Derive equations 1st, 2nd and 3rd graphically
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Answer:
Derivation of Equation of Motion
There are mainly three equations of motion which describe the relationship between velocity, time, acceleration and displacement.
First, consider a body moving in a straight line with uniform acceleration. Then, let the initial velocity be u, acceleration is denoted as a, time period is denoted as t, velocity is denoted as v, and the distance travelled is denoted as S.
The equation of motions derivation can be done in three ways which are:
Derivation of equations of motion by Simple Algebraic Method
Derivation of Motion by Graphical Method
Derivation of Motion by Calculus Method
Below, the equations of motion are derived by all the three methods in a simple and easy to understand way.
Derivation of First Equation of Motion
The first equation of motion is:
v = u + at
Derivation of First Equation of Motion by Algebraic Method
It is known that the acceleration (a) of the body is defined as the rate of change of velocity.
So, the acceleration can be written as:
a = v − ut
From this, rearranging the terms, the first equation of motion is obtained, which is:
v = u + at
Derivation of First Equation of Motion by Graphical Method
Consider the diagram of the velocity-time graph of a body below:
Derivation Of Equation Of Motion
In this, the body is moving with an initial velocity of u at point A. The velocity of the body then changes from A to B in time t at a uniform rate. In the above diagram, BC is the final velocity, i.e. v after the body travels from A to B at a uniform acceleration of a. In the graph, OC is the time t. Then, 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,BC = BD + DC
So, 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 = at + u
Derivation of First Equation of Motion by Calculus Method
It is known that,
Derivation Of Equation Of Motion
So,
Derivation Of Equation Of Motion
Derivation of Second Equation of Motion
The second equation of motion is:
S = ut + ½ a2
Derivation of Second Equation of Motion by Algebraic Method
Consider the same notations for the derivation of the second equation of motion by the simple algebraic method.
Derivation Of Second Equation Of Motion
Derivation of Second Equation of Motion by Graphical Method
Taking the same diagram used in first law derivation:
Derivation Of Equation Of Motion
In this diagram, the distance travelled (S) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD.
Now, the area of the rectangle OADC = OA × OC = ut
And, Area of triangle ABD = (1/2) × Area of rectangle AEBD = (1/2) at2 (Since, AD = t and BD = at)
Thus, the total distance covered will be:
S = ut + (1/2) at2
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 =(udt+atdt) ∫s0ds=∫t0udt+∫t0atdt s=ut+12at2
Derivation of Third Equation of Motion
The third equation of motion is:
v2 = u2 + 2aS
Derivation of Third Equation of Motion by Algebraic Method
Derivation Of Third Equation Of Motion
Derivation of Third Equation of Motion by Graphical Method
Derivation Of Equation Of Motion
The total distance travelled, S = Area of trapezium OABC.
So, S= 1/2(SumofParallelSides)×Height
S=(OA+CB)×OC
Since, OA = u, CB = v, and OC = t
The above equation becomes
S= 1/2(u+v)×t
Now, since t = (v – u)/ a
The above equation can be written as:
S= 1/2(u+v)×(v-u)/a
Rearranging the equation, we get
S= 1/2(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
It is known that,
Derivation Of Equation Of Motion
These were the detailed derivations for equations of motion in the graphical method, algebraic method and calculus method.