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Example 3.3 Obtain equations of motionfor constant acceleration using method ofcalculus |
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Answer» We know , acceleration is the change in velocity per unit time .e.g an instant time ; a = dv/dt a = dv/dt a∫dt = ∫dv now, integrate a.(t₂ - t₁) = v - u Let (t₂- t₁) = t then, at = v - u v = u + at ___ its a motion equation , _______________________________ now , v = u + at multiply both sides, with dt v.dt = u.dt + at.dt we know, v.dt = S { distance = speed * time } use this S = u(t₂-t₁) + a(t₂-t₁)²/2 hence, S = ut + 1/2at² ___motion, equation.______________________________now, S = ut + 1/2at² ___(1) v = u + atsquaring both sides, v² = u² + a²t² + 2uat v² = u² + 2a{ 1/2at² + ut } from eqn (1) v² = u² + 2aS ____motion equation, __________________________________or we know, v.dv/dx = a ∫v.dv = a∫dx integrate both sides, v² - u² = 2a( x₂ - x₁) Let ( x₂ - x₁) = S v² = u² + 2aS___motion equation |
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