KINEMATICS
Distance
Displacement
t
Origin 10m
->
Length of the
path between
1.
10
points.
:
-isstart
disp
A
quantity
S1 Unit:mele (m)
can travelled a
4km.
distance
·
of
·
2
"
bm ->
>
>
&
·
E
·
Sm
<-
<
·
↑
-
·
F
-
-
qm-
-
distance
displ
OA
10m
+10m
OB
18m
10 8
OC
20m
2
OD
32m
OE
38m
0
=
Vector
3. S1 Unit:meter (m)
4. can displaced 3 km
due south East.
-
tells
3
A
>
<
2.
sign of displ
-
<
to
·
4.
E
·
two
↑B
Scala
D<
-
points.
3.
possible
distance between
two
2.
shortest
-
about
-
2M
+
2 0
=
-12m
-
47m
OF
-
=
-
12 + 6
6 9
+
=-
=
6m
3m
+
the
position of the ·
body from the origin.
speed (V)
Rate
1.
Velocity (v)
of change
1.
Adv =
=
SI
3.
Unitim/s
4.
Scale
5.
Manual is
at
a
ms
of change
2.v
pv 1
walking
speedof
m/s
mst
average
at0.5
mys
towards
the
hit
7
1.
rebound
due
v
-
10m/s
total time
-1
Now
103
6Om
=
North start.
door.
1
x
h
=30s
>
End
A
·
"
+
Om/s
+
1
....
1.
time
velocity:total displ
B
Manual is walking
Sm/s
hit
10m/s
total
=
=
3.31 Unit:
5.
speed:total distance
average
4. Vector
quantity
quantity
0.5m/s
....
Rate
of displacement
of distance.
2.v
displacement
S:
=
rebound
as
arg. speed
arg. speed
=
di
I
00
by =arg. velocity
-
n
=
p
=
+b2
482
k 72.1m
=
=
=
2.5m/s
t
arg.neL=
Acceluation
Example #2
of change ofvelocity
Rate
=
OR
change of velocityper unittime
1
a v
Ay
a
=
=
nothi
-
final velocity
v
a
7
roll
=
0
a =
-
18
2
t 25
a
=
=
-
:.
5m/s
deceleration:5 m/s
mention"
need to
-
v=
rest
v 0-
(no
-
"sign. I
SI Unit:m/s
·
· Vector
initial velocity
quantityExample 3
t: time interval
ball hits
A
u:
a
there
3m/s
=
Om/s
+3
is
of 3 mys, every see.
3
+
with
upwards
of
v
a v
am/s
=
back
Oml. The
of
u
F
-
a8
acceleration
-
=
·
time
the
Calculate
ball.
the
+
-
X
hit
velocity
V
change in
change
u -
↑
i
-
speed up,
slowdown
on
pacvelationestration
.
lOm/s
>
10m/s
·
·
hit
#4
Example
A ball is thrown
10mm. It
rebound
and
circle
falls
is
-
elation.
aV
=
-
u
a
=
-
F
mms
u
a
y
1
20
t 55
a =
D
-
1 10)
I
a
16m/s2
=
+10m/
t
+
-10m/s2
=
..... ..
=
!
↓
->
a
Find
the accele
↑
=
=
find acceleration.
-
E
aV
3s, its speed
After
20m/s.
now
u
xm?":
upwards
vertically
and
then
rises
back down.
=
Example #1
1
=
at
moving in
10ms-
in
direction
magnitude
X
90mys
a
V
1
=
⑭
-
=
change
8ms
v+
....
V
1 10)
0.2
in
a
andrebounds
0.25
collision was
acceleration
+3
>3m/s<6m/s
ground with
speed of 10m/s
change
a
the
v=
20m/
Graph
Graphe
Distance
1.
-
time
graph
LINE
(constant gradient)
Speed-time graph
2.
generally...
m 0
·
X
-
axis
m
-i
n
a.
a
mana
i
y-axis
·
y, y,
-
=
4
m2
=
mt
=
m=
Area
=
inc
=
dec.
Area
Area
of Lxb
=
m 0
=
Area
q
-
m
xc x,
M
m
-
-
P
our
M=
=
u
m
kchanging gradient)
=
Y-axis
gradient
CURVE
of
1bh
m=
=
gradient
a,"
q
P-h ->
q
(slope
=
>measureof li es
how
vertical
gradient inc.
horizontal
gradient dec.
of trapezium 1(p a)h
+
1
dec.
Distance time
-
dim
graphs
dim
N
N
4.
1=
x-axis: time
1
A
V
(deceleration)
=
m
Y-axis: distance
gradient:speed
h
0
speed decrease
dec.
m
y Ac spedthe
=
=
=
0
Area:meaning less
>
&
*
I
t(s A
Ibh
>
&
:
=
dim
N
xd)
1(t
speed increase
v
·N
=
N
m
10
0
At rest.
m v 0
=
>
t(s
dim
N
20
At rest ahead
m v
=
0
or
=
behind
the
starting point.
0
>
&
on
Amcorpus
0
constantiain
>
&
t(s
speed
t(s
>
&
=
&
3.
(acceleration)
0=
dim
0
t(s
t(s
Example
item
dIm
350,,,,,,,,,,,,,.
250 ((()/
I
E
/I) ·
/
↑ ↑
*
=
150m
100
X
X
1
I((( 1)
=
=
②
=
I
-
E>4sI
O
p
10
-
④
-
>t/s
25
14
X
X
as
as
the
object.
I
b)
mspeed and
during journey
the
calculate
-
mispeed
->
a)
Object speeds up
Osto10s:
C)
laccelerate
10s 14s:Object
moves
to
constant
at
as
object slow down
14 to 255:
speed of the whole
journey
Average
-
speed.
the
ofjourney
Explain The motion ofthe object.
the
speed of
find the topaverage
speed
find
b)
calculate the speed during
last 10 seconds
-
grad:In
xc
Idecelerates)
-
x,
200
120
-
=
40-30
b)
speed
speed
max
m
gradient
->
->
max
gradient
b)
y, y,
=
8m/s
=
-
speed
max
m
gradient
max
->
1y =
2000:
=
=
22 x/
-
-
222 x,
10
-
0
10m/s
-
m 250
-
=
14
Vmax=37.5m/s
100
18
-
m 150
=
#
=
37.5
min
speed
x2 x,
-
I
120
-
30
-
=
Varg:d
350
=
25
Varg= 14m/s.
108
- >
v 10
=
27
gradient
->
=
=
min
grad=yc -y,
also
d
v
4
v 37.5
->
m/s
=
1)
Varg:
18
8 1m/s
=
200 5ml
=
=
speed-time Graphs
v/ms-
y=r=acceleration
grad:
↑
I
o
0
V =
&
E
vxt d
=
Area:
>
1bh
=1(tXv)
t/5
Area:distance
so
>
#
↑
~
①
v/ms-
v/ms-
②
X
v/ms-
③
X
i
const
m a 0
=
=
=
at=
m
O
O
stIS
0
At
0
rest
Constant Acceleration
(Uniform)
0 ->
v/ms-
⑤
stIS
0
constant
X
2m/s
->
4m/s ->6m/s-Omls
x
⑥ v/ms-
2.
u
on
&x
7
843x
O
0
IUniform)
acc=
ad
stIS
O
m/s-Omls -6m/s-4m/
deceasing
N=
=
m
stIS
ConstantDeceleration
10
O
Es
to
Moving at
v/ms-
④
&
Acceleration
Increasing
O_1m/sT3m/Emp-loms
+4
o
0
stIS
declaring Acceleration
075m/s-9m/s-l2mp-Rs
~
v/ms-
v/msx
⑦
⑧
O
O
stIS
0
Increasing Dealiation
15m/s ->10m/s76m/s ->3m/s
5
-
stIS
0
Decreasing Deceleration
-
x
15m/s -14m/s ->12m/s ->9m/s -5m/s
3
-
Example #1
I
-
X
20
-
maxacc
as
v(ms
2
m yc
=
-
3
-
->
4
grad.
max
y,som
-
- -
x( x,
-
m
12
=
a
=
2.4m/s
=
&
-
b)
0.
⑧
I
!
"
A
Ibh
=
1151(12)
A
=>
=
d1=30m
t/s
A
Detamine
1(p
=
a)h A 1(12 20)(10)
=
+
+
=
d2=160m
max acceleration
the
b) the distance travelled
the whole
as
during
2)
the
average
dT d du
=
+
38 168
journey
=
+
d+=190m
speed of journey
n
c)
di
arg. speed:
FT
190
=
arg speed
12.7
=
m/s
Q44.
grad-
1
acc.
↓
↓
O
-
S
&
&
/
Q45.
grad=acc 0
=
:
grad
acceleration
↑
0
12
Q48.
Q49.
a
V
/
Endarg.
it
velocity:
(04
=
7
=0.06
kmymin
start
Q50.
14
Q51.
grad:speedrec
speed 0
"rest"
->
=
et
4
Q52.
F
m v0
=
=
·
↑
grad:speed
station
·
station
~
/
15
Q53.
Q54.
grad
->
acceleration"
14.4,29)
↓ extra knowledge
acceleration at
·
26
t 35
m
yz y
:
-
=
->
(106,16)
x( x,
-
2
1 1.6
=
a
4.6m/sh
=
u
26m/s
<v
=
L
=
=
aV
=
-
U
->
t
a 2doemp
=
1
==
i
↑
o
ANSWERS
1.A
2.B
3.A
4.C
5.C
6.D
7.C
8.D
9.B
10.C
11.B
12.D
13.B
14.D
15.B
16.C
17.A
18.D
19.B
20.B
21.B
22.C
23.A
24.A
25.D
26.D
27.B
28.D
29.B
30.C
31.C
32.B
33.C
34.C
35.A
36.B
37.C
38.B
39.C
40.A
Q41.C Q42.A Q43.D Q44.C
Q51. Q52.D Q53.A Q54.B
Q45.D Q46.C Q47.D
Q48.B
Q49.A
Q50.D
16
v/mst
What is the
acceleration
X
att 35.
=
·
+
gradient.
acceleration:
5
y,
y
⑥
=
xz x,
-
+
3
>Es