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Write graphical derivation of v=u +at |
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Answer» y has an INITIAL velocity u at a 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 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. Now, Initial velocity of the body, u=OA ...(1) And, Final velocity of the body, v=BC ...(2) But from the graph BC=BD+DCTherefore, v=BD+DC ...(3) Again DC=OASo, v=BD+OANow, from equation (1), OA=uSo, v=BD+u ...(4) We should find out the value of BD now. We know the slope of a velocity-time graph is equal to the acceleration, a. Thus, Acceleration, a= slope of line ABor a= ADBD But AD=OC=t, so PUTTING t in place of AD in the above relation, we get: a= tBD or BD=atNow, putting this value of BD in equation(4), we get: v=u+atsolution |
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