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Derivation of third equation of motion |
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Answer» u: velocity at time t1 v: velocity at time t2 a: uniform acceleration of the body ALONG the straight line Displacement covered during the time interval t2 - t1 = Area ABt1t2 S = Area of TRIANGLE ABA’ + Area of rectangle AA’t2t1 = Area of trapezium S = ½ x Sum of parallel sides x Perpendicular distance begin mathsize 12px style straight s equals 1 half open PARENTHESES straight v plus straight u close parentheses straight t end style ....... (Equation 1) From the first equation of motion, v = u + at; begin mathsize 12px style straight t equals fraction numerator straight v minus straight u over denominator straight a end fraction end style Substituting in equation 1, we get begin mathsize 12px style straight s equals 1 half open parentheses straight v plus straight u close parentheses fraction numerator open parentheses straight v minus straight u close parentheses over denominator straight a end fraction end style or, v2 – u2 = 2as which is the REQUIRED equation of motion. Explanation: mark as brainliest |
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