s=ut+(1)/(2)at^2toa

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s=ut+(1)/(2)at^2toa

Answer 1

\(a=[2(s-u t)] / t^{2}\)

Answer 2

The given equation \(s=ut+\frac{1}{2}at^2\) is a common formula in physics to describe the motion of an object with a constant acceleration. However, here we don't want the formula itself, instead we are trying to isolate and solve it for acceleration (a). Physically, this represents specifying the finesse or intensity with which an object is speeding up or slowing down. So, rather than a reformulation or rearrangement of the known equations of motion, what's wanted here is a demonstration of the necessary algebraic steps to convert this equation into a formula for the acceleration 'a' when given the object's initial speed 'u', the time of travel 't', and the displacement 's' it experiences. We get this equation by rearranging the terms as follows: Subtract 'ut' from both sides of the equation: \(s - ut = 1/2 * a * t^2\). Multiply both sides of the equation by '2': \(2(s - ut) = a * t^2\). Finally, divide both sides of the equation by 't^2' to solve for acceleration 'a': \(a = [2(s - ut)] / t^2\). Therefore, the solution, expressed as an equation for 'a', will be: \(a=[2(s-u t)] / t^{2}\) It follows that if you know the object’s initial velocity (u), final position (s), and amount of time it moved (t), you could figure out the acceleration experienced during that time.

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s=ut+(1)/(2)at^2toa

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