Prove that s=ut+1/2 at²
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Answered by
185
Distance travelled = Area under the line
= ut + ½ (v-u)t
Acceleration (a) = (v-u)/t and so (v-u) = at
Therefore,
Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²
Thus,proved.
= ut + ½ (v-u)t
Acceleration (a) = (v-u)/t and so (v-u) = at
Therefore,
Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²
Thus,proved.
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Answered by
475
acceleration is defined as change in velocity per unit time.
=> (v - u) / (t - 0) = a => v = u + at
Distance traveled is = average velocity * time duration
![S = \frac{(initial + final velocity)}{2} * duration \\ \\ S = \frac{u + v}{2} * t \\ \\ S = \frac{1}{2} (u + u + a t ) * t \\ \\ S = \frac{2ut + a t^2}{2} \\ \\ S = u t + \frac{1}{2} a t^2 \\ \\ S = \frac{(initial + final velocity)}{2} * duration \\ \\ S = \frac{u + v}{2} * t \\ \\ S = \frac{1}{2} (u + u + a t ) * t \\ \\ S = \frac{2ut + a t^2}{2} \\ \\ S = u t + \frac{1}{2} a t^2 \\ \\](https://tex.z-dn.net/?f=+S++%3D+%5Cfrac%7B%28initial+%2B+final+velocity%29%7D%7B2%7D+%2A+duration+%5C%5C+%5C%5C+S+%3D+%5Cfrac%7Bu+%2B+v%7D%7B2%7D+%2A+t%C2%A0++%5C%5C+%5C%5C+S+%3D+%5Cfrac%7B1%7D%7B2%7D+%28u+%2B+u+%2B+a+t+%29+%2A+t+%5C%5C+%5C%5C+S+%3D+%5Cfrac%7B2ut+%2B+a+t%5E2%7D%7B2%7D+%5C%5C+%5C%5C+S+%3D+u+t+%2B+%5Cfrac%7B1%7D%7B2%7D+a+t%5E2+%5C%5C+%5C%5C+)
=> (v - u) / (t - 0) = a => v = u + at
Distance traveled is = average velocity * time duration
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