Derive the third equation of motion with the help of graph.
Note:- Irrelevant answers would be not allowed. Graph is compulsory. Give qualified answer.
Answers
Explanation:
Consider an object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time t = 0 and t = t are u and v respectively. Draw a line parallel to x-axis DA from point, D from where object starts moving
OD = u, OC = v and OE = DA = t
Let, the Initial velocity of the object = u
Let, the object is moving with uniform acceleration, a
Let, the object reaches at point B after time t, and its final velocity becomes v.
Draw a line parallel to x-axis DA from point, D from where object starts moving.
Draw another line BA from point B parallel to Y-axis which meets at E at y-axis.
Third Equation of Motion: The Distance covered by the object moving with uniform acceleration is given by the area of trapezium ABDOE.
∴ Area of trapezium ABDOE = ½ x (Sum of Parallel Slides + Distance between Parallel Slides)
Distance (s) = ½ (DO + BE) x OE = ½ (u + v) x t … (3)
Now from equation (2): a=v−ut,
∴ t=v−ua … 4
Now, substitute equation (4) in equation (3) we get:
s=12(u+v)×(v−ua),
s = ½a (v + u) (v – u)
2as = (v + u) (v – u)
2as = v² – u²
v² = u² + 2as