Deduce the
following equation of motion:
Vsquare =Usquare + 2as
S= ut + ½ at
Answers
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+1/2at)×t
On further simplification, the equation becomes:
s=ut+1/2at2
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+1/2at2
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
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