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Obtain equations of motion for constant acceleration using method of calculus. |
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Answer» SOLUTION :By definition `a=(d upsilon)/(dt)` `d upsilon = a dt` INTEGRATING both SIDES `int_(upsilon_(0))^(upsilon)d upsilon = int_(0)^(t) a dt` `= a int_(0)^(t) dt ""` (a is constant) `upsilon - upsilon_(0)=at` `upsilon=upsilon_(0)+at` Further,`upsilon = (dx)/(dt)` `dx = upsilon dt` Integrating both sides `int_(x_(0))^(x)dx=int_(0)^(t)upsilon dt` `=int_(0)^(t)(upsilon_(0)+at)dt` `x-x_(0)=upsilon_(0)t+(1)/(2)at^(2)` `x=x_(0)+upsilon_(0)t+(1)/(2)at^(2)` We can write `a=(d upsilon)/(dt)=(d upsilon)/(dx)(dx)/(dt)=upsilon (d upsilon)/(dx)` or `upsilon d upsilon = a dx` Integrating both sides, `int_(upsilon_(0))^(upsilon)upsilon d upsilon = int_(x_(0))^(x)a dx` `(upsilon^(2)-upsilon_(0)^(2))/(2)=a(x-x_(0))` `upsilon^(2)=upsilon_(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. t |
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