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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?
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Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

Dec 25, 2015

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Page 1: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: 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?at 3 sec?at 4½ sec?

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

Page 2: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: 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

Page 3: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 4: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 5: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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

Page 6: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

Ex5) How fast is the plane going at 1.5 sec

dist

(m)

time (sec)

Page 7: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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

Page 8: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 9: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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

- of B - of C

Page 10: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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/s3. Object D starts at 0m and moves with

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

Page 11: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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?

Page 12: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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?df = di + vt

0m = 10m + (–2m/s)(t) t = 5sec

Page 13: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 14: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

Lab5.1 – Ave vs Inst Acceleration

- due tomorrow

- Ch5 HW#2 due at beginning of period

Page 15: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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:

Page 16: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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:

Page 17: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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:

Page 18: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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:

Page 19: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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

Page 20: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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?

Page 21: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

Ex9) Describe each motion:

A:

B:

Find the acceleration of each :

A:

B:

Find the distance traveled for each :

A:

B:

Page 22: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 23: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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 = ?

Page 24: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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 = ?

2/735

/36/15sm

s

smsm

t

vva if

Page 25: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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 = ?

2/735

/36/15sm

s

smsm

t

vva if

2/35.2

)/3(/5.4sm

s

smsm

t

vva if

Page 26: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 27: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 28: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 29: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 30: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

Ch5.3 – Motion Equations

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

Page 31: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

Ch5.3 – Motion Equations

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

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

Page 32: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 33: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 34: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

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?

Page 35: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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?

Page 36: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

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?

Page 37: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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 starts at rest and speeds up at 3.5 m/s2 . How far will it have gone

when it is going 25 m/s?

Page 38: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

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

Page 39: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

Lab5.2 – Motion on an Incline Plane

- due tomorrow

- Ch5 HW#4 due at beginning of period

Page 40: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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?

Page 41: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 42: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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 =

Page 43: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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 = ?

Page 44: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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 = ?

Page 45: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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 = ?

Page 46: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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?

Page 47: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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?

Page 48: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 49: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

Lab5.3 – Measuring ‘g’

- due tomorrow

- Ch5 HW#5 due at beginning of period

- Ch5 HW#6 due tomorrow

Page 50: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 51: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 52: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 53: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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 = ?

Page 54: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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 = ?

d = ½(65+162)(10)

d = 1135 m

Page 55: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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?

Page 56: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 57: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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)

Page 58: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 59: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 60: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 61: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 62: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 63: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 64: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 65: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 66: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 67: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 68: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 69: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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)

Page 70: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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

Page 71: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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

Page 72: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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

What is her final displacement?

Page 73: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 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

What is her final displacement?

- 2 m

Page 74: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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.

Page 75: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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.

Page 76: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 77: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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?

Page 78: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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?

Page 79: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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

Page 80: Ch5 – Mathematical Models 30 25 dist (m) 20 15 10 5 1 2 3 4 5 6 time (sec) Time (sec) Position (m) 010 112 214 316 420 526 Where is the runner: at 2 sec?

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?