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Prove the formula S=ut+1|2(at sq), graphically.

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To Derive (Displacement-time equation) s=u t+\frac{1}{2} a t^{2}$ by Graphical Method:-

  • Consider the below velocity-time graph of a body. The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 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. Here draw the perpendicular CB from point C, and draw AD parallel to OC. BE is perpendicular from point B to OE.

          

 

The area under the velocity-time graph is equal to the displacement. In the time interval 0-t, the displacement s is equal to the area OABE.

s= area OABE= area of the rectangle OACE + area of the triangle ABC

{\begin{aligned} &s=(O A) \cdot(O E)+\frac{1}{2}(A C) \cdot(B C)\\ &=(O A) \cdot(O E)+\frac{1}{2}(O E) \cdot\left(\frac{B C}{A C} \cdot A C\right)\\ &=(O A) \cdot(O E)+\frac{1}{2}(O E) \cdot\left(\frac{B C}{A C} \cdot O E\right)\\ &=(O A) \cdot(O E)+\frac{1}{2}\left(\frac{B C}{A C}\right)(O E)^{2} \end{aligned}}

Now, OA=u, OE=t and BC/AC=slope=a

s=u t+\frac{1}{2} a t^{2}$

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avinash.dongre

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