of rectangle = b x h Area = 5s x 4 m/s2 ₌ 20 m/s

-10 0 10 20 30 40 50 60

0 1 2 3 4 5 6

time (s)

0 5 10 15 20 25

0 1 2 3 4 5 6

time (s)

To determine the velocity at any point in time you need to find the slope of the distance-time graph. This means that you need to find the slope of the tangent line drawn at the point of interest. By selecting two points spaced evenly on either side of the point of interest, a line can be drawn between them that has the same slope as the tangent. (shown below).

Slope between 1s and 3s shows the

velocity at 2s The velocity at 2s is p/t = (18m - 2m)/ (3s - 1s) = 8m/s

The acceleration is given by the slope of the velocity-time graph. Therefore:

Example Problem

A student is late for the school bus. She runs east

down the road at 3 m/s for 30s, then thinks that

she has dropped her calculator so stops for 10s

to check. She jogs back west at 2 m/s for 10s,

stops for 5 s then accelerates uniformly from

rest to 4 m/s east over a 10 second period.

a) Sketch the velocity-time graph of the student’s

motion

b) Determine the total distance and displacement of

the student during this time

V-t Graph for Student Going To School

-3 -2 -1 0 1 2 3 4 5

0 10 20 30 40 50 60 70

time (s)

ve

lo

ci

ty

(

m

/s

)

Total distance traveled by the student is….

d

= d

+ d

+ d

+ d

+ d

d

= s



t

+ s



t

+ s



t

+ s



t

+ s



t

d

= (3m/s)(30s) + (0m/s)(10s) + (2m/s)(10s) +

………

(0m/s)(5s) + (1/2(4m/s)(10s)

d

= 130 m

Total displacement by the student is….



d

=



d

+



d

+



d

+



d

+



d



d

=

v



t

+

v



t

+

v



t

+

v



t

+

v



t



d

= (3m/s)(30s) + (0m/s)(10s) + (-2m/s)(10s) +

………

(0m/s)(5s) + (1/2(4m/s)(10s)



d

=

+ 90 m (East)

Average velocity of the student is…..

v

=



d

/



t

= + 90m East / 65s

v

=

1.4 m/s East

V-t Graph for Student Going To School

-3 -2 -1 0 1 2 3 4 5

0 10 20 30 40 50 60 70

time (s)

ve

lo

ci

ty

(

m

/s

)

+ 90 m

- 20 m

More Kinematics Equations that Make

Sense!

d_f = d_i + v_av t

d_f = d_i + (v_i + v_f) /2 t but v_f = v_i + a_av t

d_f = d_i + (v_i + (v_i + a_av t) /2 t

df = di + vi t + 1/2 aavt2

d_f = d_i + (v_i + v_f) /2 t but t = (v_f - v_i ) / a_av

d_f = d_i + (v_i + v_f) /2 (v_f - v_i) /a_av

d = (v_i + v_f) /2 (v_f - v_i) /a_{av So… d = (v}

f2 - vi2 ) /2aav

and… _v

f2 = vi2 + 2aav d

d = v_i t + 1/2 a_avt2

or…

Free Fall

1. Free fall describes the motion of an object which is only under the influence of

gravity. I.e. a ball thrown upwards or dropped

3. Kinematics equations can be used for solving free fall problems by replacing a_av in

the expressions with g where g is 9.8 m/s2 _downwards

2. An object in free fall experiences a constant uniform acceleration of 9.8 m/s2 _{in the}

downwards direction

v_f = v_i + g t _d

v = vi t + 1/2 g t2 vf

2 _{= v}

i2 + 2gdv

v_i is positive

+

-g is ne-gative

If you define up as the positive direction, g must be negative because the velocity gets less

positive over time

-+

v_i is negative

g is positive

If you define up as the negative direction, g must be positive because the velocity gets less negative over time

4. Air resistance limits the time of free fall. Eventually a falling object will reach a

Freely Falling Bodies

Example 1

A Falling Stone

Freely Falling Bodies



d

v

a

v

f

v

i



t

Freely Falling Bodies



d

v

a

v

f

v

i



t

?

-9.80 m/s

0 m/s

3.00 s



d

= v



t + ½ g



t

= (0 m/s)(3.00s) + ½ (-9.8 m/s

)(3.00s)

Freely Falling Bodies

Example 2

How High Does it Go?

Freely Falling Bodies



d

v

a

v

f

v

i



t

Freely Falling Bodies



d

v

a

v

f

v

i



t

?

-9.80 m/s

0 m/s

+5.00

m/s

2

2

2

f

i

v

v

v

d

g



 



 







2 2

2

0 m s

5.00 m s

2 9.80 m s

v

d



 



v

f

2

= v

i

2

+ 2 g



d

v