Chapter # 2
Kinematics
(Numerical Question Solution )
( National book foundation )
List of formulas and units you need to memorize before moving
towards numericals portion .
1.First equation of motion
Vf= Vi+at
2.Second equation of motion
S= Vit+1/2at^2
3.Third equation of motion
2as = vf^2-vi^2
4.displacement = change in position / time
5.velocity (v) = displacement (d) / t
6.speed (v) = distance/ t
7.acceleration
= change in velocity / time
=(
vf-vi)/t
Units
velocity -> m/s
speed -> m/s
acceleration -> m/s^2
distance -> m
displacement -> m
FAQ
Q.1 What is the circumference of circle ?
Ans . circumference =2 * 3.14* r
Q.2 What would be displacement if starting and ending position are same ?
Ans . Displacement would be 0 .
As displacement =change in position / time taken
Q.3 Is displacement in circular motion is zero ?
Ans . yes
Q.4 what would be the sign convention of g if object move in opposite direction of g?
Ans . Negative . The g will be equal to -9.8.
Q.5 What would be the sign convention of g if object move towards g ?
Ans . Positive . The g will be equal to 9.8 .
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Q.5 A car starts its motion and accelerates at rate of 2 m/s . Find its velocity after covering distance of 500m .
Given
Vi=0m/s
a=2m/s2
s=500m
To find
Vf=?
Calculation
From second equation of motion ,
S=vit+1/2 * a*t2
500=0+0.5*2 t2
t2=500
Taking square root on both sides we have
t=10
From first equation of motion
Vf=vi+at
=0+2(10)
=20
Q.6 Consider the following speed time graph tell :
a)Which part of the graph is showing acceleration , deceleration and zero acceleration .
acceleration =OA
Zero=AB
deceleration =BC
b)Calculate covered distance from 10 seconds to 20 seconds from the graph .
Given
t1=10s
t2=20s
t= t2- t1=20-10=10s
v=50m/s
To find:
S=?
Calculation:
V=s/t
50=s/10
S=50*10
=500m
Q.7 If a ball is dropped from a high building and it reaches ground in 5 seconds . Find the velocity with which it hits the ground ? What is its average speed ? What is the height of the building?
Solution .
Given
t=5s
a=9.8
vi=0m/s
To find
Vf=?
Average speed(v)=?
Height(h)=?
Calculation
Vf= Vi+at
=0+(9.8)(5)
=9.8*5
Vf =49m/s
From second equation of motion
S= Vit+1/2at2
Replace the s by h and a by g we have
h= Vit+1/2gt2
=0(5)+0.5*(9.8)(5*5)
=0+0.5*9.8*25
=4.9*25
h=122.5m
average speed(v)=s/t
=h/t
=122.5/5
v=24.5m/s
Q.8 A ball is thrown upward with the velocity of 20m/s from ground .
Solution .
Given
Vi=20m/s
Vf = 0m/s
a=g=-9.8 as direction of ball is opposite to earth
a. In how much time , will it reach the top of its paths ?
To find:
t=?
From first equation of motion
Vf= Vi+at
0=20+(-9.8)t
-9.8t=-20
t=-20/-9.8=2s
b. It falls back on ground , what is total time of its trip?
To Find :
Total time to hit the ground = 2*t=2*2= 4s
c. How high will it rise ?
To find :
S=?
From second equation of motion
S= Vit+1/2at2
=20(2)+0.5(-9.8)(2*2)
=40-19.6
=20.4m