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Summary Motion in straight line or 1- dimension motion

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Here is summary notes of motion in straight line chapter of physics. They are easy to understand, helpful and worthy. All concepts and forumals are included in minimum area to help you revise easily and faster. Hope you will like it.

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Uploaded on July 13, 2024
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Written in 2023/2024
Type Summary

class 11th motion in straight line notes

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Motion with constant acceleration: Equations of motion
(i) v=u+at Important points abou
1 at2 analysis of motion
(ii) S =ut+-
2
Instantaneous velocity is the
Distance = Length of actual path A Person travels from A to B covers unequal distances in equal position-time curve
interval of time with constant acceleration a
Displacement = Length of then Area of v-t curve gives displa
shortest path 3S1-S2 S1 S2
initial velocity U= Slope of velocity-time curve
Distance > |displacement| 2t t t acceleration
S2-S1 A B
Acceleration a = 0
t2 Area of a-t curve gives chan

A particle moves from A to B in a circular path of
U (iii) v2=u2+2a.s
radius R covering an angle θ with uniform speed U
U The number of planks required to stop the bullet u v

( θ
Distance=AB Rθ Displacement=AB= 2RSin
( u2
=
2 N= 2 2 A car accelerates from rest
u -v
( θ2
( a constant rate β, to come
Ratio of Displacement to Distance = Sin
αβ
Time t =Rθ
The two ends of a train moving with constant acceleration pass a certain Vmax = t
θ α+β
( θ2
U 2USin
( 2
point with velocities u and v. The velocity with which the middle point of
Average Velocity = the train passes the same point is
θ u v
2 v = u 2+v2
(
( Mid
Average Acceleration = U 2Sin θ 2
2
θ u2 u 0
R Calculation of stopping distance s= s MOTION UNDER GRA
2 2a .........
Sign Convention
For uniform motion (iv) sn =u+ _
a (2n-1) a
2 (i) initial velocity
Displacement = velocity x time Ratio of distance travelled in equal interval of time in a uniformly +ve = upward motion
Average speed = |average velocity|=|instantaneous velocity| accelerated motion from rest -ve = downward motion
S1 S1 S2 S3

S1:S2:S3 = 1:3:5 (ii) Acceleration
t t t Always -ve
Time average speed A B
(iii) Displacement
s + s + s + ....+s n v 1t1 +v 2t 2 + v 3 t3 + ...... vavg u+v +ve = final position is abov
=
Total distance covered
= 1 2 3 = for uniform accelerated motion =
v av Total time elapsed t1 + t2 + t3 + ....+ tn t1 + t2 + t3 + ...... 2 -ve = final position is below
Zero = final position & init
If t1 = t2 =t3 = .....= t n
then
v + v2 + v 3 +.....+V n Different Cases v-t graph s-t graph Object is dropped from top of a
v av = 1
n (i) Ratio of displacement in eq
for v1 & v2, v v=constant s
v1+v2 vt
1. Uniform motion s= (ii) Ratio of time of covering e
Vavg= (Arithmetic mean of speeds)
2 t t t1:(t2-t1):(t3-t2):.......:(tn-
Distance average speed
Total distance covered s1 + s2 + s3 +.....+ sn v s =½ at (iii) Ratio of total distance co
2

s + s + s3 + .....+ s n 2. Uniformly accelerated motion at s
v av = = = 1 2 v=
Total time elapsed t1 + t2 +t3 + ....+ tn s1 s2 s3 s with u =0 at t=0 If a body is thrown vertically up
+ + +....
.. + n t

v1 v2 v 3 vn t

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