Kinematics JUNIOR EAMCET. Distance and Displacement O A B 4m 3m 5m O to B: distance is 7m and...
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Transcript of Kinematics JUNIOR EAMCET. Distance and Displacement O A B 4m 3m 5m O to B: distance is 7m and...
![Page 1: Kinematics JUNIOR EAMCET. Distance and Displacement O A B 4m 3m 5m O to B: distance is 7m and displacement is 5m.](https://reader036.fdocuments.net/reader036/viewer/2022062618/5513dfa45503463a298b57e8/html5/thumbnails/1.jpg)
Kinematics Kinematics
JUNIOR EAMCETJUNIOR EAMCET
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Distance and DisplacementDistance and Displacement
O A
B
4m
3m5m
O to B: distance is 7m and displacement is 5m
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Distance and DisplacementDistance and Displacement
O A
BC
4m
3m5m
O to C along OABC: distance is 11m and displacement is 3m
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Distance and DisplacementDistance and Displacement
O A
BC
4m
3m5m
O to O along OABCO: distance is 14m and displacement is zero
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Arc of a Circle Arc of a Circle
RR
A B
Arc length AB is R
AB jaw length is
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Average Speed and Average Average Speed and Average Velocity Velocity
u v
t t
v
s
u
s
Both averge speed and average velocity
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Average Speed and Average Average Speed and Average Velocity Velocity
vs
u
s
Both averge speed
Both averge velocity
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Train Crossing a BridgeTrain Crossing a Bridge
Post Train
Bridge Train
usTime of crossing
the post is
Time of crossing the bridge is
us L
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Average Speed and Averge Average Speed and Averge Velocity Velocity
ABR
Average speed =
Average velocity =
Average acceleration =
u
u
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Direction of AccelerationDirection of Acceleration
u v
u v
a
a
Direction of acceleration is v – u
Acceleration is positive if v > u
Acceleration is negative if v < u
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Sign Application Sign Application
+ g Downward direction is positive
Freely falling body S, u, v, g, h are all positive
Body projecte up Upward direction is positive a = – g
u, v, h are all positive
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Body Projected From the Top of Body Projected From the Top of a Towera Tower
u
Upward direction is positive
a = – g s = – h
u is positive
s, g are negative
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Tossing in a TrainTossing in a Train
If Vt = Vb : a = 0; the ball falls in the hand
If Vt > Vb : a > 0 the ball falls bedhind
If Vt < Vb : a < 0; the ball falls in front
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Vn – Vn-1 = a Vn = u + an
Vn-1 = u + a(n – 1)
Sn – Sn-1 = a
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u v
s
tDisplacement = average velocity × time
If u = 0
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A
B
a
u
ABa
u
A Ba
u
A travels with acceleration a, B with uniform velocity u If they start simultaneously Time of meet is
If A is ahead of B
d
dIf B is ahead of A
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One Body Projected and the One Body Projected and the Other Falling FreelyOther Falling Freely
u = 0
u = u
Time of meet is
hHeight of meet is
h = h1 + h2
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u u
h
Time taken to reach the ground is t1 when throuwn up
Time taken to reach the ground is t2 when thrown down
t1
t2
Time of free fall is
Initial velocity u =
Height h =
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Water Drops from the TapWater Drops from the Tap
h
(n – 1)t =
‘t’ is time interval with which the drops are released
Ratio of displacements of 4th, 3rd, 2nd and 1st drops is 1 : 3 : 5 : 7
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h height of the window
u velocity at the top of the windowt time of crossing the window
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If the velocity is reduced by after travelling a distnce x, then the total distance it can travel is
n
v
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If th front and back of the train cross a post with velocities u and v, the center will cross the same post with velocity ----
u
v
x
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Graphical RepresentationGraphical Representation
v
t
Uniform velocity
a
aO
-a10m/s
3 5 79
-5
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Graphical RepresentationGraphical Representation
v
t
Uniform velocity
10/3
aO
-510m/s
3 5 79
-5
15 20 10-5
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Graphical RepresentationGraphical Representation
v
tO
vmax
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v
tO
P
V – t GraphV – t GraphThe slope gives acceleration
Acceleration is positive if < 90
Acceleration is negative if > 90
Area represents displacement
Displacement is positive if area is above x - axis
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Graphical RepresentationGraphical Representation
v
t
u constant
au = 0
a
u
-a
O
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A Bsu
vDistance is 2s
Displacement is zero
Time for forward journey is
Time for return journey is
Average speed is
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u
at
u
at
d
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Equations of MotionEquations of Motion
X = a + bt + ct2
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Problem Problem
Find the acceleration
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Problem Problem