explain and derive the Laws of motion
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Derive relation F = ma from Newton 2nd Law of Motion. ... According to the Newton's 2nd Law of motion, the rate of change of linear momentum of a body is directly proportional to the applied external force and in the direction of force. It means that the linear momentum will change faster when a bigger force is applied
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Laws of motion ::
1. u = v + at
2. s = ut + 1 / 2 at²
3. v² = u² + 2as
Derivation of 1st state of motion::
Acceleration = a , initial velocity = u , final velocity = v , time = t
a= slope of AB
= PERPENDICULAR / BASE
= BC / AC
BC = v -u and AC = t
a = v - u / t
at = v- u
at +u = v
Hence derieved
Derivation of second state of motion::
Distance =s
s = area under slope of AB
s = Area of trapezium OABD
s = Area of rectangle OACD + Area pf triangle ABC
s = ( l * b ) + ( 1 / 2 * b * h )
s = ( t * u ) + ( 1 / 2 * t * ( v- u )
Using v = u + at
v - u = at
s = ut + ( 1 / 2 * t * at )
s = ut + 1 / 2 at²
Hence derived
Thirs Equation of motion::
s = area of trapezium
s = 1 / 2 * ( Sum of Paralllel sides ) * h
s = 1 / 2 * ( OA + OB ) *AC
s = 1 / 2 * ( v + u ) t
using v = u + at
v - u = at
v - u / a = t
s = 1 / 2 * ( v + u ) * ( v - u / a )
using ( a + b ) ( a - b ) = a² - b²
s = 1 / 2 ( v² - u² ) / a
as = v² - u² / 2
2as = v² - u²
u² + 2as = v²
Hence derived
1. u = v + at
2. s = ut + 1 / 2 at²
3. v² = u² + 2as
Derivation of 1st state of motion::
Acceleration = a , initial velocity = u , final velocity = v , time = t
a= slope of AB
= PERPENDICULAR / BASE
= BC / AC
BC = v -u and AC = t
a = v - u / t
at = v- u
at +u = v
Hence derieved
Derivation of second state of motion::
Distance =s
s = area under slope of AB
s = Area of trapezium OABD
s = Area of rectangle OACD + Area pf triangle ABC
s = ( l * b ) + ( 1 / 2 * b * h )
s = ( t * u ) + ( 1 / 2 * t * ( v- u )
Using v = u + at
v - u = at
s = ut + ( 1 / 2 * t * at )
s = ut + 1 / 2 at²
Hence derived
Thirs Equation of motion::
s = area of trapezium
s = 1 / 2 * ( Sum of Paralllel sides ) * h
s = 1 / 2 * ( OA + OB ) *AC
s = 1 / 2 * ( v + u ) t
using v = u + at
v - u = at
v - u / a = t
s = 1 / 2 * ( v + u ) * ( v - u / a )
using ( a + b ) ( a - b ) = a² - b²
s = 1 / 2 ( v² - u² ) / a
as = v² - u² / 2
2as = v² - u²
u² + 2as = v²
Hence derived
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