English, asked by pagalapyaar, 6 months ago

what is meant by motion derived???​

Answers

Answered by Anonymous
1

Motion is the phenomenon in which an object changes its position over time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and time. ... One can also speak of motion of images, shapes and boundaries.

Answered by Anonymous
2

Equations of motion of kinematics describe the basic concept of the motion of an object such as the position, velocity or the acceleration of an object at various times. These three equations of motion govern the motion of an object in 1D, 2D and 3D. The derivation of the equations of motion is one of the most important topics in Physics. In this article, we will show you how to derive the first, second and third equation of motion by graphical method, algebraic method and calculus method.

Derivation of Second Equation of Motion by Algebraic Method

Velocity is defined as the rate of change of displacement. This is mathematically represented as:

Velocity=DisplacementTime

Rearranging, we get

Displacement=Velcoity×Time

If the velocity is not constant then in the above equation we can use average velocity in the place of velocity and rewrite the equation as follows:

Displacement=(InitialVelocity+FinalVelocity2)×Time

Substituting the above equations with the notations used in the derivation of the first equation of motion, we get

s=u+v2×t

From the first equation of motion, we know that v = u + at. Putting this value of v in the above equation, we get

s=u+(u+at))2×t

s=2u+at2×t

s=(2u2+at2)×t

s=(u+12at)×t

On further simplification, the equation becomes:

s=ut+12at2

Derivation of Second Equation of Motion by Graphical Method

Derivation of Equation of Motion

From the graph above, we can say that

Distance travelled (s) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD

s=(12AB×BD)+(OA×OC)

Since BD = EA, the above equation becomes

s=(12AB×EA)+(u×t)

As EA = at, the equation becomes

s=12×at×t+ut

On further simplification, the equation becomes

s=ut+12at2

Derivation of Second Equation of Motion by Calculus Method

Velocity is the rate of change of displacement.

Mathematically, this is expressed as

v=dsdt

Rearranging the equation, we get

ds=vdt

Substituting the first equation of motion in the above equation, we get

ds=(u+at)dt

ds=(u+at)dt=(udt+atdt)

On further simplification, the equation becomes:

∫s0ds=∫t0udt+∫t0atdt

=ut+12at2

Derivation of Third Equation of Motion

Derivation of Third Equation of Motion by Algebraic Method

We know that, displacement is the rate of change of position of an object. Mathematically, this can be represented as:

Displacement=(InitialVelocity+FinalVelocity2)×t

Substituting the standard notations, the above equation becomes

s=(u+v2)×t

From the first equation of motion, we know that

v=u+at

Rearranging the above formula, we get

t=v−ua

Substituting the value of t in the displacement formula, we get

s=(v+u2)(v−ua)

s=(v2−u22a)

2as=v2−u2

Rearranging, we get

v2=u2+2as

Derivation of Third Equation of Motion by Graphical Method

From the graph, we can say that

The total distance travelled, s is given by the Area of trapezium OABC.

Hence,

S = ½ (Sum of Parallel Sides) × Height

S=(OA+CB)×OC

Since, OA = u, CB = v, and OC = t

The above equation becomes

S= ½ (u+v) × t

Now, since t = (v – u)/ a

The above equation can be written as:

S= ½ ((u+v) × (v-u))/a

Rearranging the equation, we get

S= ½ (v+u) × (v-u)/a

S = (v2-u2)/2a

Third equation of motion is obtained by solving the above equation:

v2 = u2+2aS

Derivation of Third Equation of Motion by Calculus Method

We know that acceleration is the rate of change of velocity and can be represented as:

a=dvdt (1)

We also know that velocity is the rate of change of displacement and can be represented as:

v=dsdt (2)

Cross multiplying (1) and (2), we get

adsdt=vdvdt

∫s0ads=∫vuvds

as=v2−u22

v2=u2+2as

This is how we derive the three equations of motion by algebraic, graphical and calculus method.

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