Average speed, average velocity, accelerationp. 177

Average speed is the total distance moved per total time. (scalar)

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

Please do not use the formulas in the book.

Average velocity is the rate of change in position. (vector)

v = Δ𝑥

Δ𝑡

Top green block on pg. 178: Not average speed, but average velocity.

Wayde runs 300 m in 30 seconds during practice as indicated in the diagram. What are Wayde’s average speed and velocity respectively?

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 300

30

= 10 m.s-1

v = Δ𝑥

Δ𝑡

= 100

30

= 3,33 m.s-1; west

km·h-1 and m·s-1

𝑘𝑚

ℎ

× 1000

÷ 3600

𝑚

𝑠

÷ 3,6

× 3,6

Homework

pg. 185nos. 1, 4.1 – 4.3, 5.1 – 5.5, 7, 8, 10

1. Luke rides on his bicycle and covers 900 m in a straight line, turns around and covers another 2 100 m. He completes the full round in 400 s.Answer each of the following questions, measured from the origin.

1.1 What distance did Luke travel?

-12

00

-90

0

-60

0

-30

0

0 30

0

60

0

90

0

Distance = 900 + 2 100= 3 000 m

1. Luke rides on his bicycle and covers 900 m in a straight line, turns around and covers another 2 100 m. He completes the full round in 400 s.Answer each of the following questions, measured from the origin.

1.2 Calculate Luke’s average speed.

-12

00

-90

0

-60

0

-30

0

0 30

0

60

0

90

0

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 3 000

400

= 7,5 m.s-1

1. Luke rides on his bicycle and covers 900 m in a straight line, turns around and covers another 2 100 m. He completes the full round in 400 s.Answer each of the following questions, measured from the origin.

1.3 What was Luke’s total displacement?

-12

00

-90

0

-60

0

-30

0

0 30

0

60

0

90

0

Δx = -1200 m= 1 200 m; west

Choose east as positive

1. Luke rides on his bicycle and covers 900 m in a straight line, turns around and covers another 2 100 m. He completes the full round in 400 s.Answer each of the following questions, measured from the origin.

1.4 Calculate Luke’s average velocity.

-12

00

-90

0

-60

0

-30

0

0 30

0

60

0

90

0

= −1 200

400

= -3 m.s-1

v = Δ𝑥

Δ𝑡

= 3 m.s-1; west

Choose east as positive

1. Luke rides on his bicycle and covers 900 m in a straight line, turns around and covers another 2 100 m. He completes the full round in 400 s.Answer each of the following questions, measured from the origin.

1.5 Now explain the difference between average speed and average velocity,referring to 1.2 and 1.4.

-12

00

-90

0

-60

0

-30

0

0 30

0

60

0

90

0

Average speed is the rate of change in distance. (scalar)Average velocity is the rate of change in position. (vector)

4. An athlete runs at a constant speed from point A, clockwise around a circular track with circumference 220 m and radius 35 m. He completes three rounds in 240 s.

4.1 Determine the athlete’s average speed during the first 40 s.

A

B

C

D

N

O

S

W

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 110

40

= 2,75 m.s-1

240 s for 3 rounds40 s for 0,5 rounds

4. An athlete runs at a constant speed frompoint A, clockwise around a circular trackwith circumference 220 m and radius 35 m. He completes three rounds in 240 s.

4.2 What will the athlete’s speed be at anyplace or time around the track during the240 s he runs around the track? Explain.

A

B

C

D

N

O

S

W

2,75 m.s-1

The athlete runs at a constant speed.

4. An athlete runs at a constant speed from point A, clockwise around a circular track with circumference 220 m and radius 35 m. He completes three rounds in 240 s.

4.3 Calculate the athlete’s average velocityduring the first 40 s.

A

B

C

D

N

O

S

W

= 70

40

= 1,75 m.s-1; south

v = Δ𝑥

Δ𝑡

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles). 5.1 What distance had the horse run when it reached

the following points?5.1.1 Point C?

200 m

A

B

C D

N

200m

E

F

127,4m

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles). 5.1 What distance had the horse run when it reached

the following points?5.1.2 Point F?

600 m

A

B

C D

N

200m

E

F

127,4m

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles). 5.1 What distance had the horse run when it reached

the following points?5.1.3 Back at point A?

800 m

A

B

C D

N

200m

E

F

127,4m

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles). 5.2 What was the displacement of the horse when it

reached the following points?5.2.1 Point C?

127,4 m; south

A

B

C D

N

200m

E

F

127,4m

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles). 5.2 What was the displacement of the horse when it

reached the following points?5.2.2 Point F?

200 m; east

A

B

C D

N

200m

E

F

127,4m

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles). 5.2 What was the displacement of the horse when it

reached the following points?5.2.3 Back at A?

0 m

A

B

C D

N

200m

E

F

127,4m

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles). 5.3 How long (in s) does it take the horse to

complete one round?

A

B

C D

N

200m

E

F

127,4m

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

20 = 800

𝑡𝑖𝑚𝑒

Time = 40 s

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles). 5.4 Calculate average velocity of the horse, from

A to F.A

B

C D

N

200m

E

F

127,4m

= 200

30

= 6,67 m.s-1; east

v = Δ𝑥

Δ𝑡

Choose east as positive

5. The diagram below represents a horse-racing track. A jockey on his horse, starting from point A, completes the race at a constant speed of 20 m.s-1 and finishes at A again. (ABC and DEF are semi-circles).5.5 What is the average velocity of the horse for one

complete round around the track?A

B

C D

N

200m

E

F

127,4m

0 m.s-1

7. Points A, B and C are on the same straight railway line, with AB = 60 m andBC = 80 m. A locomotive travels from A to B at an average speed of 10 m.s-1, then from B to C in a time of 4 s and then returns to B. The average speed for the complete trip is 5 m.s-1.

7.1 Calculate the average speed of the locomotive in the second part of its trip (BC).

A B C

60m 80m

10 m.s-1 4s

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 80

4

= 20 m.s-1

7. Points A, B and C are on the same straight railway line, with AB = 60 m andBC = 80 m. A locomotive travels from A to B at an average speed of 10 m.s-1, then from B to C in a time of 4 s and then returns to B. The average speed for the complete trip is 5 m.s-1.

7.2 Calculate the average speed of the locomotive for the motion from A to C.

A B C

60m 80m

10 m.s-1 4sAB: Speed =

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

10 = 60

𝑡𝑖𝑚𝑒

Time = 6 sAC: Speed =

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 140

10

= 14 m.s-1

7. Points A, B and C are on the same straight railway line, with AB = 60 m andBC = 80 m. A locomotive travels from A to B at an average speed of 10 m.s-1, then from B to C in a time of 4 s and then returns to B. The average speed for the complete trip is 5 m.s-1.

7.3 Calculate the time the locomotive will take to travel the third section (CB) of itsmotion.

A B C

60m 80m

10 m.s-1 4sABCB: Speed =

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

5 = 220

𝑡𝑖𝑚𝑒

Time = 44 s

Thus time for CB = 44 – 10 = 34 s

7. Points A, B and C are on the same straight railway line, with AB = 60 m andBC = 80 m. A locomotive travels from A to B at an average speed of 10 m.s-1, then from B to C in a time of 4 s and then returns to B. The average speed for the complete trip is 5 m.s-1.

7.4 What was the displacement of the locomotive for the three sections of its motion?

A B C

60m 80m

10 m.s-1 4s60 m; east

Choose original direction of motion as positive

7. Points A, B and C are on the same straight railway line, with AB = 60 m andBC = 80 m. A locomotive travels from A to B at an average speed of 10 m.s-1, then from B to C in a time of 4 s and then returns to B. The average speed for the complete trip is 5 m.s-1.

7.5 What was the average velocity of the locomotive for the three sections of its motion?

A B C

60m 80m

10 m.s-1 4s

= 60

44

= 1,36 m.s-1; east

v = Δ𝑥

Δ𝑡

Choose original direction of motion as positive

8. During a trip a vehicle moves downhill along 1 km from A to B at an average speed of 90 km.h-1 and uphill along 1 km from B to C at an average speed of60 km.h-1. The shortest distance betweenA and C is 1,875 km, as shown on the right.

8.1 Calculate the average speed of thevehicle from A to C.

A

B

C1,875km

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

25 = 1 000

𝑡𝑖𝑚𝑒

Time = 40 s

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

16,67 = 1 000

𝑡𝑖𝑚𝑒

Time = 60 s

AC:

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 2 000

100

= 20 m.s-1

AB:

Speed = 90

3,6= 25 m.s-1

BC:

Speed = 60

3,6= 16,67 m.s-1

8. During a trip a vehicle moves downhill along 1 km from A to B at an average speed of 90 km.h-1 and uphill along 1 km from B to C at an average speed of60 km.h-1. The shortest distance betweenA and C is 1,875 km, as shown on the right.

8.2 Calculate the average velocity of the vehicle from A to C.

A

B

C1,875km

= 1 875

100

= 18,75 m.s-1; right

v = Δ𝑥

Δ𝑡

Choose right as positive

10. A mouse leaves its hole and walks 4 m East in 20 s, then 3 m North in 12 s and eventually walks 5 m back to its hole in 28 s as shown in the diagram.

10.1 Calculate the average speed of the mouse for the complete route.

D1= 4m

t1 = 20s

D2= 3m

t2 = 12s

D3= 5m

t3 = 28s

Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 12

60

= 0,2 m.s-1

10. A mouse leaves its hole and walks 4 m Eastin 20 s, then 3 m North in 12 s and eventually walks 5 m back to its hole in 28 s as shown in the diagram.

10.2 Calculate the average speed of the mouse for each of the three sections of the route.

D1= 4m

t1 = 20s

D2= 3m

t2 = 12s

D3= 5m

t3 = 28s

Speed (D1) = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 4

20

= 0,2 m.s-1

Speed (D2) = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 3

12

= 0,25 m.s-1

Speed (D3) = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

= 5

28

= 0,18 m.s-1

10. A mouse leaves its hole and walks 4 m East in 20 s, then 3 m North in 12 s and eventually walks 5 m back to its hole in 28 s as shown in the diagram.

10.3 Calculate the average speed of the threeaverages you calculated in Question 10.2.

D1= 4m

t1 = 20s

D2= 3m

t2 = 12s

D3= 5m

t3 = 28s

Average speed = 0,20 + 0,25 + 0,18

3

= 0,21 m.s-1

10. A mouse leaves its hole and walks 4 m East in 20 s, then 3 m North in 12 s and eventually walks 5 m back to its hole in 28 s as shown in the diagram.

10.4 Compare the average speed for the complete route with the average of theaverages for the three separate sectionsof the route. Explain your answer.

D1= 4m

t1 = 20s

D2= 3m

t2 = 12s

D3= 5m

t3 = 28s

The values are not equal, because the mouse does not take the same time to walk each of the three sections.

10. A mouse leaves its hole and walks 4 m East in 20 s, then 3 m North in 12 s and eventually walks 5 m back to its hole in 28 s as shown in the diagram.

10.5 What is the average velocity of the mousefor the complete route? Explain.

D1= 4m

t1 = 20s

D2= 3m

t2 = 12s

D3= 5m

t3 = 28s

0 m.s-1

No change in position

Instantaneous speed and velocity

Instantaneous velocity is the rate of change in displacement, i.e. the displacement divided by a very short time interval or the velocity at a specific moment. (vector)

Instantaneous speed is the magnitude of the instantaneous velocity. (scalar)

Accelerationp. 195

Acceleration is the rate of change in velocity. (vector)

a = Δ𝑣

Δ𝑡

Positive acceleration:

An object moving in the positive direction experiences an increase in speed.

An object moving in the negative direction experiences a decrease in speed.

Negative acceleration:

An object moving in the positive direction experiences a decrease in speed.

An object moving in the negative direction experiences an increase in speed.

Deceleration: An object experiences a decrease in speed.

(Do not use this expression!)

Homework

pg. 196nos. 2, 3, 4, 5

2. The following statements each describe the motion of different objects. Read through each statement carefully and decide whether it involves acceleration or not.

2.1 A truck approaches a bend and moves through it at 75 km.h-1

Yes, there is change in direction

2. The following statements each describe the motion of different objects. Read through each statement carefully and decide whether it involves acceleration or not.

2.2 A motorcycle rides at a uniform velocity of 130 km.h-1 on a straight road.

No, there is no acceleration

2. The following statements each describe the motion of different objects. Read through each statement carefully and decide whether it involves acceleration or not.

2.3 A satellite moves at a constant velocity in a circular orbit around the earth.

Yes, there is acceleration since the direction changes constantly.

2. The following statements each describe the motion of different objects. Read through each statement carefully and decide whether it involves acceleration or not.

2.4 A motorist applies brakes as he approaches a stop street.

Yes, acceleration is negative

2. The following statements each describe the motion of different objects. Read through each statement carefully and decide whether it involves acceleration or not.

2.5 A rugby player kicks a ball to the goal-posts.

Yes, it accelerates just after the player hits the ball with his foot, and thereafter accelerates down towards the earth.

3. The equation to calculate the acceleration is:

a = Δ𝑣

Δ𝑡= 𝑣𝑓−𝑣𝑖

∆𝑡Complete the following table for this equation:

Symbol Meaning SI unit Scalar/Vector

a

Δv

Δt

vf

vi

Acceleration m.s-2 vector

Change in velocity m.s-1 vector

Change in time s scalar

Final velocity m.s-1 vector

Initial velocity m.s-1 vector

4.1 A truck leaves from rest and reaches a velocity of 12 m.s-1 in 3 seconds. Calculate theacceleration of the truck if it was travelling east initially.

a = Δ𝑣

Δ𝑡= 𝑣𝑓−𝑣𝑖

∆𝑡

Vi = Δt =

a = ?

Vf =0 12 m.s-13s

= 12−0

3

= 4 m.s-2; eastwards

4.2 An object moving at 12 m.s-1 North, reaches a velocity of 53 m.s-1 in 5 s. Calculate theacceleration of the object.

a = Δ𝑣

Δ𝑡= 𝑣𝑓−𝑣𝑖

∆𝑡

Vi = Δt =

a = ?

Vf =12 m.s-1 53 m.s-15s

= 53−12

5

= 8,2 m.s-2; north

5.1 A train travels at 28 m.s-1 on a straight track in an easterly direction, after which it brakes and reaches a velocity of 15 m.s-1 in 5 s. Calculate the acceleration of the train.

a = Δ𝑣

Δ𝑡= 𝑣𝑓−𝑣𝑖

∆𝑡

Vi = Δt =

a = ?

Vf =28 m.s-1 15 m.s-15s

= 15−28

5

= −2,6 m.s-2

= 2,6 m.s-2; west

5.2 A motorcycle travelling North along a straight road at 30 m.s-1, applies brakes and comesto a halt after 6 s. Calculate the acceleration of the motorcycle.

a = Δ𝑣

Δ𝑡= 𝑣𝑓−𝑣𝑖

∆𝑡

Vi = Δt =

a = ?

Vf =30 m.s-1 0 m.s-16s

= 0−30

6

= −5 m.s-2

= 5 m.s-2; south