derive the formola[s=ut + 1/2at^2]?

Answer 1

How detailed a derivation do you want?
Best way is with calculus:
start with the definition of acceleration
a = ( v - u ) / t

and rearrange to get v = u + at

now integrate with respect to time over the limits 0 to t

integral(v)dt = s
integral(u + at)dt = ut + 1/2at^2

hence s = ut + 1/2at^2

The alternative is to do it graphically:
(Integration has the same effect as finding the area under a graph)

Draw v and t axes. Then draw a straight line that intercepts the v axis at some non-zero point.
Label the point of intersection u (initial speed)
Now draw a vertical line down at some time t.
You will have formed a trapezium.
Divide the area into a square and a triangle.
You should already know that finding the area under a speed-time graph gives the distance travelled.
Area of square = ut
Area of triangle = 1/2t*(v-u)
Now substitute for (v - u) using the definition of acceleration from above. This gives Area of triangle = 1/2at^2

Therefore total distance travelled = s = ut + 1/2at^2

Hope this helps.

Answer 2

There is an unfproof of this formula. The best explanation is derived by Physics theorist Milton W. Monson in a publication called "Physics is constipated"which explains some of the misconceptions about The world of physics.

Answer 3

draw one of those velocity time graphs, starting at a velociy u, then with constant acceleration (y=mx type).

So the area under the graph is the area of a trapezium
1/2. (u+v).t
but this is the distance travelled
s= (u+v).t / 2
v is the speed attained by the acceleration.
but v= u+ at (from the gradient of the graph)
so
s= (u+u+at).t / 2
s= (2ut + at^2)/2
s= ut + 1/2 at^2