MIND
MAP
+1 PHYSICS
,MEASUREMENT OF MASS & TIME
SIGNIFICANT FIGURES
The digits in a measured quantity which are reliable and confidence
UNITS & MEASUREMENTS
in our measurement + the digit which is uncertain.
MASS
•Unified atomic mass unit(amu) is used to measure ERRORS IN MEASUREMENT
mass of atoms & molecules RULES FOR SIGNIFICANT FIGURES
Dimensional Analysis
•1amu =(1/12)th mass of 1. All non-zero digits are significant. For example, 42.3 has three Difference between true value
one C12 atom significant figures; 243.4 has four significant figures; and 24.123 has & measured value of a quantity
five significant figures. Dimensions of a physical quantity are power to which units of base quantity
•1amu = 1.66×10-27 kg
•Electron mass- 10-30 kg 2. A zero becomes significant figure if it appears between two are raised. Eg: [M]a [L]b [T]c [A]d [K]e
non-zero digits. For example, 5.03 has three significant figures; Systematic Errors Random Errors
•Earth mass : 1025 kg 5.604 has four significant figures; and 4.004 has four significant Errors which tend to occur Irregular and at random
APPLICATIONS
only in one direction, in magnitude & direction
figures. either positive or negative
•Observable Universe 1055 kg
3. Leading zeros or the zeros placed to the left of the number are
TIME never significant. For example,0.543 has three significant figures;
Instrumental Personal
0.045 has two significant figures; and 0.006 has one significant figure. checking the correctness of conversion of one system Deducing relation
Experimental
•SI unit is second (based on caesium clock with an Due to inbuilt defect Limitations in Due to individual
uncertainity less than 1 part in 10-13 4. Trailing zeros or the zeros placed to the right of the number are
various formulae of unit into another among physical of measuring instrument experimental bias,Lack of proper
technique
ie,3μs loss every year) significant. For example, 4.330 has four signif icant figures; 433.00 Eg: If Z=A+B,[Z]=[A]=[B] n1u1=n2u2 quantity
setting of apparatus
has five significant figures; and 343.000 has six significant figures. Eg: n1[M1A L1B T1C] = n2[M2A L2B T2C] • Least count error is the smallest value that can be measured by
•Timespan of unstable particle: 10-24 s instrument (occurs with random & systematic errors)
•Age of universe: 1017 s 5. In exponential notation, the numerical portion gives the number of M2 A L2 B T2 C
significant figures. For example,1.32 x 10-² has three significant n1= n2 [ ] [ ] [ ] • Absolute Error :- Δa = ai-amean , amean= a1+a2+a3+ ....+an
figures and 1.32 x 104 has three significant figures. M1 L1 T1 n
Δamean Δa1+Δa2+Δa3+ ....+Δan
• Relative Error:- Δamean=
amean n
MEASUREMENT OF LENGTH
RULES FOR ROUNDING OF A MEASUREMENT Δamean
•Large distance is measured by p DIMENSIONAL FORMULA INSTRUMENTS • Percentage Error:-
amean
x 100
parallax method 1. If the digit to be dropped is less than 5, then the preceding digit is
left unchanged. For example,x = 7.82 is rounded off to 7.8 and Least Count:
BASIS b
•Parallax angle= DISTANCE
=x x x
x = 3.94 is rounded off to 3.9.
1) Pressure=stress=Young‛s modulus=ML-1 T-2 Smallest quantity an instrument can
2. If the digit to be dropped is more than 5, then the preceding digit
measure COMBINATION OF ERRORS
•1O=1.745 x 10-2 rad is raised by one. For example, x = 6.87 is rounded off to 6.9 and 2) Work=Energy=Torque=M L2 T-2
•1‛=2.91×104 rad. x = 12.78 is rounded off to 12.8.
3) Power P=M L2 T-3
b mm scale vernier scale screw gauge
•1"=4.85×106 rad. 3. If the digit to be dropped is 5 followed by digits other than zero, 4) Gravitational constant G=M-1 L3 T-2 ↓ ↓ ↓ Absolute
1mm
Relative Percentage error
•For very small sizes, optical microscope, then the preceding digit is raised by one. For example, x = 16.351 is 0.1mm 0.01mm
Operations Formula Z
error Δ Z error ΔZ/Z 100 x Δ Z/ Z
rounded off to 16.4 and x = 6.758 is rounded off to 6.8. 5) Force constant=Spring constant=M T-2
tunneling microscope, electron microscope ΔA+ΔB ΔA+ΔB
ΔA+ ΔB
were used. 4. If the digit to be dropped is 5 or 5 followed by zeros, then the 6) Coefficient of viscosity=M L-1 T-1 Sum A+B
A+B A+B
x 100
•1 AU = 1.496×10 m 11 preceding digit, if it is even, is left unchanged. For example,
x = 3.250 becomes 3.2 on rounding off and x = 12.650 becomes 12.6
7) Latent heat L=L2 T-2 VERNIER CALIPERS Difference A-B ΔA+ ΔB ΔA+ΔB ΔA+ΔB
x 100
A-B A-B
•1 ly = 9.46 × 1015 m on rounding off. 8 Least Count = 1 MSD - 1VSD
I Multiplication AxB AΔB+ BΔA
ΔA
+
ΔB
( AA+ BB( x 100
Δ Δ
•1parsec= 3.08 x 10 16
m 5. If the digit to be dropped is 5 or 5 followed by zeros, then the μ0 If n VSD Coincides with (n-1)
A B
A
9 =M L2 T-3 A-2 BΔA+ AΔB ΔA ΔB
( AA+ BB(x 100
ε
Δ Δ
preceding digit, if it is odd, is raised by one. For example, MSD, Division +
•Size of proton: 10-15 m x = 3.750 is rounded off to 3.8, again x = 16.150 is rounded off 0 then (n-1) MSD= n VSD
B
B2 A B
ΔA ΔA x 100
•Radius Of Earth: 10 m 7 to 16.2. Power An n A n - 1 ΔA n n
10) Capacitance=M-1 L-2 T-4 A2 1VSD = n-1 MSD
n
A A
11) Permittivity ε0=M-1 L-3 T4 A2
1 A 1/n-1 ΔA
•Distance to Boundary Of RULES FOR ROUNDING OF A MEASUREMENT Root A 1/n
1 ΔA 1 ΔA x 100
n-1 n n A n A
Observable Universe : 1026 m 12) Angular momentum = planck‛s constant Least Count = 1MSD - n MSD = 1MSD
n
ADDITION & SUBTRACTION =M1 L2 T-1 Total Reading = Main Scale Reading + coinciding General rule:
In addition or subtraction, the final result should be reported Vernier Scale division x least count If Z = APBq
SI SYSTEM
,Then the maximum fractional relative
to the same number of decimal Places as that of the original
13) M= hc L= hG T= hG Cr error in Z will be:
number with minimum number of decimal places G 2
c
5
c
ΔZ =p ΔA +q ΔB
In a vernier calipers, one main scale division is x cm +r ΔC
3.1421
7 Base units and 2 supplementary units l m R & n division of vernier scale coincide with n-1 divisions Z A B C
0.241 Tα g
α k
α g of the main scle. the least count (in cm) of the
Base Units +0.09 (has two decimal places)
NO. Quantity Unit Symbol (Answer should be reported to two decimal
Time period L
= RC = LC
calipers is.
3.4731 n-1 nx x x
places after rounding off) R a) ( n ) x b) c) d)
1 Length meter m (n-1) (n-1) n
2 Mass kilogram kg In an expirement four quantities a,b,c
Answer = 3.47
3 Time second s and d are measured with percentage
4 Temperature kelvin K
MULTIPLICATION & DIVISION DIMENSIONLESS SCREW GAUGE error1%, 2%, 3% and 4% respectievely.
5 Electric current ampere A Quantity P is calculated as follows:
6 Luminous intensity candela cd When numbers are multiplied or divided, the number of QUANTITIES
7 Amount of mole mol significant figures in the answer equals the smallest number Main Scale Reading
substance
of significant figures in any of the original numbers 1) Strain Pitch = a2b2
No.of rotations P=
NO.
Quantity
Supplementary Units
Unit Symbol
51.028 2) Refractive index cd
x 1.31 (Three significant figures)
1 Plane angle radian rad 3) Relative density pitch
Solid angle steradian sr 66.84668 (Answer should have three significant figures Least Count =
Total no.of divisions on
(a) 14% (b) 10%
2
after rounding off) 4) Plane angle
circlular scale (c) 7% (d) 4%
Answer = 66.8 5) Solid angle Total Reading = Linear Scale Reading + circular scale
reading x least count
Unit of permittivity of free space ε0 is In SI Units, the dimensions of ε 0
If L=2.331cm, B= 2.1cm,then L+B = ? μ0
The least count of the main scale of a screw gauge
(a) coloumb/newton-metre is 1mm the minimum no.of divisions on its circular
is:
(b) newton-metre2 /coloumb² (a) 4.431 cm (b) 4.43 cm a)A-1 T M L3 b)A T2 M-1L-1 scale required to measure 5μm diameter of wire is
(c) coloumb²/newton-metre2 (c) 4.4 cm (d) 4 cm c)A T-3 M L3/2 d)A2 T3 M-1 L-2 a) 200 b) 50 c) 400 d) 100
(d) coloumb2/(newton-metre)2
, Motion with constant acceleration: Equations of motion
MOTION
(i) v=u+at Important point 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 v-t curve area 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
t 2 a-t curve area 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 = Rθ Displacement = 2RSin u2
2 N= A car accelerates from rest at a constant rate α for some time, after which it decelerates at
(
u2-v2 a constant rate β, to come to rest. If the total time elapsed is t, the maximum velocity
= 2Sin θ
Ratio of Displacement to Distance
(
2 attained αβ
Vmax = t αβ
(
Time t =Rθ θ
The two ends of a train moving with constant acceleration pass a certain
α+β Total Distance = 1 t2 v A
(
2 α+β
max
point with velocities u and v. The velocity with which the middle point of
(
U (
θ the train passes the same point is
Average Velocity = 2USin B
2 u v O t1 t2 t
θ v= u +v
2 2
(
Average Acceleration = 2U2Sin θ
( Mid
2
2 u2 u 0
Rθ Calculation of stoping distance s= s MOTION UNDER GRAVITY
2a .........
Sign Convension
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
vavg u+v +ve = final position is above initial position
Total distance covered
=
s1 + s2 + s3 + ....+s n v 1t1 +v 2t 2 + v 3 t3 + ...... 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
If t1 = t2 =t3 = .....= t n u=0
then Different Cases v-t graph s-t graph Object is dropped from top of a tower
v + v2 + v 3 +.....+V n v1+v2 (Arithmetic mean of speeds)
h
v av = 1 = 2 (i) Ratio of displacement in equal interval of time S1:S2:S3....=1:3:5....
n v v=constant s
vt
1. Uniform motion s= (ii) Ratio of time of covering equal distance
Distance average speed t t t1:(t2-t1):(t3-t2):.......:(tn-tn-1)= 1: ( 2- 1):( 3- 2):...:( n- n-1
Total distance covered s1 + s2 + s3 +.....+ sn s + s + s3 + .....+ s n v s =½ at (iii) Ratio of distance covered at the end of time t:2t:3t:....=12:22:32....
2
s
v av = = = 1 2 2. Uniformly accelerated motion at
v= H
Total time elapsed t1 + t2 +t3 + ....+ tn s1 s2 s3 s with u =0 at t=0
+ + +....
.. + n t If a body is thrown vertically up with a velocity u in the uniform gravitational field (neglecting
If s1= s2= s 3= ...... sn s s .... t v1 v2 v 3
t ...... vn t
air resistance) then u
s u2
then n 2V1V2 v s=ut+½at2 (i) Maximum height attained H =
v av = = (Harmonic mean of speeds) 3. Uniformly accelerated with u+
at 2g
(ii) Time of ascent = time of descent u
1 1 1 1 V1+V2 u = 0 at t=0 v=
u
+ + + .....
v1 v2 v 3
.+
vn t g
2u
s (iii) Total time of flight = g
v s=s0+ut+½at2
4. Uniformly accelerated motion u+
at (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 v s downward journey. If a body thrown upwards crosses a point in time t 1 & t2 h
Instantaneous Acceleration a= ∆ v =∫adt s=ut-½at2
dt 5. Uniformly retarded motion till u v=u
-a respectively then
t
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 or x = f(t) V= f(x) t=f(x) 8
dv d2x dV then v s
t1+t2= 2u v 1
a= = a= V a=-(double diff. of t w.r.to x) V3 6. Uniformly retarded then g v
X u
dt dt2 dx accelerated in opposite direction t0 A body is thrown upward, downward & horizontally with same speed 3
t t v 2
Differentiation Differentiation t0 takes time t1, t2 & t3 respectively to reach the ground then
Displacement Velocity Acceleration t3 = t t & height from where the particle was throw is h= ½ gt1 t2
Integration Integration 1 2
MAP
+1 PHYSICS
,MEASUREMENT OF MASS & TIME
SIGNIFICANT FIGURES
The digits in a measured quantity which are reliable and confidence
UNITS & MEASUREMENTS
in our measurement + the digit which is uncertain.
MASS
•Unified atomic mass unit(amu) is used to measure ERRORS IN MEASUREMENT
mass of atoms & molecules RULES FOR SIGNIFICANT FIGURES
Dimensional Analysis
•1amu =(1/12)th mass of 1. All non-zero digits are significant. For example, 42.3 has three Difference between true value
one C12 atom significant figures; 243.4 has four significant figures; and 24.123 has & measured value of a quantity
five significant figures. Dimensions of a physical quantity are power to which units of base quantity
•1amu = 1.66×10-27 kg
•Electron mass- 10-30 kg 2. A zero becomes significant figure if it appears between two are raised. Eg: [M]a [L]b [T]c [A]d [K]e
non-zero digits. For example, 5.03 has three significant figures; Systematic Errors Random Errors
•Earth mass : 1025 kg 5.604 has four significant figures; and 4.004 has four significant Errors which tend to occur Irregular and at random
APPLICATIONS
only in one direction, in magnitude & direction
figures. either positive or negative
•Observable Universe 1055 kg
3. Leading zeros or the zeros placed to the left of the number are
TIME never significant. For example,0.543 has three significant figures;
Instrumental Personal
0.045 has two significant figures; and 0.006 has one significant figure. checking the correctness of conversion of one system Deducing relation
Experimental
•SI unit is second (based on caesium clock with an Due to inbuilt defect Limitations in Due to individual
uncertainity less than 1 part in 10-13 4. Trailing zeros or the zeros placed to the right of the number are
various formulae of unit into another among physical of measuring instrument experimental bias,Lack of proper
technique
ie,3μs loss every year) significant. For example, 4.330 has four signif icant figures; 433.00 Eg: If Z=A+B,[Z]=[A]=[B] n1u1=n2u2 quantity
setting of apparatus
has five significant figures; and 343.000 has six significant figures. Eg: n1[M1A L1B T1C] = n2[M2A L2B T2C] • Least count error is the smallest value that can be measured by
•Timespan of unstable particle: 10-24 s instrument (occurs with random & systematic errors)
•Age of universe: 1017 s 5. In exponential notation, the numerical portion gives the number of M2 A L2 B T2 C
significant figures. For example,1.32 x 10-² has three significant n1= n2 [ ] [ ] [ ] • Absolute Error :- Δa = ai-amean , amean= a1+a2+a3+ ....+an
figures and 1.32 x 104 has three significant figures. M1 L1 T1 n
Δamean Δa1+Δa2+Δa3+ ....+Δan
• Relative Error:- Δamean=
amean n
MEASUREMENT OF LENGTH
RULES FOR ROUNDING OF A MEASUREMENT Δamean
•Large distance is measured by p DIMENSIONAL FORMULA INSTRUMENTS • Percentage Error:-
amean
x 100
parallax method 1. If the digit to be dropped is less than 5, then the preceding digit is
left unchanged. For example,x = 7.82 is rounded off to 7.8 and Least Count:
BASIS b
•Parallax angle= DISTANCE
=x x x
x = 3.94 is rounded off to 3.9.
1) Pressure=stress=Young‛s modulus=ML-1 T-2 Smallest quantity an instrument can
2. If the digit to be dropped is more than 5, then the preceding digit
measure COMBINATION OF ERRORS
•1O=1.745 x 10-2 rad is raised by one. For example, x = 6.87 is rounded off to 6.9 and 2) Work=Energy=Torque=M L2 T-2
•1‛=2.91×104 rad. x = 12.78 is rounded off to 12.8.
3) Power P=M L2 T-3
b mm scale vernier scale screw gauge
•1"=4.85×106 rad. 3. If the digit to be dropped is 5 followed by digits other than zero, 4) Gravitational constant G=M-1 L3 T-2 ↓ ↓ ↓ Absolute
1mm
Relative Percentage error
•For very small sizes, optical microscope, then the preceding digit is raised by one. For example, x = 16.351 is 0.1mm 0.01mm
Operations Formula Z
error Δ Z error ΔZ/Z 100 x Δ Z/ Z
rounded off to 16.4 and x = 6.758 is rounded off to 6.8. 5) Force constant=Spring constant=M T-2
tunneling microscope, electron microscope ΔA+ΔB ΔA+ΔB
ΔA+ ΔB
were used. 4. If the digit to be dropped is 5 or 5 followed by zeros, then the 6) Coefficient of viscosity=M L-1 T-1 Sum A+B
A+B A+B
x 100
•1 AU = 1.496×10 m 11 preceding digit, if it is even, is left unchanged. For example,
x = 3.250 becomes 3.2 on rounding off and x = 12.650 becomes 12.6
7) Latent heat L=L2 T-2 VERNIER CALIPERS Difference A-B ΔA+ ΔB ΔA+ΔB ΔA+ΔB
x 100
A-B A-B
•1 ly = 9.46 × 1015 m on rounding off. 8 Least Count = 1 MSD - 1VSD
I Multiplication AxB AΔB+ BΔA
ΔA
+
ΔB
( AA+ BB( x 100
Δ Δ
•1parsec= 3.08 x 10 16
m 5. If the digit to be dropped is 5 or 5 followed by zeros, then the μ0 If n VSD Coincides with (n-1)
A B
A
9 =M L2 T-3 A-2 BΔA+ AΔB ΔA ΔB
( AA+ BB(x 100
ε
Δ Δ
preceding digit, if it is odd, is raised by one. For example, MSD, Division +
•Size of proton: 10-15 m x = 3.750 is rounded off to 3.8, again x = 16.150 is rounded off 0 then (n-1) MSD= n VSD
B
B2 A B
ΔA ΔA x 100
•Radius Of Earth: 10 m 7 to 16.2. Power An n A n - 1 ΔA n n
10) Capacitance=M-1 L-2 T-4 A2 1VSD = n-1 MSD
n
A A
11) Permittivity ε0=M-1 L-3 T4 A2
1 A 1/n-1 ΔA
•Distance to Boundary Of RULES FOR ROUNDING OF A MEASUREMENT Root A 1/n
1 ΔA 1 ΔA x 100
n-1 n n A n A
Observable Universe : 1026 m 12) Angular momentum = planck‛s constant Least Count = 1MSD - n MSD = 1MSD
n
ADDITION & SUBTRACTION =M1 L2 T-1 Total Reading = Main Scale Reading + coinciding General rule:
In addition or subtraction, the final result should be reported Vernier Scale division x least count If Z = APBq
SI SYSTEM
,Then the maximum fractional relative
to the same number of decimal Places as that of the original
13) M= hc L= hG T= hG Cr error in Z will be:
number with minimum number of decimal places G 2
c
5
c
ΔZ =p ΔA +q ΔB
In a vernier calipers, one main scale division is x cm +r ΔC
3.1421
7 Base units and 2 supplementary units l m R & n division of vernier scale coincide with n-1 divisions Z A B C
0.241 Tα g
α k
α g of the main scle. the least count (in cm) of the
Base Units +0.09 (has two decimal places)
NO. Quantity Unit Symbol (Answer should be reported to two decimal
Time period L
= RC = LC
calipers is.
3.4731 n-1 nx x x
places after rounding off) R a) ( n ) x b) c) d)
1 Length meter m (n-1) (n-1) n
2 Mass kilogram kg In an expirement four quantities a,b,c
Answer = 3.47
3 Time second s and d are measured with percentage
4 Temperature kelvin K
MULTIPLICATION & DIVISION DIMENSIONLESS SCREW GAUGE error1%, 2%, 3% and 4% respectievely.
5 Electric current ampere A Quantity P is calculated as follows:
6 Luminous intensity candela cd When numbers are multiplied or divided, the number of QUANTITIES
7 Amount of mole mol significant figures in the answer equals the smallest number Main Scale Reading
substance
of significant figures in any of the original numbers 1) Strain Pitch = a2b2
No.of rotations P=
NO.
Quantity
Supplementary Units
Unit Symbol
51.028 2) Refractive index cd
x 1.31 (Three significant figures)
1 Plane angle radian rad 3) Relative density pitch
Solid angle steradian sr 66.84668 (Answer should have three significant figures Least Count =
Total no.of divisions on
(a) 14% (b) 10%
2
after rounding off) 4) Plane angle
circlular scale (c) 7% (d) 4%
Answer = 66.8 5) Solid angle Total Reading = Linear Scale Reading + circular scale
reading x least count
Unit of permittivity of free space ε0 is In SI Units, the dimensions of ε 0
If L=2.331cm, B= 2.1cm,then L+B = ? μ0
The least count of the main scale of a screw gauge
(a) coloumb/newton-metre is 1mm the minimum no.of divisions on its circular
is:
(b) newton-metre2 /coloumb² (a) 4.431 cm (b) 4.43 cm a)A-1 T M L3 b)A T2 M-1L-1 scale required to measure 5μm diameter of wire is
(c) coloumb²/newton-metre2 (c) 4.4 cm (d) 4 cm c)A T-3 M L3/2 d)A2 T3 M-1 L-2 a) 200 b) 50 c) 400 d) 100
(d) coloumb2/(newton-metre)2
, Motion with constant acceleration: Equations of motion
MOTION
(i) v=u+at Important point 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 v-t curve area 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
t 2 a-t curve area 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 = Rθ Displacement = 2RSin u2
2 N= A car accelerates from rest at a constant rate α for some time, after which it decelerates at
(
u2-v2 a constant rate β, to come to rest. If the total time elapsed is t, the maximum velocity
= 2Sin θ
Ratio of Displacement to Distance
(
2 attained αβ
Vmax = t αβ
(
Time t =Rθ θ
The two ends of a train moving with constant acceleration pass a certain
α+β Total Distance = 1 t2 v A
(
2 α+β
max
point with velocities u and v. The velocity with which the middle point of
(
U (
θ the train passes the same point is
Average Velocity = 2USin B
2 u v O t1 t2 t
θ v= u +v
2 2
(
Average Acceleration = 2U2Sin θ
( Mid
2
2 u2 u 0
Rθ Calculation of stoping distance s= s MOTION UNDER GRAVITY
2a .........
Sign Convension
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
vavg u+v +ve = final position is above initial position
Total distance covered
=
s1 + s2 + s3 + ....+s n v 1t1 +v 2t 2 + v 3 t3 + ...... 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
If t1 = t2 =t3 = .....= t n u=0
then Different Cases v-t graph s-t graph Object is dropped from top of a tower
v + v2 + v 3 +.....+V n v1+v2 (Arithmetic mean of speeds)
h
v av = 1 = 2 (i) Ratio of displacement in equal interval of time S1:S2:S3....=1:3:5....
n v v=constant s
vt
1. Uniform motion s= (ii) Ratio of time of covering equal distance
Distance average speed t t t1:(t2-t1):(t3-t2):.......:(tn-tn-1)= 1: ( 2- 1):( 3- 2):...:( n- n-1
Total distance covered s1 + s2 + s3 +.....+ sn s + s + s3 + .....+ s n v s =½ at (iii) Ratio of distance covered at the end of time t:2t:3t:....=12:22:32....
2
s
v av = = = 1 2 2. Uniformly accelerated motion at
v= H
Total time elapsed t1 + t2 +t3 + ....+ tn s1 s2 s3 s with u =0 at t=0
+ + +....
.. + n t If a body is thrown vertically up with a velocity u in the uniform gravitational field (neglecting
If s1= s2= s 3= ...... sn s s .... t v1 v2 v 3
t ...... vn t
air resistance) then u
s u2
then n 2V1V2 v s=ut+½at2 (i) Maximum height attained H =
v av = = (Harmonic mean of speeds) 3. Uniformly accelerated with u+
at 2g
(ii) Time of ascent = time of descent u
1 1 1 1 V1+V2 u = 0 at t=0 v=
u
+ + + .....
v1 v2 v 3
.+
vn t g
2u
s (iii) Total time of flight = g
v s=s0+ut+½at2
4. Uniformly accelerated motion u+
at (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 v s downward journey. If a body thrown upwards crosses a point in time t 1 & t2 h
Instantaneous Acceleration a= ∆ v =∫adt s=ut-½at2
dt 5. Uniformly retarded motion till u v=u
-a respectively then
t
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 or x = f(t) V= f(x) t=f(x) 8
dv d2x dV then v s
t1+t2= 2u v 1
a= = a= V a=-(double diff. of t w.r.to x) V3 6. Uniformly retarded then g v
X u
dt dt2 dx accelerated in opposite direction t0 A body is thrown upward, downward & horizontally with same speed 3
t t v 2
Differentiation Differentiation t0 takes time t1, t2 & t3 respectively to reach the ground then
Displacement Velocity Acceleration t3 = t t & height from where the particle was throw is h= ½ gt1 t2
Integration Integration 1 2
