Force and Motion Motion 5 5 1 Motion (Graded E.g.) 5.1Distance and DisplacementDistance and Displacement 5.2Speed and velocitySpeed and velocity 5.3AccelerationAcceleration.
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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
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