Consider the velocity-time graph of a body shown in the Figure. The body has an initial velocity u at the point and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a Uniform acceleration a from A to B and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C and draw AD parallel to OC. BE is the perpendicular from point B to OE.
1. To Derive v = u + at by Graphical Method:-
→
The initial velocity of the body, u = OA
The final velocity of the body, v = BC
BC = BD+DC → v = BD + AO → v = BD + u
We should find out the value of BD now. We know that the slope of a velocity-time graph is equal to acceleration, a.
→ a = BD/AD
→ a = BD/t
→ BD = at
Now, putting this value of BD we get:
v = at + u → v = u + at
2. To Derive s = ut + ½ at² by Graphical method:-
→
The distance travelled by the body is given by the area of the space between the velocity-time graph AB and the time axis OC, which is equal to the area of the figure OABC. Thus:
→ Distance = Area of OABC
→ Distance = Area of rectangle OADC + area of triangle ABD
→ Distance = (OA x OC) + (1/2 x AD x BD)
here, OA = u, OC = t, AD = 0C = t and BD = at
→ Distance, s = (u x t) + ( 1/2 x t x at )
→ s = ut + 1/2 x at² → s = ut + 1/2 at²
3. To Derive v² = u² + 2as by Graphical Method
→
The distance traveled s by the body in time t is given by the area of the figure OABC, which is a trapezium.
→ Distance s = Area of trapezium OABC
→ Distance s = ½ x (Sum of parallel sides) x Height
→ s = ½ x (OA + CB) x OC
here, OA = u and CB = v i.e. OA+CB = u + v, OC = t
→ s = ½ (v+u) x t
We now want to eliminate t from the above
equation.
v = u +at → (v-u)/a = t (First equation of Motion)
Now, putting this value → s = (v + u) x (v - u) / 2 x a
→ 2as = v² - u² [because (v + u) (v - u) = v² - u²]
→ v² = u² + 2as
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