1 / 74

# Physics - PowerPoint PPT Presentation

Physics. Unit 1 Motion Graphs. 5 Main Branches of Physics. - Mechanics. KINEMATICS. A “description” of motion. - Electricity. DYNAMICS. - Magnetism. A study of what “causes” motion. - Optics. - Waves.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Physics' - gagan

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Physics

Unit 1 Motion Graphs

- Mechanics

KINEMATICS

A “description” of motion

- Electricity

DYNAMICS

- Magnetism

A study of what “causes”

motion

- Optics

- Waves

The purpose of this chapter is to learn the 1st step of Mechanics (the study of motion) which is KINEMATICS (the study of motion with no regards to what is causing the motion). The study of what is “causing” the motion is known as dynamics, and we will study this in a later chapter.

• Scalar – a quantity that has a magnitude only, no direction.

• Ex: time (5 hours)

age (17 years)

temperature (20˚C)

distance (20 miles)

* YES, scalars have units.

• Vector – a quantity that has both magnitude AND a direction.

• Ex: displacement (10 m [S])

force (5 N [E])

• Distance (d) – the length of the path followed by an object

(scalar)

* If an object’s path is straight, the distance is the length of the straight line between start and finish.

** If an object’s path is NOT straight, the distance is the

length of the path if you were to “straighten it out” and

measure it the way you would measure the length of a

curved shoelace.

finish

start

finish

start

C

A

meters

-3 -2 -1 0 1

Using the number line above, what would be the distance travelled if an object travelled from …..

1m

- A to B

- A to C

- A to C and then back to A

- C to B, passing through A

4m

4m + 4m = 8m

4m + 1m = 5m

• Displacement – the change in an object’s position during a time interval. (vector)

• OR the length of a straight line from start to finish.

• Displacement ≠ distance. However, sometimes the magnitude will be the same.

• It doesn’t matter what path you take from your house to school, displacement will never change but distance will.

*Displacement must have both a magnitude (size) and a

direction (right, left, up, down, north, south, etc).

C

A

meters

-3 -2 -1 0 1

Using the number line above, find the distance travelled and the displacement in moving from

1m, 1m [right]* *notice a direction is given for displacement

- A to B

- C to A

- A to C and then back to A

- C to B, passing through A

Dx = 1 – (1m) = 0m

4m, 4m [left]*

8m, 0m no direction needed here

4m, 3m [left]*

+ or – can be used instead of R and L

Dx = (-2) – (1m) = -3m OR 3m [left]

• Average Speed (s) – the distance travelled during a time interval divided by the elapsed time. (scalar)

s = dist/Dt

(or s=dist/t)

C

A

miles

-3 -2 -1 0 1

Larry walks from point B to point C, and then goes directly to point A. If he walks at an average speed of 6 mph, how long does the trip take him?

d = 3mi + 4mi = 7mi

s = 6 mi/h

s = d/t  t = d/s = (7mi)/(6mi/h)=1.17h

1 h, 10 min

Use appropriate units

C

A

km

-3 -2 -1 0 1

Larry runs from point A to point B in 5 minutes and then proceeds to jog directly to point C, taking his time in 30 additional minutes. Find…

• Larry’s average speed during the first portion of the trip.

• The average speed during the second portion of the trip.

• Larry’s average speed for the entire trip.

s = d/t = (1km)/(5min) = 0.2 km/min = 12 km/h

s = d/t = (3km)/(30min) = 0.1 km/min = 6 km/h

s = d/t = (4km)/(35min) = 0.114 km/min = 6.86 km/h

• Average Velocity ( v ) – the displacement of an object divided by the elapsed time. (vector)

• Avg. velocity is a change in position over a change in time.

• Since displacement ≠ distance, velocity ≠ speed.

v = displace/Dt

(or v=d/t)

D

Sam runs the 400m dash. He starts and finishes at point A, travelling one complete circuit around the track. Each section of the track is 100m long. His average speed during each interval are as follows.

AB: 7 m/s

BC: 8 m/s

CD: 6 m/s

DA: 7.5 m/s

Find Sam’s avg. speed and avg. velocity for the entire trip.

C

B

s = d/t  t = d/s = 100m/7sec = 14.286 sec

100m/8sec = 12.5 sec

100m/6sec = 16.667 sec

100m/7.5sec = 13.333 sec

s = d/t = (400m)/(56.786s) = 7.04 m/sec

Avg Velocity = 0 since Dx = 0 for the entire trip. He ended in the exact location he started!!

D

Sam runs the 400m dash. He starts and finishes at point A, travelling one complete circuit around the track. Each section of the track is 100m long. His average speeds during each interval are as follows.

AB: 7 m/s, 14.286 sec

BC: 8 m/s, 12.5 sec

CD: 6 m/s, 16.667 sec

DA: 7.5 m/s, 13.333 sec

31.831 m

C

• Find Sam’s average speed and average velocity for the 1st half of the race.

B

s = d/t  t = d/s = 200m/(14.286+12.5s) = 7.47 m/s

v = Dx/t = (104.94m )/(14.286+12.5s) = 3.92 m/sec

Use Pythagorean theorem to determine the displacement from A to C

104.94

100

• What scalars have we learned about thus far?

• distance speed time

• What vectors have we learned about thus far?

• Displacement velocity

Scalars vs. Vectors

1

2

B

A

Displacement:

has magnitude & direction (example: 15 cm east)

Distance:

has a magnitude only (example: 6 ft)

Displacement is NEVER greater than distance traveled!

Scalars vs. Vectors (continued)

2

7 km

25 km

16o

1

24 km

has magnitude & direction (example: 15 mi/h North)

Velocity:

Speed:

has a magnitude only (example: 30 km/h)

Total time for the trip from 1 to 2: 1 hr

Speed = 31 km/h

Velocity = 25 km/h at 16o NE

If an object STARTS & STOPS at the same point, the velocity is ZERO! (since the displacement is zero)

d (m)

F

E

120

100

50

30

D

C

SLOPE

B

Speed on a d-t graph can be found by taking the _______________.

This gives us the change in distance of an object over a change in time.

sAB = rise/run = (30-0m) / (10-0s) = 3 m/s

t (sec)

A

sCD = rise/run = (100-50m) / (20-15s) = 10 m/s

10 15 20 27 36

Constant speed

Speeding UP –notice how more distance is covered each second

Constant Speed (but faster than AB)

Slowing Down–less dist. covered each second

d (m)

At rest –no distance covered, but time goes by

F

E

120

100

50

30

D

C

B

t (sec)

A

10 15 20 27 36

FYI -d-t graphs CANNOT have sharp points. That would mean you came to a stop instantaneously without slowing down first.

minutes

520 – 170yd = 350 yd (approximately)

Like d-t graphs)

(displacement)

x (m)

x2

x1

x3

C

B

D

t (sec)

A

t1 t2 t3

Constant speed (Constant + velocity, or constant velocity in the + direction)

Slow down, speed up, slow down, speed up

2 moments where the object is “at rest” (for a moment) imagine slowing your car to a stop, then going in reverse

How to get the displacement/position (d) at a certain time (t) off an d-t graph. Sometimes I’ll refer to this as x-t

d(m)

30

20

10

0

24m

t (s)

10 20 30 40 50

Example:

What is the position at t = 30 seconds?

Go over to t = 30.

Find the pt on the curve.

Find the x value for this time.

x(m)

17 m

30

20

10

0

10 m

t (s)

10 20 30 40 50

Example:

What is the displacement from t = 10 to t = 40?

Find xi

Find xf

Use D x = xf- xi

= + 7 m

x(m)

30

20

10

0

10 m

17 m

t (s)

10 20 30 40 50

Example:

What is the distance traveled from t = 10 to t = 40?

Find the distance traveled in the + direction to the highest point.

Find the distance traveled in the – direction from the highest point.

(27 m)

x(m)

30

20

10

0

10 m

17 m

t (s)

10 20 30 40 50

Example:

What is the average speed from t = 10 to t = 40 seconds?

dist10-40 = 27 m

(previous slide)

Avg. Speed = dist/ Dt

= 27m / 30 sec

= 0.9 m/s

x(m)

30

20

10

0

Dx10-40 = + 7 m (previous slide)

Avg. Velocity = slope

= Dx/ Dt

= + 7 / 30 sec

= + 0.23 m/s

t (s)

10 20 30 40 50

Example:

What is the average velocity from t = 10 to t = 40 seconds?

Notice the + sign. It indicates direction.

Will avg. velocity EVER be equal to avg. speed?

YES!!! When the path travelled was one-way, in a straight line.

NO!!!

Negative Average Velocity? x-t graph.

x(m)

30

20

10

0

Avg. vel. = slope = rise/run= -7 m / 20

= -.35 m/s

t (s)

10 20 30 40 50

Example:

What is the average velocity from t = 20 to t = 40 seconds?

Since the objects displacement is in the NEGATIVE direction, so is its average velocity.

Graph Example x-t graph.

-10 m

2)

3)

4)

avg velocity = slope = -15m / 6sec = -2.5 m/s

s = |v| = 2.5 m/s

At rest at t = 0 and t = 12 sec

5) x-t graph.

6)

Speeding up, const negative vel, slowing down, speeding up,

const positive velocity(slow), speeding up, constant positive velocity (fast)

Dx = x2 – x1 = (-10m) – (10m) = -20m (approximately)

Definition x-t graph.

• Instantaneous Velocity (v)– the velocity of an object at a precise moment in time.

Dt

Dt

Dt

Dt

Finally, “in the limit” that the time interval is infinitely small (or approximately zero), we find the velocity at a single moment in time.

 Hence the term

“instantaneous velocity”

To find the average velocity between two points in time, we find the slope of the line connecting these two points, thus finding the change in position (rise) over the change in time (run).

Dt

As the two points move EVEN CLOSER together, we find the average velocity for an EVEN SMALLER time interval.

As the two points move closer together, we find the average velocity for a smaller time interval.

To find instantaneous velocities, we still use the concept of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

Definition

Tangent to a Curve– a line that intersects a given curve at exactly one point.

Good Tangents of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

x(m) of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

30

20

10

0

Draw the tangent to the curve at the point in question. Then, find the slope of the tangent.

Slope = rise/run = 15 m / 9 s

= 1.7 m/s (approx)

t (s)

10 20 30 40 50

How to find the instantaneous velocity of a specific time interval from an x-t graph …

Example:

What is the instantaneous velocity at t = 20 seconds?

(24, 30)

(15, 15)

YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!!

x(m) of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

30

20

10

0

If the pt lies on a segment, find the slope of the segment.

Slope = 5 m / 10 s = 0.5 m/s

t (s)

10 20 30 40 50

How to find the instantaneous velocity of a specific time interval from an x-t graph …

Example:

What is the instantaneous velocity at t = 5?

YOU MUST SPECIFY WHICH POINTS YOU USED WHEN FINDING THE SLOPE!!!!

(10,10)

(0,5)

x(m) of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

30

20

10

0

Draw the tangent to the curve at the point in question. Then, find the slope of the tangent.

Slope = 0 (object at rest)

t (s)

10 20 30 40 50

How to find the instantaneous velocity of a specific time interval from an x-t graph …

Example:

What is the instantaneous velocity at t = 25 seconds?

x-t graphs of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

x (m)

x2

x1

x3

2

1

3

t (sec)

0

t1 t2 t3

Slope of line segment

Slope of line segment

Open to in your Unit 1 packet of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

1

(0,6)

(33,2)

(11,-20)

(13.5,-20)

Tangent to the curve has a slope of +22m / 22sec = 1 m/s

Tangent to the curve has a slope of -26m / 13.5s = -1.93 m/s

THEREFORE, v = -1.93 m/s and s = 1.93 m/s (approximately)

Definition of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

• Average Acceleration ( a )– the change in an object’s velocity in a given time interval……..IN OTHER WORDS, the rate of change of an object’s velocity.

• When you stop a car, you actually push the break petal a few seconds before coming to a complete stop. The velocity gradually slows.

a = vf-viOR a = Dv/t

tf-ti

Find the acceleration of each object of slope. But we find the slope OF THE TANGENT TO THE CURVE at the time in question

Slows down

Negative accel

• An object is moving at 40 m/s when it slows down to 20 m/s over a 10 second interval.

• An object is moving at -40 m/s, and 5 seconds earlier it was moving at -10 m/s.

• An object travelling at -10 in/min is moving at +20 in/min 2 minutes later.

4) An object moving at -30 mph is moving at -20 mph 10 hours later.

a = Dv/Dt = (20 – 40m/s) / 10sec = -2 m/s2

Speeds up

Negative accel

a = Dv/Dt = [-40 – (-10m/s)] / 5sec = -6 m/s2

a = Dv/Dt = [20 – (-10 in/min)] / 2min = +15 in/min2

Speeds up

+ Accel

Slows down

+ Accel

a = Dv/Dt = [-20 – (-30 mi/h)] / 10hr = + 1mi/h2

How can something have a negative acceleration when traveling in a positive direction?

• When a train, traveling in a positive direction (right) slows as it approaches the next station, velocity can still be +, but acceleration will be negative because initial velocity is larger than final velocity. ∆v is negative.

• But be careful…

• Negative acceleration doesn’t always mean deceleration. Think of a train moving in a negative direction (in reverse, or just to the left). Acceleration would be negative when the train gained speed and positive when the train lost speed to enter a station.

Did the last line confuse you? traveling in a positive direction?How can something slow down and have a positive acceleration?This example may help

An object moving at -30 mph is moving at -20 mph 10 hours later.

Its speed (30mph vs. 20 mph) clearly decreases.

* remember, speed is |v|

As time marches on, the velocity become MORE positive (b/c -20mph is more positive than - 30mph)  THEREFORE, Dv is +

What do the “units” of acceleration mean? traveling in a positive direction?

m/s/s

m/s2

m/s per second

3 m/s2 means that your velocity increased by 3 m/s every second.

-0.1 km/min2 means that your velocity decreased by 0.1 km/min every minute that you were moving.

The Key Equations traveling in a positive direction?

Displacement: ∆d = df- di

Really just tangents to the curve at a point.

Velocity:

Acceleration:

AVERAGE

INSTANTANEOUS

1 m/s traveling in a positive direction?

-1 m/s

4 m/s

+4 m/s

9 m/s

+9 m/s

14 m/s

+14 m/s

19 m/s

+19 m/s

Call right + and left – traveling in a positive direction?

vi= 5 m/s right = + 5 m/s vf= 4.8 m/s left = -4.8 m/s

a = (vf– vi) / t = (-4.8 – 5) / .002sec = -4,900 m/s2 = 4,900 m/s2 left

vi= 60 mi/h v2 = 0 mi/h a = -5 mi/h2

a = (vf– vi) / t  -5 = (0 – 60) / t  t = 12 s

x-t traveling in a positive direction?

v-t

END OF TODAY’S LECTURE

UNIFORM Velocity traveling in a positive direction?

Speed increases as slope increases

x

x

x

x

x

x

x

t

t

t

t

t

t

t

Object at REST

Moving forward or backward

x-t‘s

Object Positively Accelerating

Changing Direction

Object Speeding up

Object Negatively Accelerating

x traveling in a positive direction?

x

x

x

t

t

t

t

POSITIVE OR NEGATIVE ACCEL? SPEEDING UP OR SLOWING DOWN?

• Slope of the tangent gives vinst

• Getting more sloped  speeding up

• Getting more + sloped  + Accel

• Getting less sloped  slowing down

• Slopes are getting less +  - Accel

• Getting more sloped

•  speeding up

• Slopes are getting more –

•  - Accel

• Getting less sloped

•  slowing down

• Slopes are getting less –

•  + Accel

An easy way to remember it traveling in a positive direction?



I’m Positive!!!

I’m Negative!!!

Find the acceleration in each case. traveling in a positive direction?

v1 = 10 m/s, v2 = 20 m/s, Dt = 5sec

v1 = 10 m/s, v2 = -20 m/s, Dt = 10sec

v1 = -9 km/h, v2 = -27 km/h, Dt = 3 h

v1 = -9 km/h, v2 = 6 km/h, Dt = 3 h

v traveling in a positive direction?

v

v

v

v

t

t

t

t

t

UNIFORM Positive (+) Acceleration

Acceleration increases as slope increases

Changing Direction

v-t‘s

UNIFORM Velocity

(no acceleration)

Object at REST

UNIFORM Negative (-) Acceleration

v(m/s) traveling in a positive direction?

8

6

4

2

0

-2

-4

t (s)

2 4 6 8 10 12

v-t graphs

Constant + accel (slowing down)

At rest

Constant - Vel

Constant negative accel (speeding up)

Constant negative accel (slowing down)

Constant + Vel (constant speed)

Constant + accel (speeding up)

v(m/s)

8

6

4

2

0

-2

-4

t (s)

2 4 6 8 10 12

-2 m/s

Example:

What is the velocity at t = 8 seconds?

Go over to t = 8.

Find the pt on the graph.

Find the v value for this time.

v(m/s)

8

6

4

2

0

-2

-4

2 4 6 8 10 12

Example:

What is the average acceleration between 0 & 2, 2 & 4, and 4 & 10 seconds?

a0-2 = (vf– vi) / Dt

= rise / run

= +4/2 = +2 m/s2

A2-4 = (vf– vi) / Dt

= rise / run

= 0 m/s2

A4-10 = (vf– vi) / Dt

= rise / run

= -7 / 6 = -1.17 m/s2

v-t graphs graph

The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration

Slope of any segment is the AVERAGE acceleration

v (m/s)

t (sec)

t0 t1

3 graph

Open to in your Unit 1 packet

30

20

10

0

-10

-20

-30

v = -30 m/s s = 30 m/s

a = slope = (-30 m/s) / 16sec = -1.875 m/s2

a = slope = (+57 m/s) / 32sec = +1.78 m/s2 (approx)

3 graph

Open to in your Unit 1 packet

30

20

10

0

-10

-20

-30

3)

4)

5)

You can’t say. You know its speed at the start, but not where it is 

Object is at rest whenever it crosses the t-axis  t = 0, 36, 80 sec

Const – accel (object speeds up), const – vel, const + accel (slows down), const + accel (speeds up), const – accel (slows down) END OF LECTURE

UNIFORM Velocity graph

Speed increases as slope increases

x

x

x

x

x

x

x

t

t

t

t

t

t

t

Object at REST

Moving forward or backward

x-t‘s

Object Positively Accelerating

Changing Direction

Object Speeding up

Object Negatively Accelerating

v graph

v

v

v

v

t

t

t

t

t

UNIFORM Positive (+) Acceleration

Acceleration increases as slope increases

Changing Direction

v-t‘s

UNIFORM Velocity

(no acceleration)

Object at REST

UNIFORM Negative (-) Acceleration

A Quick Review graph

Average velocity

Inst. velocity

Avg. accel.

Inst. accel.

The slope between 2 points on an x-t graph gets you the _______________.

The slope at a single point (the slope of the tangent to the curve) on an x-t graph gets you the ____________.

The slope between 2 points on a v-t graph gets you the ____________.

The slope at a single point (the slope of the tangent to the curve) on a v-t graph gets you the ____________.

NEW CONCEPT graph

v

v

t

t

The “area under the curve” is really the area between the graph and the t-axis.

This is NOT the area under the curve 

When you find the area “under the curve” on a v-t graph, this gets you the displacement during the given time interval.

v graph

t

The displacement during the first 4 seconds is -20m

A = ½ (4)(-10) = -20m

A = ½ (2)(-10) = -10m

The displacement during the next 2 seconds is -10m

A = ½ (4)(15) = 30m

The displacement during the next 4 seconds is 30m

15

-10

A = -20 + (-10) + 30

= 0m

4 6 10

The OVERALL displacement from 0 to 10 seconds is zero (its back to where it started)

Find the area under the curve from ….

• 0-4 seconds.

• 4-6

• 6-10

• 0-10

How to find the graph displacement from one time to another from a v-t graph

v(m/s)

8

6

4

2

0

-2

-4

12 m + 4 m = 16 m

Find the positive area bounded by the “curve”

t (s)

2 4 6 8 10 12

Find the negative area bounded by the “curve”

(-2.25 m) + (-4.5 m) = - 6.75 m

Example:

What is the displacement from t = 2 to t = 10?

Add the positive and negative areas together 

Dx = 16 m + (-6.75 m) = 9.25 m

How to find the graph distance traveled from one time to another from a v-t graph

v(m/s)

8

6

4

2

0

-2

-4

12 m + 4 m = 16 m

Find the positive area bounded by the “curve”

t (s)

2 4 6 8 10 12

Find the negative area bounded by the “curve”

(-2.25 m) + (-4.5 m) = - 6.75 m

Example:

What is the distance traveled from t = 2 to t = 10?

Add the MAGNITUDES of these two areas together 

distance = 16 m + 6.75 m = 22.75 m

How to find the graph average velocity during a time interval on a v-t graph

v(m/s)

8

6

4

2

0

-2

-4

12 m + 4 m = 16 m

The DISPLACEMENT is simply the area “under” the curve.

Dx = 16 m + (-6.75 m)

= 9.25 m

t (s)

2 4 6 8 10 12

(-2.25 m) + (-4.5 m) = - 6.75

Example:

What is the average velocity from t = 2 to t = 10?

The AVG velocity = Dx / Dt = 9.25 m / 8 s = 1.22 m/s

How to find the graph final position of an object using a v-t graph (and being given the initial position)

v(m/s)

8

6

4

2

0

-2

-4

4 m 12 m + 4 m = 20 m

The DISPLACEMENT during the 1st 10 sec is simply the area “under” the curve.

Dx = 20 m + (-6.75 m)

= 13.25 m

t (s)

2 4 6 8 10 12

(-2.25 m) + (-4.5 m) = - 6.75

Example:

What is the final position after t = 10 seconds if xi = 40 m?

Dx = xf– xi xf= Dx + xi= 13.25 m + 40 m = 53.25 m

v-t graphs graph

The slope of the tangent to the curve at any point is the INSTANTANEOUS acceleration

The area under the curve between any two times is the CHANGE in position (the displacement) during that time period.

Slope of any segment is the AVERAGE acceleration

v (m/s)

t (sec)

t0 t1

3 graph

Open to in your Unit 1 packet

30

20

10

0

-10

-20

-30

6)

7)

Distance travelled = |area| = | ½ (16)(-30) | = 240m

s = d/t = 240 m / 16 sec = 15 m/s

Displacement = |area| = ½ (16)(-30) + 8 (-30) = -480m

v = Dx/t = -480 m / 24 sec = -20 m/s

3 graph

Open to in your Unit 1 packet

30

20

10

0

-10

-20

-30

8)

Find all the areas “under the curve” from 0 to 44 sec

Area = ½ (16)(-30) + 12(-30) + ½ (8)(-30) + ½ (8)(30) = - 600 m

Area = Dx = - 600 m  Dx = xf– xi -600m = xf– (-16m)  xf= - 616m

4 graph

Open to in your Unit 1 packet

+3.3 m/s2

+10 m/s

0 m/s

+75 m

10 graph

-10

5

-2 m/s2

0 m

50 m

30 m

9) graph

10)

11)

12)

13)

14 & 34 sec

+.35 m/s2

+ 8 m/s

+ 2 m/s2

approx 0.8 m/s2