ch5 – mathematical models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) time (sec) position (m)...

80
Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 0 10 1 12 2 14 3 16 4 20 5 26 Where is the runner: at 2 sec? at 3 sec? at 4½ sec? At what time was the runner at the 15m mark? What is his speed at 2 sec?

Upload: clifford-strickland

Post on 25-Dec-2015

214 views

Category:

Documents


0 download

TRANSCRIPT

Ch5 – Mathematical Models

30

25dist (m)

20

15

10

5

1 2 3 4 5 6 time (sec)

Time (sec) Position (m)0 101 122 143 164 20 5 26

Where is the runner:at 2 sec?at 3 sec?at 4½ sec?

At what time was the runner at the 15m mark?What is his speed at 2 sec?

Ch5 – Mathematical Models

30

25dist (m)

20

15

10

5

1 2 3 4 5 6 time (sec)

Time (sec) Position (m)0 101 122 143 164 20 5 26

Where is the runner:at 2 sec? 14mat 3 sec? 16m at 4½ sec? 23m

At what time was the runner at the 15m mark? 2.5 secWhat is his speed at 2 sec?

Velocity (speed) = the slope of the line on a distance vs time graph.

m/s 5.22

5))((

s

m

run

risemslopev

Ex2) Describe each person’s motion. (What is their speed?)

1.

2.

3. 1 2 3 4 5 time (sec)

10

8dist (m)

6

4

2

Ex3) Write an equation that describes the motion of the airplane graphed,

then use it to find its position at t = 2.5 sec.

200

160dist (m)

120

80

40 1 2 3 4 5

time (sec)

Ex4) a car starts 200m west of town square and moves at a constant vel of 15m/s east.

A) Write an eqn to represent motion B) Where is the car after 2 min?C) When will the car reach town square?

v = 15m/s Town

Ex5) How fast is the plane going at 1.5 sec

dist

(m)

time (sec)

Ex5) How fast is the plane going at 1.5 sec

dist

(m)

r

time (sec)

velocity = slope = = 6 m/s

If the line curves, find the slope of the tangent line.

Ch5 HW#1 1 – 4

rise = 6m

run = 1sec

s

m

1

6

Ch5 HW#1 1 – 5

1) How far is:

- object A at t=3:

t=8:

-object B at t=4:

t=8:

-object C at t=5: t=10:

2) What is the velocity- of A

- of B - of C

Ch5 HW#1 1 – 5

1) How far is:

- object A at t=3: 3m

t=8: 8m

-object B at t=4: 6m

t=8: 4m

-object C at t=5: 7m t=10:

7m

2) What is the velocity- of A

- of B - of C

Ch5 HW#1 1 – 5

1) How far is:

- object A at t=3: 3m

t=8: 8m

-object B at t=4: 6m

t=8: 4m

-object C at t=5: 7m t=10:

7m

2) What is the velocity- of A 1m/s

- of B -½m/s - of C 0m/s3. Object D starts at 0m and moves with

constant velocity of 2m/s. What is its positionafter 4 sec?

Ch5 HW#1 1 – 5

1) How far is:

- object A at t=3: 3m

t=8: 8m

-object B at t=4: 6m

t=8: 4m

-object C at t=5: 7m t=10:

7m

2) What is the velocity- of A 1m/s

- of B -½m/s - of C 0m/s

3. Object D starts at 0m and moves withconstant velocity of 2m/s. What is its positionafter 4 sec?

df = di + vt 0m + (2m/s)(4s) = 8m

4. Object E starts 10m away and moves at –2m/s. When will it reach the start line?

Ch5 HW#1 1 – 5

1) How far is:

- object A at t=3: 3m

t=8: 8m

-object B at t=4: 6m

t=8: 4m

-object C at t=5: 7m t=10:

7m

2) What is the velocity- of A 1m/s

- of B -½m/s - of C 0m/s

3. Object D starts at 0m and moves withconstant velocity of 2m/s. What is its positionafter 4 sec?

df = di + vt 0m + (2m/s)(4s) = 8m

4. Object E starts 10m away and moves at –2m/s. When will it reach the start line?df = di + vt

0m = 10m + (–2m/s)(t) t = 5sec

Ch5.2 Velocity Graphs

Ex1) Explain the velocity of B:

Ex2) Explain the velocity of A:

Ex3) If B travels at 5m/s for 6 secs,

how far did it travel?

Ex4) Find the displacement of C after 4 sec.

Ex5) Find the displacement of A after 3 sec.

Ch5 HW#2 5 – 8

10 9 8 7 vel 6 (m/s) 5 4 3 2 1

1 2 3 4 5 6 7 8 9 10 time (sec)

Lab5.1 – Ave vs Inst Acceleration

- due tomorrow

- Ch5 HW#2 due at beginning of period

Ch5 HW#2 5 – 8

5) vel A at t=2 v=___ m/s

vel B at t=4 v=___ m/s

vel C at t=6 v=___ m/s

6) Displacement of A at t=5 sec:

7) Displacement of B at t=8 sec:

8) Displacement of C at t=10 sec:

Ch5 HW#2 5 – 8

5) vel A at t=2 v=_4_ m/s

vel B at t=4 v=_3_ m/s

vel C at t=6 v=_2_ m/s

6) Displacement of A at t=5 sec:

7) Displacement of B at t=8 sec:

8) Displacement of C at t=10 sec:

Ch5 HW#2 5 – 8

5) vel A at t=2 v=_4_ m/s

vel B at t=4 v=_3_ m/s

vel C at t=6 v=_2_ m/s

6) Displacement of A at t=5 sec:

d = (area) = l.w = (5sec)(4m/s) = 20m

7) Displacement of B at t=8 sec:

8) Displacement of C at t=10 sec:

Ch5 HW#2 5 – 8

5) vel A at t=2 v=_4_ m/s

vel B at t=4 v=_3_ m/s

vel C at t=6 v=_2_ m/s

6) Displacement of A at t=5 sec:

d = (area) = l.w = (5sec)(4m/s) = 20m

7) Displacement of B at t=8 sec:

d = (area) = ½b.h = ½(8sec)(8m/s)

= 32m

8) Displacement of C at t=10 sec:

Ch5 HW#2 5 – 8

5) vel A at t=2 v=_4_ m/s

vel B at t=4 v=_3_ m/s

vel C at t=6 v=_2_ m/s

6) Displacement of A at t=5 sec:

d = (area) = l.w = (5sec)(4m/s) = 20m

7) Displacement of B at t=8 sec:

d = (area) = ½b.h = ½(8sec)(8m/s)

= 32m

8) Displacement of C at t=10 sec:

d = (area) – (area) = ½b.h – ½b.h

= ½(8sec)(8m/s) – ½(2sec)(2m/s)

= 32m – 2m = 30m

Ch 5.2 - Acceleration

t

vva

t

va if

or Units:

HW#9) A race car’s velocity increases from 4.0m/s to 36m/s in 4 sec.

What is it average acceleration?

Ex9) Describe each motion:

A:

B:

Find the acceleration of each :

A:

B:

Find the distance traveled for each :

A:

B:

Ex2) Describe the sprinter’s velocity :

Describe the accl :

What is the instantaneous accl

a. t= 1 sec?

b. t = 5 sec?

What is the average accl from t = 0 – 10 sec?

Ch5 HW#3 9 – 14

10 9 8 7 vel 6 (m/s) 5 4 3 2 1

1 2 3 4 5 6 7 8 9 10 time (sec)

Ch5 HW#3 9 – 14 9. In class

10. A race car slows from 36 m/s to 15 m/s over 35sec. What was its average acceleration?

vi = 36 m/s

vf = 15 m/s

t = 35 sec

a = ?

11) A car is coasting backward downhill at a speed of 3 m/s when the driver gets the engine started. After 2.5 sec, the car is moving uphill at 4.5 m/s.

If uphill is (+) what is the car’s average acceleration?

vi = –3 m/s

vf = +4.5 m/s

t = 2.5 sec

a = ?

Ch5 HW#3 9 – 14 9. In class

10. A race car slows from 36 m/s to 15 m/s over 35sec. What was its average acceleration?

vi = 36 m/s

vf = 15 m/s

t = 35 sec

a = ?

11) A car is coasting backward downhill at a speed of 3 m/s when the driver gets the engine started. After 2.5 sec, the car is moving uphill at 4.5 m/s.

If uphill is (+) what is the car’s average acceleration?

vi = –3 m/s

vf = +4.5 m/s

t = 2.5 sec

a = ?

2/735

/36/15sm

s

smsm

t

vva if

Ch5 HW#3 9 – 14 9. In class

10. A race car slows from 36 m/s to 15 m/s over 35sec. What was its average acceleration?

vi = 36 m/s

vf = 15 m/s

t = 35 sec

a = ?

11) A car is coasting backward downhill at a speed of 3 m/s when the driver gets the engine started. After 2.5 sec, the car is moving uphill at 4.5 m/s.

If uphill is (+) what is the car’s average acceleration?

vi = –3 m/s

vf = +4.5 m/s

t = 2.5 sec

a = ?

2/735

/36/15sm

s

smsm

t

vva if

2/35.2

)/3(/5.4sm

s

smsm

t

vva if

12) A bus is moving at 25 m/s when the driver steps on the breaks and brings the bus to a stop in 3.0 sec.

a. What is the average acceleration?

b. If it took twice as long to stop, what would be the accl?

vi = 25 m/s

vf = 0 m/s

t = 3.0 sec

a = ?

13)Look at the graph:

a. Where is the speed constant?

b. Over what intervals is accl positive?

c. Over what intervals is accl negative?

14) Find average acceleration at:

a. 0 to 5 sec

b. 15 to 20 sec

c. 0 to 40 sec

10

8

6

4

2

10 20 30 40 50 time (sec)

vel (m/s)

12) A bus is moving at 25 m/s when the driver steps on the breaks and brings the bus to a stop in 3.0 sec.

a. What is the average acceleration?

b. If it took twice as long to stop, what would be the accl?

vi = 25 m/s

vf = 0 m/s

t = 3.0 sec

a = ?

13)Look at the graph:

a. Where is the speed constant?

b. Over what intervals is accl positive?

c. Over what intervals is accl negative?

14) Find average acceleration at:

a. 0 to 5 sec

b. 15 to 20 sec

c. 0 to 40 sec

2/3.80.3

/25/0sm

s

smsm

t

vva if

2/1.40.6

/25/0sm

s

smsm

t

vva if

10

8

6

4

2

10 20 30 40 50 time (sec)

vel (m/s)

12) A bus is moving at 25 m/s when the driver steps on the breaks and brings the bus to a stop in 3.0 sec.

a. What is the average acceleration?

b. If it took twice as long to stop, what would be the accl?

vi = 25 m/s

vf = 0 m/s

t = 3.0 sec

a = ?

13)Look at the graph:

a. Where is the speed constant? 5-15sec

b. Over what intervals is accl positive? 0-5

c. Over what intervals is accl negative? 15-20s,

25-40s

14) Find average acceleration at:

a. 0 to 5 sec

b. 15 to 20 sec

c. 0 to 40 sec

2/3.80.3

/25/0sm

s

smsm

t

vva if

2/1.40.6

/25/0sm

s

smsm

t

vva if

10

8

6

4

2

10 20 30 40 50 time (sec)

vel (m/s)

12) A bus is moving at 25 m/s when the driver steps on the breaks and brings the bus to a stop in 3.0 sec.

a. What is the average acceleration?

b. If it took twice as long to stop, what would be the accl?

vi = 25 m/s

vf = 0 m/s

t = 3.0 sec

a = ?

13)Look at the graph:

a. Where is the speed constant? 5-15sec

b. Over what intervals is accl positive? 0-5

c. Over what intervals is accl negative? 15-20s,

25-40s

14) Find average acceleration at:

a. 0 to 5 sec (slope) a = 2m/s2

b. 15 to 20 sec (slope) a = -6/5 m/s2

c. 0 to 40 sec

2/3.80.3

/25/0sm

s

smsm

t

vva if

2/1.40.6

/25/0sm

s

smsm

t

vva if

2/040

/0/0sm

s

smsm

t

vva if

10

8

6

4

2

10 20 30 40 50 time (sec)

vel (m/s)

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

2. vf = vi + a∙t (accelerating, independent of distance)

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

2. vf = vi + a∙t (accelerating, independent of distance)

3. d = ½(vi +vf ).t (accelerating, independent of the accl)

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

2. vf = vi + a∙t (accelerating, independent of distance)

3. d = ½(vi +vf ).t (accelerating, independent of the accl)

4. df = vit + ½a.t2 (accelerating, independent of vf)

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

2. vf = vi + a∙t (accelerating, independent of distance)

3. d = ½(vi +vf ).t (accelerating, independent of the accl)

4. df = vit + ½a.t2 (accelerating, independent of vf)

5. vf2 = vi

2 + 2.a.d (accelerating, independent of time)

To solve motion problems, must be given 3 variables then solve for a 4th.

Ex1) What is the final velocity of a car that accelerates from a stop at 3.5m/s2

for 4 sec?

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

2. vf = vi + a∙t (accelerating, independent of distance)

3. d = ½(vi +vf ).t (accelerating, independent of the accl)

4. df = vit + ½a.t2 (accelerating, independent of vf)

5. vf2 = vi

2 + 2.a.d (accelerating, independent of time)

Ex2) A car is going 20m/s when it uniformly accelerates to 25m/s in 2 sec.

What distance does it travel?

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

2. vf = vi + a∙t (accelerating, independent of distance)

3. d = ½(vi +vf ).t (accelerating, independent of the accl)

4. df = vit + ½a.t2 (accelerating, independent of vf)

5. vf2 = vi

2 + 2.a.d (accelerating, independent of time)

HW#15) A golf ball rolls up hill.

a. If it starts at 2.0m/s and shows at a constant . 50 m/s2 what is its velocity

after 2.0 sec?

b. If the ball continues for 6 sec, what is its velocity?

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

2. vf = vi + a∙t (accelerating, independent of distance)

3. d = ½(vi +vf ).t (accelerating, independent of the accl)

4. df = vit + ½a.t2 (accelerating, independent of vf)

5. vf2 = vi

2 + 2.a.d (accelerating, independent of time)

Ex2) A car starts at rest and speeds up at 3.5 m/s2 . How far will it have gone

when it is going 25 m/s?

Ch5.3 – Motion Equations

1. df = di + v∙t (constant velocity, a = 0)

2. vf = vi + a∙t (accelerating, independent of distance)

3. d = ½(vi +vf ).t (accelerating, independent of the accl)

4. df = vit + ½a.t2 (accelerating, independent of vf)

5. vf2 = vi

2 + 2.a.d (accelerating, independent of time)

HW#19) A race car traveling at 44m/s slows at a constant rate to

a velocity of 22m/s over 11sec. How far does it move over that time?

Ch5 HW#4 15 – 22

Lab5.2 – Motion on an Incline Plane

- due tomorrow

- Ch5 HW#4 due at beginning of period

Ch5 HW#4 15 – 22

15. In class

16. A bus travelling at 30 m/s, speeds up at a constant rate of 3.5 m/s2. What is its

velocity 6.8 sec later?

vi =

vf =

t =

a =

17. If a car accelerates from rest at a constant 5.5 m/s2, how long will it need to

reach 28 m/s?

Ch5 HW#4 15 – 22

15. In class

16. A bus travelling at 30 m/s, speeds up at a constant rate of 3.5 m/s2. What is its

velocity 6.8 sec later?

vi = 30 m/s vf = vi + a ∙ t

vf = ?

t = 6.8 sec

a = 3.5 m/s2

17. If a car accelerates from rest at a constant 5.5 m/s2, how long will it need to

reach 28 m/s?

vi = 0 m/s vf = vi + a ∙ t

vf = 28 m/s

a = 5.5 m/s2

t = ? or d = ? vf2 = vi

2 + 2.a.d

Ch5 HW#4 15 – 22

18. A car slows from 22 m/s to 3 m/s at a constant rate of 2.1 m/s2 . How much time

did this take?

vi =

vf =

a =

t =20. A car accelerates at a constant rate from 15 m/s to 25 m/s while it travels

125m. How much time does this take?

vi =

vf =

d =

t =

Ch5 HW#4 15 – 22

18. A car slows from 22 m/s to 3 m/s at a constant rate of 2.1 m/s2 . How much time

did this take?

vi = 22 m/s vf = vi + a ∙ t

vf = 3 m/s

a = –2.1 m/s2

t = ? 20. A car accelerates at a constant rate from 15 m/s to 25 m/s while it travels

125m. How much time does this take?

vi = 15 m/s d = ½(vi +vf ).t

vf = 25 m/s

d = 125 m

t = ?

21.A bike rider accelerates constantly to a velocity of 7.5 m/s during 4.5 sec. The bikes displacement during that accl is 19m. What is the initial velocity of the bike?vi = ? vf =t =d =

22. An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2

for 30 seconds a. How far did it move? b. How fast was it going?

vi =a =t =d = ?vf = ?

21.A bike rider accelerates constantly to a velocity of 7.5 m/s during 4.5 sec. The bikes displacement during that accl is 19m. What is the initial velocity of the bike?vi = ? d = ½(vi +vf ).tvf = 7.5 m/s

t = 4.5s d = 19m

22. An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2

for 30 seconds a. How far did it move? b. How fast was it going?

vi = 0 m/s df = vit + ½a.t2 vf = vi + a ∙ ta = 3.00 m/s

2 t = 30 secd = ?vf = ?

Ch 5.4 – Free Fall

Neglecting air resistance, all objects fall at the same rate. (Same acceleration.)

Doesn’t matter if objects are thrown upward, thrown downward, across, etc, their acceleration is toward the center of the earth.

a = g = 9.8 m/s2 (You are authorized to use 10 m/s

2.)

Ex1) Demon drop falls freely for 1.5 sec, after starting from rest.

a. What is its velocity at 1.5 sec? b. How far does it fall?

Ex2) A bullet is fired straight up at 250 m/s.

a. How high does it go?

b. How much time is it in the air?

c. How fast is it going when it comes back down?

Graphs Ex3) A ball is dropped from a hovering helicopter.

Graph its velocity and displacement for the 1st 5 sec.

vf = vi + a ∙ t d = ½(vi +vf ).t

t = 1

t = 2

t = 3

t = 4

t = 5

80

70

60

50

40

30

20

10

1 2 3 4 5 time (sec)

50

40

30

20

10

1 2 3 4 5 time (sec)

Ch5 HW#5 23-25

vel (m/s)

dist (m)

Lab5.3 – Measuring ‘g’

- due tomorrow

- Ch5 HW#5 due at beginning of period

- Ch5 HW#6 due tomorrow

Ch5 HW#5 23-25

23) A brick is dropped, a. What is its velocity after 4 sec?

b. How far does it fall? vf = vi + a ∙ t df = vit + ½a.t2

vi = 0 m/s

a = 9.8 m/s2

t = 4 sec

vf = ? d = ?

24) Tennis ball thrown straight up with an initial speed of 22.5 m/s.

a. How high does it rise? vf = 0 d = ? ttotal = ?

b. How long till caught again? a = –9.8 m/s2

vi = 22.5 m/s

Ch5 HW#5 23-25

23) A brick is dropped, a. What is its velocity after 4 sec?

b. How far does it fall? vf = vi + a ∙ t df = vit + ½a.t2

vi = 0 m/s

a = 9.8 m/s2 df = ½g.t2

t = 4 sec

vf = ? d = ? vf = 39.2 m/s df = 78.4 m

24) Tennis ball thrown straight up with an initial speed of 22.5 m/s.

a. How high does it rise? vf = 0 d = ? ttotal = ?

b. How long till caught again? a = –9.8 m/s2

vi = 22.5 m/s

Ch5 HW#5 23-25

23) A brick is dropped, a. What is its velocity after 4 sec?

b. How far does it fall? vf = vi + a ∙ t df = vit + ½a.t2

vi = 0 m/s

a = 9.8 m/s2 df = ½g.t2

t = 4 sec

vf = ? d = ? vf = 39.2 m/s df = 78.4 m

24) Tennis ball thrown straight up with an initial speed of 22.5 m/s.

a. How high does it rise? vf = 0 d = ? ttotal = ?

b. How long till caught again? a = –9.8 m/s2

vi = 22.5 m/s

vf2 = vi

2 + 2.a.d vf = vi + a∙tup

02 = 22.52 + 2(–9.8)d 0 = 22.5 + (–9.8) ∙tup

d = 25.8 m tup = 2.3 sec x 2

ttotal = 4.6 sec

25) A spaceship accelerates uniformly from 65.0 m/s to 162.0 m/s in 10.0 sec.

How far does it move?vi = 65.0 m/s vf = 162.0 m/s t = 10 sec d = ½(vi +vf ).td = ?

25) A spaceship accelerates uniformly from 65.0 m/s to 162.0 m/s in 10.0 sec.

How far does it move?vi = 65.0 m/s vf = 162.0 m/s t = 10 sec d = ½(vi +vf ).td = ?

d = ½(65+162)(10)

d = 1135 m

Ch5 HW#6 - Motion Equations WS 26. What is the velocity of a car that starts at rest,

and accelerates at a constant 5 m/s2 for 5 sec?27. What is the velocity of a car that starts at rest,

and accelerates at a constant 5 m/s2 for 20 m?28. What is the displacement of an object that starts

with an initial velocity of 10 m/s and undergoes a constant acceleration of 2 m/s2 for 4 sec?

29. If a bicyclist accelerates at a steady rate to finish the sprint at 25 m/s and had an average velocity of 18 m/s over the sprint, what was his initial velocity?

30. An object is dropped from the top of a building 35 m tall. If it accelerates at 9.8 m/s2, neglecting air friction, how long will it take for it to hit the ground?

31. In the last problem, how fast is the object going right before it hits the ground?

32. If you are curious about the depth of a mine shaft, old timers will tell you to drop a rock in the shaft and time until you hear the rock hit the bottom. If the acceleration of gravity, g, is 9.8 m/s2 and the rock falls for 3.5 sec, roughly how deep is the hole?

Ch5 HW#6 - Motion Equations WS 26. What is the velocity of a car that starts at rest,

and accelerates at a constant 5 m/s2 for 5 sec?27. What is the velocity of a car that starts at rest,

and accelerates at a constant 5 m/s2 for 20 m?28. What is the displacement of an object that starts

with an initial velocity of 10 m/s and undergoes a constant acceleration of 2 m/s2 for 4 sec?

29. If a bicyclist accelerates at a steady rate to finish the sprint at 25 m/s and had an average velocity of 18 m/s over the sprint, what was his initial velocity?

30. An object is dropped from the top of a building 35 m tall. If it accelerates at 9.8 m/s2, neglecting air friction, how long will it take for it to hit the ground?

31. In the last problem, how fast is the object going right before it hits the ground?

32. If you are curious about the depth of a mine shaft, old timers will tell you to drop a rock in the shaft and time until you hear the rock hit the bottom. If the acceleration of gravity, g, is 9.8 m/s2 and the rock falls for 3.5 sec, roughly how deep is the hole?

26. vf= vi+at 27. vf

2 = vi

2 + 2as

28. d = vi

.t + ½at2 29. vave = ½(vi +vf ) 30. d = vi

.t + ½at2

31. vf= vi+at 32. d = vi

.t + ½at2

Ch5 – Graphing Worksheet1. Graph of velocity vs time for a car on a straight road. a. Inst vel at: t=2:___, t=3.5:___, t=8:___ b. Car’s displacement at: t=3:___, t=4:___, t=5:___,

t=7:___, t=9:___ c. What is the inst accl at: t=2:___, t=3.5:___, t=4.5:___,

t=6:___, t=8:___

40 35vel 30 (m/s) 25 20 15 10 5

1 2 3 4 5 6 7 8 9 10 time (sec)

Ch5 – Graphing Worksheet1. Graph of velocity vs time for a car on a straight road. a. Inst vel at: t=2: 6, t=3.5: 10, t=8: 10 b. Car’s displacement at: t=3:___, t=4:___, t=5:___,

t=7:___, t=9:___ c. What is the inst accl at: t=2:___, t=3.5:___, t=4.5:___,

t=6:___, t=8:___

40 35vel 30 (m/s) 25 20 15 10 5

1 2 3 4 5 6 7 8 9 10 time (sec)

Ch5 – Graphing Worksheet1. Graph of velocity vs time for a car on a straight road. a. Inst vel at: t=2: 6, t=3.5: 10, t=8: 10 b. Car’s displacement at: t=3: 15, t=4: 25, t=5: 40,

t=7: 80, t=9: 100 c. What is the inst accl at: t=2:___, t=3.5:___, t=4.5:___,

t=6:___, t=8:___

40 35vel 30 (m/s) 25 20 15 10 5

1 2 3 4 5 6 7 8 9 10 time (sec)

Ch5 – Graphing Worksheet1. Graph of velocity vs time for a car on a straight road. a. Inst vel at: t=2: 6, t=3.5: 10, t=8: 10 b. Car’s displacement at: t=3: 15, t=4: 25, t=5: 40,

t=7: 80, t=9: 100 c. What is the inst accl at: t=2: +3.3, t=3.5: 0, t=4.5: +10,

t=6: 0, t=8: +10

From a vel-time graph, you can get 3 things:1. velocity: read it2. displacement: area3. acceleration: slope

40 35vel 30 (m/s) 25 20 15 10 5

1 2 3 4 5 6 7 8 9 10 time (sec)

2. Make a displacement vs time graph for a person walks 10m in 5s, pauses for 10s to tie a shoe, runs 20m in 5s, pauses for 10s to tie other shoe, then runs back to start in 10s.

3. Velocity at each time interval:

0-5sec: __________ 5-10sec: __________10-15sec: __________15-20sec: __________20-25sec: __________25-30sec: __________30-35sec: __________35-40sec: __________

40 35dist 30 (m) 25 20 15 10 5

5 10 15 20 25 30 35 40 time (sec)

2. Make a displacement vs time graph for a person walks 10m in 5s, pauses for 10s to tie a shoe, runs 20m in 5s, pauses for 10s to tie other shoe, then runs back to start in 10s.

3. Velocity at each time interval:

0-5sec: __________ 5-10sec: __________10-15sec: __________15-20sec: __________20-25sec: __________25-30sec: __________30-35sec: __________35-40sec: __________

40 35dist 30 (m) 25 20 15 10 5

5 10 15 20 25 30 35 40 time (sec)

2. Make a displacement vs time graph for a person walks 10m in 5s, pauses for 10s to tie a shoe, runs 20m in 5s, pauses for 10s to tie other shoe, then runs back to start in 10s.

3. Velocity at each time interval:

0-5sec: _2________ 5-10sec: _0________10-15sec: _0________15-20sec: _4________20-25sec: _0________25-30sec: _0________30-35sec: _-3________35-40sec: _-3________

4. Graph the velocities: 4 3 2 1

vel 0 5 10 15 20 25 30 35 40 t(sec)-

(m/s) -1 -2 -3 -4

40 35dist 30 (m) 25 20 15 10 5

5 10 15 20 25 30 35 40 time (sec)

2. Make a displacement vs time graph for a person walks 10m in 5s, pauses for 10s to tie a shoe, runs 20m in 5s, pauses for 10s to tie other shoe, then runs back to start in 10s.

3. Velocity at each time interval:

0-5sec: _2________ 5-10sec: _0________10-15sec: _0________15-20sec: _4________20-25sec: _0________25-30sec: _0________30-35sec: _-3________35-40sec: _-3________

4. Graph the velocities: 4 3 2 1

vel 0 5 10 15 20 25 30 35 40 t(sec)-

(m/s)-1 -2 -3 -4

40 35dist 30 (m) 25 20 15 10 5

5 10 15 20 25 30 35 40 time (sec)

Ch5 – Practice Graphs1. Graph of dist v time. Car starts at home, where is it at: t=2s? ____

t=6s? ____t=8s? ____

d. Speed at t=1? ____ t=3? ____ t=4.5? ____ t=5.5? ____ t=7? ____

2. Graph of velocity v time. Velocity at t=1s? ____ t=2.5s? ____ t=3.5? ____

t=4.5s? ____ t=6s? ____ t=7.5s? ____t=8? ____

What is magnitude of displacement at: t=2s? ____ t=3s? ____ t=4? ____ t=5s? ____ t=7s? ____ t=8s? ____

100 90 80 70dist 60 (m) 50 40 30 20 10

1 2 3 4 5 6 7 8 9 10 time (sec)

100 90 80 70vel 60 (m/s) 50 40 30 20 10 0 -10 -20

1 2 3 4 5 6 7 8 9 10 time (sec)

Ch5 – Practice Graphs1. Graph of dist v time. Car starts at home, where is it at: t=2s? _40_

t=6s? _50_t=8s? _0__

d. Speed at t=1? ____ t=3? ____ t=4.5? ____ t=5.5? ____ t=7? ____

2. Graph of velocity v time. Velocity at t=1s? ____ t=2.5s? ____ t=3.5? ____

t=4.5s? ____ t=6s? ____ t=7.5s? ____t=8? ____

What is magnitude of displacement at: t=2s? ____ t=3s? ____ t=4? ____ t=5s? ____ t=7s? ____ t=8s? ____

100 90 80 70dist 60 (m) 50 40 30 20 10

1 2 3 4 5 6 7 8 9 10 time (sec)

100 90 80 70vel 60 (m/s) 50 40 30 20 10 0 -10 -20

1 2 3 4 5 6 7 8 9 10 time (sec)

Ch5 – Practice Graphs1. Graph of dist v time. Car starts at home, where is it at: t=2s? _40_

t=6s? _50_t=8s? _0__

d. Speed at t=1? _20_ t=3? _0__ t=4.5? _10_ t=5.5? _0__ t=7? _-25_

2. Graph of velocity v time. Velocity at t=1s? ____ t=2.5s? ____ t=3.5? ____

t=4.5s? ____ t=6s? ____ t=7.5s? ____t=8? ____

What is magnitude of displacement at: t=2s? ____ t=3s? ____ t=4? ____ t=5s? ____ t=7s? ____ t=8s? ____

100 90 80 70dist 60 (m) 50 40 30 20 10

1 2 3 4 5 6 7 8 9 10 time (sec)

100 90 80 70vel 60 (m/s) 50 40 30 20 10 0 -10 -20

1 2 3 4 5 6 7 8 9 10 time (sec)

Ch5 – Practice Graphs1. Graph of dist v time. Car starts at home, where is it at: t=2s? _40_

t=6s? _50_t=8s? _0__

d. Speed at t=1? _20_ t=3? _0__ t=4.5? _10_ t=5.5? _0__ t=7? _-25_

2. Graph of velocity v time. Velocity at t=1s? _10_ t=2.5s? _20_ t=3.5? _10_

t=4.5s? _0__ t=6s? _-10_ t=7.5s? _-20_t=8? _-20_

What is magnitude of displacement at: t=2s? ____ t=3s? ____ t=4? ____ t=5s? ____ t=7s? ____ t=8s? ____

100 90 80 70dist 60 (m) 50 40 30 20 10

1 2 3 4 5 6 7 8 9 10 time (sec)

100 90 80 70vel 60 (m/s) 50 40 30 20 10 0 -10 -20

1 2 3 4 5 6 7 8 9 10 time (sec)

Ch5 – Practice Graphs1. Graph of dist v time. Car starts at home, where is it at: t=2s? _40_

t=6s? _50_t=8s? _0__

d. Speed at t=1? _20_ t=3? _0__ t=4.5? _10_ t=5.5? _0__ t=7? _-25_

2. Graph of velocity v time. Velocity at t=1s? _10_ t=2.5s? _20_ t=3.5? _10_

t=4.5s? _0__ t=6s? _-10_ t=7.5s? _-20_t=8? _-20_

What is magnitude of displacement at: t=2s? _20_ t=3s? _40_ t=4? _50_ t=5s? _50_ t=7s? _30_ t=8s? _10_

100 90 80 70dist 60 (m) 50 40 30 20 10

1 2 3 4 5 6 7 8 9 10 time (sec)

100 90 80 70vel 60 (m/s) 50 40 30 20 10 0 -10 -20

1 2 3 4 5 6 7 8 9 10 time (sec)

3. Make 2 graphs: A person walks West (+) at 1 m/s for 2s. She pauses for 1s. She then continues West at 2 m/s for 2s. She pauses for 1s. She turns and walks East (–) at 1 m/s for 8s. Graph d vs t , and vel vs t.

6 5 4 3 2 1

d 0 1 2 3 4 5 6 7 8 9 10 11 12 t(sec)-

(m) -1 -2 -3 -4

6 5 4 3 2 1

vel 1 2 3 4 5 6 7 8 9 10 11 12 t(sec)-

(m/s)-1 -2 -3 -4

3. Make 2 graphs: A person walks West (+) at 1 m/s for 2s. She pauses for 1s. She then continues West at 2 m/s for 2s. She pauses for 1s. She turns and walks East (–) at 1 m/s for 8s. Graph d vs t , and vel vs t.

6 5 4 3 2 1

d 0 1 2 3 4 5 6 7 8 9 10 11 12 t(sec)-

(m) -1 -2 -3 -4

6 5 4 3 2 1

vel 1 2 3 4 5 6 7 8 9 10 11 12 t(sec)-

(m/s)-1 -2 -3 -4

3. Make 2 graphs: A person walks West (+) at 1 m/s for 2s. She pauses for 1s. She then continues West at 2 m/s for 2s. She pauses for 1s. She turns and walks East (–) at 1 m/s for 8s. Graph d vs t , and vel vs t.

6 5 4 3 2 1

d 0 1 2 3 4 5 6 7 8 9 10 11 12 t(sec)-

(m) -1 -2 -3 -4

6 5 4 3 2 1

vel 1 2 3 4 5 6 7 8 9 10 11 12 t(sec)-

(m/s)-1 -2 -3 -4

What is her final displacement?

3. Make 2 graphs: A person walks West (+) at 1 m/s for 2s. She pauses for 1s. She then continues West at 2 m/s for 2s. She pauses for 1s. She turns and walks East (–) at 1 m/s for 8s. Graph d vs t , and vel vs t.

6 5 4 3 2 1

d 0 1 2 3 4 5 6 7 8 9 10 11 12 t(sec)-

(m) -1 -2 -3 -4

6 5 4 3 2 1

vel 1 2 3 4 5 6 7 8 9 10 11 12 t(sec)-

(m/s)-1 -2 -3 -4

What is her final displacement?

- 2 m

Ch5 Rev HW 33 – 37 33. A supersonic jet flying at 145m/s is accelerated uniformly at a rate of 23 m/s

2 for 20.0s. a. What is its final velocity? b. The speed of sound in air is 331 m/s. How many times the speed of sound is the plane traveling?

34. Determine the displacement of a plane that accelerates 66m/s to 88m/s in 12s.

Ch5 Rev HW 33 – 37 33. A supersonic jet flying at 145m/s is accelerated uniformly at a rate of 23 m/s

2 for 20.0s. a. What is its final velocity? b. The speed of sound in air is 331 m/s. How many times the speed of sound is the plane traveling?

vf = ? vf = vi + a ∙ tvi = 145m/sa = 23 m/s

2 vf = 145m/s + 23 m/s2 ∙ 20s

t = 20s vf = 607m/s Roughly 2X speed of sound

34. Determine the displacement of a plane that accelerates 66m/s to 88m/s in 12s.

Ch5 Rev HW 33 – 37 33. A supersonic jet flying at 145m/s is accelerated uniformly at a rate of 23 m/s

2 for 20.0s. a. What is its final velocity? b. The speed of sound in air is 331 m/s. How many times the speed of sound is the plane traveling?

vf = ? vf = vi + a ∙ tvi = 145m/sa = 23 m/s

2 vf = 145m/s + 23 m/s2 ∙ 20s

t = 20s vf = 607m/s Roughly 2X the speed of sound

34. Determine the displacement of a plane that accelerates 66m/s to 88m/s in 12s.d = ?vi = 66m/s d = ½(vi +vf ).tvf = 88m/st = 12s d = ½(66m/s + 88m/s).12s

d = 924m

35. Car moves at 12 m/s and coasts up a hill w/ uniform acceleration of -1.6 m/s2

a. How far has it traveled after 6 seconds? b. How far after 9 seconds?

36. An engineer must design a runway to accommodate airplane that must reach a velocity of 61 m/s before taking off. The planes accelerate at 2.5 m/s

2 a. How long will it take the plane to reach take off speed?b. What must be the minimum length of the runway?

35. Car moves at 12 m/s and coasts up a hill w/ uniform acceleration of -1.6 m/s2

a. How far has it traveled after 6 seconds? b. How far after 9 seconds? df = vit + ½a.t2

df = ? = (12m/s)(6s) + ½(-1.6 m/s2).(6s)2

vi = +12m/s df = 43m a = - 1.6 m/s

2 = (12m/s)(9s) + ½(-1.6 m/s2).(9s)2

t = 6s then 9s df = 36. An engineer must design a runway to accommodate airplane that must reach a velocity of 61 m/s before taking off. The planes accelerate at 2.5 m/s

2 a. How long will it take the plane to reach take off speed?b. What must be the minimum length of the runway?

35. Car moves at 12 m/s and coasts up a hill w/ uniform acceleration of -1.6 m/s2

a. How far has it traveled after 6 seconds? b. How far after 9 seconds? df = vit + ½a.t2

df = ? vi = +12m/s = (12m/s)(6s) + ½(-1.6 m/s

2).(6s)2 = (12m/s)(9s) + ½(-1.6 m/s2).(9s)2

a = - 1.6 m/s2 df = 43m df =

t = 6s then 9s36. An engineer must design a runway to accommodate airplane that must reach a velocity of 61 m/s before taking off. The planes accelerate at 2.5 m/s

2 a. How long will it take the plane to reach take off speed?b. What must be the minimum length of the runway?vi = 0m/s vf = vi + a ∙ t d = ½(vi +vf ).tvf = 61m/sa = +2.5m/s

2 61m/s = 0 + (+2.5m/s2)∙(t) = ½(0 +61).(24)

t = ?d = ? t = 24s d = 774m

37. Bag is dropped from a hovering helicopter. When the bag has fallen for 2 sec,a. What’s its velocity?b. How far has it fallen?