# Derive s = ut + 1/2gt^2

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The formula s= ut + 1/2gt^2

Represents the total distance "s" travelled by an object in time "t" which has an initial speed of "u" and is accelerating at the uniform rate of "g". It is customary to use the symbol "g" for acceleration due to gravity. However this equation is valid for any value of uniform acceleration.

We can derive the equation as follows.

The total distance travelled by a uniformly accelerating object is equal to average speed multiplied by time

That is: s = (average speed)*t ... (1)

And, average speed = (initial speed + final speed at the end of period t)/2

And final speed = u + gt

Therefore: Average speed = (u + u + gt) = 2u + gt

Substituting this value of average speed in equation (1) we get:

s = (2u + gt)/2]*t = ut +1/2gt^2

Let u be the initial velocity of an object.

Under constant acceleration g, the velocity of the object after time t is u+at.

Let us represent this in a graph. Let t be on the X axis and velocity be shown along the y axis.

Mark the origin, (0,0) as O.

At time 0, velocity of the object is u. Mark the (0,u) and let this point be A on Y axis.

At time t, the velocity is u+gt because of constant acceleration g.Mark the point B. Also mark the point (t , 0) on X axis as P.

Join the line PB

Draw a parallel line to X axis from A to meet PB at A'.

Now see that the displacement of the object due to the variable velocity is represented by the area under the line AB and and x axis and enclosed between the ordinates OA and PB.

Under uniform velocity the object would have made a displacement equal to the area OAA'P = (ordinate length u)(OP)= ut.

Under the effect of acceleration it has an additional displacement represented by the area of the triangle AA'B= (1/2)(AA')(A'P)

=(1/2)(OP)(PB-PA')=(1/2)(t)(gt)=(1/2)gt^2.

Thus the displacemet, s of a body with an initial velocity u and constant acceleration g, in time t is given by s= ut+(1/2)gt^2. Proved geometrically.

Hope this helps.

s=ut+1/2gt^2