Derive the equation s= ut +1/2at^2
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Answers
s = ut + ½ at²
So far we have the equations: v = u + at and s = ½ (u+v) t
The second can be useful for finding displacement. However, if we don't know the final velocity, it's useless.
We could first work out v from "v = u + at". But let's be radical and substitute "v = u + at" into the second equation!
s = ½ (u + v) × t
s = ½ (u + u + at) × t
s = ut + ½ at²
A third equation, useful for finding displacement when you know the initial velocity, time and acceleration.
Hope you understand
Hi I tried to derive the distance traveled by a body at contact acceleration from the definition of acceleration (increase in speed every sec), but the ended with a different result. Can you see what I am doing wrong. u = initial speed t = time taken S = {distance in 1st sec} + {distance in 2nd sec } + {distance in 3rd sec) + ............... + {distance in t sec} S = {u } + {u + a} + {u + 2a } + ................... + {u + (t - 1)a} S = u * t + {a + 2a + .........+ (t - 1)a } S = ut + a( 1 + 2 + 3 ........+ (t-1)) S = ut + a * t * (t-1/2)
The distance moved is based on average velocity during each period, not the final velocity at the end of each time period: S = {u + (1/2)a } + {u + (3/2)a} + {u + (5/2)a } + ... + {u + ((2t-1)/2)a}
= s= ut+1/2at^2