Motion with constant acceleration: Equations of motion
MOTION
(i) v=u+at Important points about graphical
1 at2 analysis of motion
(ii) S =ut+-
2
Instantaneous velocity is the slope of [ [
v=
dx
dt ALONG A
A Person travels from A to B covers unequal distances in equal
[∆ x=∫vdt[
Distance = Length of actual path position-time curve
STRAIGHT
interval of time with constant acceleration a
Displacement = Length of then Area of v-t curve gives displacement.
shortest path 3S1-S2
[a = dt [ LINE
S1 S2
initial velocity U= dv
Slope of velocity-time curve = instantaneous
Distance > |displacement| 2t t t acceleration
S2-S1
[ ∆ v =∫adt[
A B
Acceleration a = 0
t2 Area of a-t curve gives change in velocity.
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 at a constant rate α for some time, after which it decelerates at
u -v
( θ2
( a constant rate β, to come to rest. If the total time elapsed is t, then
Ratio of Displacement to Distance = Sin
αβ
t
(
Time t =Rθ θ
The two ends of a train moving with constant acceleration pass a certain Vmax =
α+β Total Distance = 1
αβ
t2
( vmax A
( θ2 2 α+β
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 B
θ u v O
v
t1 t2 t
2 = u 2+v2
(
( Mid
Average Acceleration = U 2Sin θ 2
2
θ u2 u 0
R Calculation of stopping distance s= s MOTION UNDER GRAVITY
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 above initial position
=
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 initial position
Zero = final position & initial position are at same level
1
n
If t1 = t2 =t3 = .....= t n
u=0
then
v + v2 + v 3 +.....+V n Different Cases v-t graph s-t graph Object is dropped from top of a tower
v av = 1 h
n (i) Ratio of displacement in equal interval of time S1:S2:S3....=1:3:5....
for v1 & v2, v v=constant s
v +v vt
1. Uniform motion s= (ii) Ratio of time of covering equal distance
Vavg= 1 2 (Arithmetic mean of speeds)
2 t t t1:(t2-t1):(t3-t2):.......:(tn-tn-1)= 1: ( 2- 1):( 3- 2):...:( n- n-1
Distance average speed
Total distance covered s1 + s2 + s3 +.....+ sn v s =½ at (iii) Ratio of total distance covered at the end of time t:2t:3t:....=12:22:32....
2
s + s + s3 + .....+ s n 2. Uniformly accelerated motion a t s
v av = = = 1 2 v= H
Total time elapsed t1 + t2 +t3 + ....+ tn s1 s2 s3 s with u =0 at t=0 If a body is thrown vertically up with a velocity u in the uniform gravitational field (neglecting
+ + +....
.. + n t
If s1= s2= s 3= ...... sn v v v v t
air resistance) then
1 2 3 n
u
s u2
then
n
v s=ut+½at2 (i) Maximum height attained H =
for v1 & v2, 3. Uniformly accelerated with u+
at 2g
v av = 2V1V2 v= u
1 1 1 1 u = 0 at t=0 & s=0 at t=0 u (ii) Time of ascent = time of descent
+ + + ..... .+ Vavg=
V1+V2
(Harmonic mean of speeds)
t g
v1 v2 v 3 vn
(iii) Total time of flight = g
v
s
s=s0+ut+½at2 2u
4. Uniformly accelerated motion u+a
t
(iv) Velocity of fall at the point of projection = u (downwards)
dx with u=0 and s=s0 at t=0 v= t1
Instantaneous Velocity v = ∆ x =∫vdt u
H t2
dt t At any point on its path the body will have same speed for upward journey and
dv u h
Instantaneous Acceleration a = dt ∆ v =∫adt v
v=
s s=ut-½at2 downward journey. If a body thrown upwards crosses a point in time t1 & t2
5. Uniformly retarded motion till u u-
at respectively then
Case 1 Case 2 Case 3 velocity becomes zero 1
t0
t
t0 t height of point h=½ gt1t2 Maximum height H = g(t1 +t2) 2
v=f(t) or x=f(t) V= f(x) t=f(x) 8
dv d2x dV then v s
s=ut-½at2 Time of flight = t1+t2= 2u
g v1
a= V a=-(double diff. of t w.r.t. x) 6. Uniformly retarded then v3
a= = 2 X V 3
u v=
dt dt dx accelerated in opposite direction
u-
at A body is thrown upward, downward & horizontally with same speed v2
t t
Differentiation Differentiation
t0 t0 takes time t1, t2 & t3 respectively to reach the ground then h
Displacement Velocity Acceleration t3 = t t & height from where the particle was throw is h= ½ gt1 t2
Integration Integration 1 2
MOTION
(i) v=u+at Important points about graphical
1 at2 analysis of motion
(ii) S =ut+-
2
Instantaneous velocity is the slope of [ [
v=
dx
dt ALONG A
A Person travels from A to B covers unequal distances in equal
[∆ x=∫vdt[
Distance = Length of actual path position-time curve
STRAIGHT
interval of time with constant acceleration a
Displacement = Length of then Area of v-t curve gives displacement.
shortest path 3S1-S2
[a = dt [ LINE
S1 S2
initial velocity U= dv
Slope of velocity-time curve = instantaneous
Distance > |displacement| 2t t t acceleration
S2-S1
[ ∆ v =∫adt[
A B
Acceleration a = 0
t2 Area of a-t curve gives change in velocity.
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 at a constant rate α for some time, after which it decelerates at
u -v
( θ2
( a constant rate β, to come to rest. If the total time elapsed is t, then
Ratio of Displacement to Distance = Sin
αβ
t
(
Time t =Rθ θ
The two ends of a train moving with constant acceleration pass a certain Vmax =
α+β Total Distance = 1
αβ
t2
( vmax A
( θ2 2 α+β
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 B
θ u v O
v
t1 t2 t
2 = u 2+v2
(
( Mid
Average Acceleration = U 2Sin θ 2
2
θ u2 u 0
R Calculation of stopping distance s= s MOTION UNDER GRAVITY
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 above initial position
=
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 initial position
Zero = final position & initial position are at same level
1
n
If t1 = t2 =t3 = .....= t n
u=0
then
v + v2 + v 3 +.....+V n Different Cases v-t graph s-t graph Object is dropped from top of a tower
v av = 1 h
n (i) Ratio of displacement in equal interval of time S1:S2:S3....=1:3:5....
for v1 & v2, v v=constant s
v +v vt
1. Uniform motion s= (ii) Ratio of time of covering equal distance
Vavg= 1 2 (Arithmetic mean of speeds)
2 t t t1:(t2-t1):(t3-t2):.......:(tn-tn-1)= 1: ( 2- 1):( 3- 2):...:( n- n-1
Distance average speed
Total distance covered s1 + s2 + s3 +.....+ sn v s =½ at (iii) Ratio of total distance covered at the end of time t:2t:3t:....=12:22:32....
2
s + s + s3 + .....+ s n 2. Uniformly accelerated motion a t s
v av = = = 1 2 v= H
Total time elapsed t1 + t2 +t3 + ....+ tn s1 s2 s3 s with u =0 at t=0 If a body is thrown vertically up with a velocity u in the uniform gravitational field (neglecting
+ + +....
.. + n t
If s1= s2= s 3= ...... sn v v v v t
air resistance) then
1 2 3 n
u
s u2
then
n
v s=ut+½at2 (i) Maximum height attained H =
for v1 & v2, 3. Uniformly accelerated with u+
at 2g
v av = 2V1V2 v= u
1 1 1 1 u = 0 at t=0 & s=0 at t=0 u (ii) Time of ascent = time of descent
+ + + ..... .+ Vavg=
V1+V2
(Harmonic mean of speeds)
t g
v1 v2 v 3 vn
(iii) Total time of flight = g
v
s
s=s0+ut+½at2 2u
4. Uniformly accelerated motion u+a
t
(iv) Velocity of fall at the point of projection = u (downwards)
dx with u=0 and s=s0 at t=0 v= t1
Instantaneous Velocity v = ∆ x =∫vdt u
H t2
dt t At any point on its path the body will have same speed for upward journey and
dv u h
Instantaneous Acceleration a = dt ∆ v =∫adt v
v=
s s=ut-½at2 downward journey. If a body thrown upwards crosses a point in time t1 & t2
5. Uniformly retarded motion till u u-
at respectively then
Case 1 Case 2 Case 3 velocity becomes zero 1
t0
t
t0 t height of point h=½ gt1t2 Maximum height H = g(t1 +t2) 2
v=f(t) or x=f(t) V= f(x) t=f(x) 8
dv d2x dV then v s
s=ut-½at2 Time of flight = t1+t2= 2u
g v1
a= V a=-(double diff. of t w.r.t. x) 6. Uniformly retarded then v3
a= = 2 X V 3
u v=
dt dt dx accelerated in opposite direction
u-
at A body is thrown upward, downward & horizontally with same speed v2
t t
Differentiation Differentiation
t0 t0 takes time t1, t2 & t3 respectively to reach the ground then h
Displacement Velocity Acceleration t3 = t t & height from where the particle was throw is h= ½ gt1 t2
Integration Integration 1 2
