derive an equation of motion using graphical method (all these equation)
Answers
Explanation:
Equation of motion by graphical method.
Derivation of v=u +at. Initial velocity u at A =OA. ...
Derivation of S = ut +1/2 x at 2 The distance travelled by the body is given by area of the space between velocity time graph AB and time axis OC , which is equal to area of figure OABC. ...
Derivation of v2 = u2 + 2as.
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.
Derivation of Equation of Motion
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 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 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
