Kinemetics mind map for jee main and neet

This is best mind map for kinemetics

1. MOTION
ALONG A
STRAIGHT
LINE
Different Cases
1. Uniform motion
2. Uniformly accelerated motion
with u =0 at t=0
3. Uniformly accelerated with
u = 0 at t=0 & s=0 at t=0
4. Uniformly accelerated motion
with u=0 and s=s
0
at t=0
5. Uniformly retarded motion till
velocity becomes zero
6. Uniformly retarded then
accelerated in opposite direction
v-t graph s-t graph
v s
s=vt
t t
v=constant
v
v
u
v=at
v=u+at
t
t
t
t
t
t
t
t
0
t
0 t
0
t
0
t
s
s
s =½
s=ut+½at
2

at
2
v
v
v
u
u
u
v=u+at
v=u-at
v=u-at
s
s
s
s=s
0
+ut+½at
2

s=ut-½at
2

s=ut-½at
2

Motion with constant acceleration: Equations of motion
(i) v=u+at
S
2
=ut+
1
at
2
-
The number of planks required to stop the bullet
The two ends of a train moving with constant acceleration pass a certain
point with velocities u and v. The velocity with which the middle point of
the train passes the same point is
Calculation of stopping distance
A car accelerates from rest at a constant rate α for some time, after which it decelerates at
a constant rate β, to come to rest. If the total time elapsed is t, then
V
max
=
u v
u
2
u
2
-v
2
N=
u
2
2a
s=
= u
2
+v
2
2
Mid
u v
u
a
s
0
.........
S
1
S
1 S
2
t t
A B
t
S
3
Ratio of distance travelled in equal interval of time in a uniformly
accelerated motion from rest
S
1
:S
2
:S
3
= 1:3:5
A Person travels from A to B covers unequal distances in equal
interval of time with constant acceleration a
then
3S
1
-S
2
2t
U=
S
1
t
A
0
B
t
S
2
initial velocity
S
2
-S
1
t
2
=Acceleration a
t
1 t
2
B
A
v
max
O t
αβ
α+β Total Distance
αβ
α+β
1
2
= (
(
t
2t
for uniform accelerated motion
v
avg
u+v
2
=
H
u
h
h
u
t
1 t
2H
1
2
3
v
v
v
If a body is thrown vertically up with a velocity u in the uniform gravitational field (neglecting
air resistance) then
(i) Maximum height attained
(ii) Time of ascent = time of descent
(iii) Total time of flight =
(iv) Velocity of fall at the point of projection = u (downwards)
At any point on its path the body will have same speed for upward journey and
downward journey. If a body thrown upwards crosses a point in time t
1
& t
2
respectively then
h=½ gt
t
1
+t
2
=
2u
1
t
2
A body is thrown upward, downward & horizontally with same speed
takes time t
1
, t
2
& t
3
respectively to reach the ground then

& height from where the particle was throw is

H=
u
2
2g
u
g
2ug
h=½ gt
1t
2
g(t
1
+ t
2
)
2
1
8
=t
3t
1
t
2
Maximum height H =
h
u=0
MOTION UNDER GRAVITY
Sign Convention
(i) initial velocity
(ii) Acceleration
(iii) Displacement
+ve = upward motion
-ve = downward motion
+ve = final position is above initial position
-ve = final position is below initial position
Zero = final position & initial position are at same level
Always -ve
Object is dropped from top of a tower
(i) Ratio of displacement in equal interval of time S
1
:S
2
:S
3
....=1:3:5....
(ii) Ratio of time of covering equal distance
t
1
:(t
2
-t
1
):(t
3
-t
2
):.......:(t
n
-t
n-1
)= 1: ( 2- 1):( 3- 2):...:( n- n-1
(iii) Ratio of total distance covered at the end of time t:2t:3t:....=1
2
:2
2
:3
2
....
Important points about graphical
analysis of motion
Instantaneous velocity is the slope of
position-time curve
Slope of velocity-time curve = instantaneous
acceleration
Area of v-t curve gives displacement.
Area of a-t curve gives change in velocity.
=v
dx
dt
[ [
a=
dv
dt[ [
x∆=∫vdt[ [
v∆=∫adt[ [
v∆=∫adt
For uniform motion
Displacement = velocity x time
Average speed = |average velocity|=|instantaneous velocity|
av
Total distance
v
=
Total time elapsed

1 2 3 n
1 2 3 n
sss....
ttt
++++s
=
++++

+ + +
=
+++
11 22 33
1 2 3
vtvtvt......
ttt......
If
1 2 3 n
ttt====
then
1 2 3 n
av
vvv
v
n
++
=
(Arithmetic mean of speeds)
2

v
1
+v
2
V
1
+V
2
2V
1
V
2
covered
....t
.....t
+.....+V
Total timeelapsed
av
v=
1 2 3 n
1 2 3 n
ss
tt
++++
=
++++

+++ +
=
+++ +
1 2 3 n
1 2 3 n
1 2 3 n
sss.....
sss s
......
vvv v
If
1 2 3 n
sss......==
then
=
+++ +
av
1 2 3 n
n
v
111 1
......
vvv v
(Harmonic mean of speeds)
Total distancecovered s s.....
t t....
s
= s
Time average speed
Distance average speed
Distance > |displacement|
Instantaneous Acceleration
Instantaneous Velocity
t=f(x)
then
a=-(double diff. of t w.r.t. x) X V
3
v=f(t) or x=f(t)
a=
dvd
2
x
dt
2
dt
V= f(x)
a=V
dV
dx
=
A particle moves from A to B in a circular path of
radius R covering an angle θwith uniform speed U
Distance= RθDisplacement=AB= 2RSin
θ
2
2
(
(
(
(
Ratio of Displacement to Distance
Time t =Rθ
Average Velocity
Average Acceleration
U
U
U
= Sin
θ
2
θ
2
θ
2
θ
2 (
(
2USin
θ
2 (
(
= U
2
Sin
θ
=v
dx
dt
x∆=∫vdt
a=
dv
dt
v∆=∫adt
Case 1 Case 2 Case 3
Displacement Velocity Acceleration
Differentiation
Integration
Differentiation
Integration
(ii)
v
2
=u
2
+2a.s(iii)
s
n=u+
a_
2
(2n-1)(iv)
height of point
Time of flight =
g
Distance = Length of actual path
Displacement = Length of
shortest path

AB =
1
n
=
R
for v
1
& v
2
,
for v
1
& v
2
,
V
avg
=
V
avg
=
v