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
(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
