InterviewSolution
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InterviewSolution
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Deduce the equations of unifromly accelerated motion in one dimension by following calculus method. |
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Answer» By definition `a = (dv)/(dt)` `dv = a dt` Integrating both sides `int_(v_(0))^(v) dv = int_(0)^(t) a dt` `= a int_(0)^(t) dt` (a is constant) `v - v_(0) = at` `v = v_(0) + at` Further, `v = (dx)/(dt)` ltbr. `dx = v dt` Integrating both sides `int_(x_(0))^(x) dx = int_(0)^(t) v dt` `= int_(0)^(t) (v_(0) + at) dt` `x - x_(0) = v_(0) t + (1)/(2) a t^(2)` `x = x_(0) + v_(0) t + (1)/(2) a t^(2)` We can write `a = (dv)/(dt) = (dv)/(dx) (dx)/(dt) = v (dv)/(dx)` or, `v dv = a dx` Integrating both sides `int_(v_(0))^(v) vdv = int_(x_(0))^(x) a dx` `(v^(2) - v_(0)^(2))/(2) = a (x - x_(0))` `v^(2) = v_(0)^(2) + 2a (x - x_(0))` The advantage of this method is that it can be used for motion with non-uniform acceleration also. Now, we shall use these equations to some important cases. |
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