Force and Motion

Motion 5

5

1

Motion (Graded E.g.)

5.1 Distance and Displacement

5.2 Speed and velocity

5.3 Acceleration

5.4 Motion graphs

5.5 Equations of uniformly accelerated motion

5.6 Vertical motion under gravity

Force and Motion

Motion 5

1 O to A 200 m east (+) OA = +200 m

2

D Displacement along a straight line (Case 1)

200 m150 m

east (+)

O A

Force and Motion

Motion 5

3

1 O to A 200 m east (+) OA = +200 m

2 A to B 350 m west (−) AB = −350 m

200 m150 m

east (+)

AB

Force and Motion

Motion 5

4

1 O to A 200 m east (+) OA = +200 m

2 A to B 350 m west (−) AB = −350 m

overall O to A to B 550 m west (−) OB = −150

m

200 m150 m

east (+)

AB O

Force and Motion

Motion 5

1 O to A 200 m west (-) OA = -200 m

5

D Displacement along a straight line (Case 2)

200 m150 m

west (+)

O A

Force and Motion

Motion 5

6

1 O to A 200 m west (-) OA = -200 m

2 A to B 350 m east (+) AB = +350 m

200 m150 m

ABwest (+)

Force and Motion

Motion 5

7

1 O to A 200 m west (-) OA = -200 m

2 A to B 350 m east (+) AB = +350 m

overall O to A to B 550 m east (+) OB = +150

m

200 m150 m

AB Owest (+)

Force and Motion

Motion 5

8

• Trip: O → A → C

• Overall displacement

OC = OA + AC

E Displacement in a plane

O A200 m

C

350 m

N

Force and Motion

Motion 5

9

• Tip-to-tail method

1.

2.

50 m

50 m

O A200 m

E Displacement in a plane

Force and Motion

Motion 5

10

4.

200 m

350 m

3.

50 m

50 m

O A

C

O A 200 m

C

350 m

θ

Force and Motion

Motion 5

11

• Pythagoras’ theorem

∴ OC is 403 m (N29.7°E)

O A 200 m

C

350 m

θ

Force and Motion

Motion 5

• Speed at a certain instant

11 m s−1 at 0 s

14 m s−1 at 3 s

16 m s−1 at 5 s

N

12

Instantaneous speed

5.2 Speed and velocityA Speed

Force and Motion

Motion 5

• Total distance travelled / total time taken

N

13

Average speed

total distance travelled = 65 m total time taken = 5

s∴ average speed = 65/5 = 13 m s−1

65 m

Force and Motion

Motion 5

14

1 km h−1 (6 km, 6 h)

3 km h−1 (6 km, 2 h)

average speed = 2 km h−1 ?

A B

• Question:

total distance travelled = 6 + 6 = 12 km

total time taken = 6 + 2 = 8 h

∴ average speed = 12/8 ≠≠ 2 km h−1

Force and Motion

Motion 5

15

• Velocity at a certain instant

• magnitude = the instantaneous speed

Instantaneous velocity

11 m s−1 (N45°E) at 0 s

14 m s−1 (S45°E) at 3 s

16 m s−1 (S45°E) at 5 s

N

Are they the same

velocities?No. They are different in magnitude and

direction

Force and Motion

Motion 5

• Overall displacement / total time taken

N

16

Average velocity

overall displacement = 30 m (east)total time taken = 5

s∴ average velocity = 30/5 = 6 m s−1 (east)

30 m (east)

same direction

It depends on the initial and final positions

ONLY.

Force and Motion

Motion 5

17

• average velocity:

magnitude ≠≠ average speed

• e.g. going round

time for a cycle = T

average velocity = 0

average speed =

but instantaneous velocity (magnitude)

= instantaneous speed

Force and Motion

Motion 5

• equal change in position for every second

18

Uniform motion

time

change in position

0 s 1 s 2 s 3 s

+0 m

10 m 20 m 30 m

+10 m

+10 m

+10 m

Force and Motion

Motion 5

19

5.3 Acceleration

• How fast velocity changes

• Vector

• change in velocity

1. change in magnitude2. change in

direction

(unit: m s−2)

speeding up

slowing down

Force and Motion

Motion 5

• equal change in velocity for every second

• constant acceleration

20

Uniformly accelerated motion

0 s

1 s

2 s

+

0 2 m s−1 4 m s−1

+2 m s−1

+2 m s−1

a = accelerationv = final velocityu = initial velocityt = time taken

3 s

6 m s−1

+2 m s−1

time

Force and Motion

Motion 5

• distance travelled in every second is longer than

the previous second

21

time

distance travelled

0 s

1 s

2 s

0 m

1 m 4 m

1 m

5 m

3 s

9 m

3 m

Force and Motion

Motion 5

22

a = 4 m s−2

1 s

2 s

1 s

a = 2 m s−2

• Speeding up with greater acceleration

⇒ reaching the same speed in a shorter time.

+

2 m s−1 4 m s−1

4 m s−1

B Magnitude of acceleration

Force and Motion

Motion 5

23

a = 4 m s−2

total distance travelled

1 s

2 s

1 m 4 m

1 s

2 s

2 m 8 m

a = 2 m s−2

• Speeding up with greater acceleration

⇒ gains a larger speed every second

⇒ travels a longer distance every second

Example 6

+

Force and Motion

Motion 5

24

• Moving against the wind

C Direction of acceleration

stage 1

stage 2

stage 3

t = 0

10 m s−1

t = 5 s

5 m s−1

t = 15 s

v = −5 m s−1

t = 20 s

v = −10 m s−1

t = 10 s

v = 0

a = −1 m s−2

sign of v sign of a motion

stage 1 + − slowing down

+

Force and Motion

Motion 5

25

• Moving against the wind

stage 2

stage 3

t = 0

10 m s−1

t = 5 s

5 m s−1

t = 15 s

v = −5 m s−1

t = 10 s

v = 0

a = −1 m s−2

+

sign of v sign of a motion

stage 2 v = 0 − momentarily at rest

t = 20 s

v = −10 m s−1

stage 1

Force and Motion

Motion 5

26

• Moving against the wind

sign of v sign of a motion

stage 3 − − speeding up

stage 2

stage 3

t = 0

10 m s−1

t = 5 s

5 m s−1

t = 15 s

v = −5 m s−1

t = 10 s

v = 0

a = −1 m s−2

+

t = 20 s

v = −10 m s−1

stage 1

Force and Motion

Motion 5

27

• To sum up

stage 2

stage 3

+

+

−

v = 0

a = −1 m s−2

+

−

stage 1

sign of v

sign of a

motion

stage 1 + − slowing down

stage 2 v = 0 − momentarily at rest

stage 3 − − speeding up

Force and Motion

Motion 5

28

5.4 Motion graphs

• Displacement–time (s–t) graph

• Velocity–time (v–t) graph

• Acceleration–time (a–t) graph

Force and Motion

Motion 5

29

A From motion to motion graphs

Uniform motion

0 m

+10 m

+20 m

+30 m

+40 m

s / m

t / s

10

20

30

40

01 2 3 4

time t

+

dis

pla

cem

en

t s

0 s

1 s

2 s

3 s

4 s

Force and Motion

Motion 5

30

a / m s−2

t / s01 2 3 4

v / m s−1

t / s

10

01 2 3 4

constant velocity

zero acceleration

• Uniform motion ⇒ constant v

Force and Motion

Motion 5

+2.5 m

31

0 m

+10 m

+22.5 m

+40 m

s / m

t / s

10

20

30

40

01 2 3 4+

dis

pla

cem

en

t s

Uniformly accelerated motion

time t

0 s

1 s

2 s

3 s

4 s

Force and Motion

Motion 5

32

a / m s−2

t / s01 2 3 4

v / m s−1

t / s01 2 3 4

5

10

15

20

5

velocity changes at fixed rate

constant acceleration

• Uniform accelerated motion ⇒ constant a

Force and Motion

Motion 5

33

B Relations between motion graphs

Relations between s–t and v–t graphs

v / m s−1

t / s01 2 3 4

s / m

t / s01 2 3 4

10

20

30

40 10

Force and Motion

Motion 5

34

v / m s−1

t / s01 2 3 4

10

s / m

t / s01 2 3 4

10

20

30

40

area

Force and Motion

Motion 5

35

Relations between v–t and a–t graphs

a / m s−2

t / s01 2 3 4

v / m s−1

t / s01 2 3 4

5

10

15

20 5

Force and Motion

Motion 5

36

a / m s−2

t / s01 2 3 4

5

v / m s−1

t / s01 2 3 4

5

10

15

20

area

Force and Motion

Motion 5

37

Negative area under a graph

v / m s−1

t / s

−5

0

5

10

−10

1 2 3 4

+

0 s

1 s

2 s

3 s

4 s

a = −5 m s−2

+10 m s−1

+5 m s−1

−5 m s−1

−10 m s−1

v = 0

Force and Motion

Motion 5

38

v / m s−1

t / s

−5

0

5

10

−10

1 2 3 4

0 s

1 s

2 s

3 s

4 s

0 m+

Δs = +7.5 m

Δs = +2.5 m

Δs = −2.5 m

Δs = −7.5 m

+ve area ⇒ Δs > 0

−ve area ⇒ Δs > 0

a = −5 m s−2

Force and Motion

Motion 5

39

General cases

Force and Motion

Motion 5

40

To analyse a complex motion graph,

•divide it into sections

•identify the motion in each section

C Analysing motion graphs

Force and Motion

Motion 5

41

5.5 Equations of uniformlyaccelerated motion

u va

t0 s s

initial state

final state

uniformly accelerated motion ⇒ constant a

Force and Motion

Motion 5

42

• Summary

• These equations are called the equations of

uniformly accelerated motion.

(1)

(2)

(3)

(4)

a is absent

s is absent

v is absent

t is absent

Force and Motion

Motion 5

Chapter 5

Example 5.7

Force and Motion

Motion 5Example 7 A ball rolling up a slope

(M1)

44

2 m

s−1

Force and Motion

Motion 5

45

(a) Find the acceleration of the ball.

2 m

s−1

Force and Motion

Motion 5

(a) Take the direction up

the plane as positive.

Given u = +2 m s−1

momentarily at rest velocity ⇒ v = 0

∴ v = 0 at t = 0.8 s

The acceleration is 2.5 m s−2 (down the plane)

Solution

46

v = 0t = 0.8 s

u = 2 m s−1

t = 0

(+)

Force and Motion

Motion 5

47

(b) Find the velocity of the ball 1.2 s after the

projection.

a = −2.5 m s−2

(+)

u = 2 m s−1

t = 0

Force and Motion

Motion 5

(b) By v = u + at,

The velocity

= 2 + (−2.5)(1.2)

= −1 m s−1

∴ The velocity is 1 m s −1 (down the plane)

Solution

48

v = –1 m s –1

t = 1.2 s

a = −2.5 m s−2(+)

u = 2 m s−1

t = 0

Force and Motion

Motion 5Example 7 A ball rolling up a slope

(M2)

49

2 m

s−1

Force and Motion

Motion 5

50

(a) Find the acceleration of the ball.

2 m

s−1

Force and Motion

Motion 5

(a) Take the direction up

the plane as positive.

Given u = -2 m s−1

momentarily at rest velocity ⇒ v = 0

∴ v = 0 at t = 0.8 s

The acceleration is 2.5 m s−2 (down the plane)

Solution

51

v = 0t = 0.8 s

u = -2 m s−1

t = 0

(-)

)/(5.28.0

)2(0sm

t

uva

Force and Motion

Motion 5

52

(b) Find the velocity of the ball 1.2 s after the

projection.

a = +2.5 m s−2

(-)

u = -2 m s−1

t = 0

Force and Motion

Motion 5

(b) By v = u + at,

The velocity

= -2 + (+2.5)(1.2)

= +1 m s−1

∴ The velocity is 1 m s −1 (down the plane)

Solution

53

v = +1 m s –1

t = 1.2 s

a = +2.5 m s−2(-)

u = -2 m s−1

t = 0

Force and Motion

Motion 5

The End

54

Force and Motion

Motion 5

55

5.6 Vertical motion under gravity

• Falling in air

A Do heavier objects fall faster?

air resistance

Force and Motion

Motion 5

56

• Without air resistance

fall with the same acceleration

Force and Motion

Motion 5

57

• Near the Earth’s surface:

g = 9.81 m s−2

with air resistance

without air resistance

coilfeather

same acceleration

acceleration due to gravity g

Force and Motion

Motion 5

displacement

where

u = 0 & a = +9.81 m

s−2

58

B Free fall

Free fall from rest

+

time

0 s1 s

2 s

3 s

t / s s / m

0 0

1 +4.95

2 +19.62

3 +44.145

Force and Motion

Motion 5

• Velocity

velocity

v = u + at

where

u = 0 & a = +9.81 m

s−2

59

+

time

0 s1 s

2 s

3 s

t / s v / m s−1

0 0

1 +9.81

2 +19.62

3 +29.43

Force and Motion

Motion 5

Summary

60

time t / s

displacement s / m

velocity v / m s−1

acceleration a / m s−2

0 0 0

+9.81(constant)

1 +4.95 +9.81

2 +19.62 +19.62

3 +44.145 +29.43s / m

t / s

0 1 2 3

20

40

v / m s−1

slope= 9.81 m s−2

0

t / s

10

20

1 2 3

a / m s−2

9.81constant

0

t / s

1 2 3

Force and Motion

Motion 5

61

Free fall with initial velocity

u = +29.43 m s−1

a = −9.81 m s−2+

time

4 s

5 s

6 s

3 s

2 s

1 s

0 s

displacement

Force and Motion

Motion 5

62

+

time

4 s

5 s

6 s

3 s

2 s

1 s

0 s

u = +29.43 m s−1

a = −9.81 m s−2

velocity

v = u + at

• Velocity

Force and Motion

Motion 5

63

+

time

4 s

5 s

6 s

3 s

2 s

1 s

0 s

u = +29.43 m s−1

a = −9.81 m s−2

velocity

v = u + at

• Velocity

+ −

v = 0 (momentarily at rest)

Force and Motion

Motion 5

64

+

time

4 s

5 s

6 s

3 s

2 s

1 s

0 s

u = +29.43 m s−1

a = −9.81 m s−2

velocity

v = u + at

• Same level ⇒ same magnitude of velocity

Force and Motion

Motion 5

Summary

•At highest position, a = −9.81 m s−2 but v = 0

•Same level ⇒ same magnitude of velocity

65

time t / sdisplacement s /

mvelocity v / m

s−1

acceleration a / m s−2

0 0 +29.43

−9.81(constant)

1 +24.525 +19.62

2 +39.24 +9.81

3 +44.145 0

4 +39.24 −9.81

5 +24.525 −19.62

6 0 −29.43

Force and Motion

Motion 5

• Motion graphs

66

s / m

t / s0 2 4 6

20

40

1 3 5

turning point

return to the initial position

1. moving up, losing speed

3. falling down, gaining speed

4

v / m s−1

slope= − 9.81 m s−2−3

0

t / s

−15

0

+15

+30

2 61 3 5

turning point

2. momentarily

at rest

1. 3.

2.

Force and Motion

Motion 5

• Motion graphs

67

1. 3.

2.

s / m

t / s0 2 4 6

20

40

1 3 5

turning point

return to the initial position

a / m s−2

−9.81

constant

0t / s

2 4 61 3 5

turning point

1. moving up, losing speed

3. falling down, gaining speed

2. momentarily

at rest

Force and Motion

Motion 5

Summary

68

4

v / m s−1

slope= − 9.81 m s−2

−30

t / s

−15

0

15

30

2 61 3 5

a / m s−2

−9.81

constant

0t / s

2 4 61 3 5

s / m

t / s

0 2 4 6

20

40

1 3 5

stage

time t sign of v motion

1 0 < t < 3 s + rising, losing speed

2 t = 3 s v = 0 momentarily at rest

33 s < t < 6

s−

falling, gaining speed