Prove algebraically and graphical equation of motion.
Please give a solution in a image .
Class 9 , Physics.
Answers
Explanation:
Derivation of First Equation of Motion by Algebraic Method
We know that the acceleration of the body is defined as the rate of change of velocity.
Mathematically, acceleration is represented as follows: a=v−ut
where v is the final velocity and u is the initial velocity.
Rearranging the above equation, we arrive at the first equation of motion as follows: v=u+at
Derivation of First Equation of Motion by Graphical Method
The first equation of motion can be derived using a velocity-time graph for a moving object with an initial velocity of u, final velocity v, and acceleration a.
In the above graph,
The velocity of the body changes from A to B in time t at a uniform rate.
BC is the final velocity and OC is the total time t.
A perpendicular is drawn from B to OC, a parallel line is drawn from A to D, and another perpendicular is drawn from B to OE (represented by dotted lines).
Following details are obtained from the graph above:
The initial velocity of the body, u = OA
The final velocity of the body, v = BC
From the graph, we know that
BC = BD + DC
Therefore, v = BD + DC
v = BD + OA (since DC = OA)
Finally,
v = BD + u (since OA = u) (Equation 1)
Now, since the slope of a velocity-time graph is equal to acceleration a,
So,
a = slope of line AB
a = BD/AD
Since AD = AC = t, the above equation becomes:
BD = at (Equation 2)
Now, combining Equation 1 & 2, the following is obtained:
v = u + at
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+v/2×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+at/2×t
s=(2/u2+at/2)×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
From the graph above, we can say that
Distance travelled (s) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD
s=(1/2AB×BD)+(OA×OC)
Since BD = EA, the above equation becomes
s=(1/2AB×EA)+(u×t)
As EA = at, the equation becomes
s=1/2×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+v/2)×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+u/2)(v−u/a)
s=(v/2−u2/2a)
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
