# Derive the equation : s=ut 1/2at2

Actually,  s  is  not  equal  to  ut 1 / 2at2.

s  =  ut  +  1/2  at2

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v=u+at

s=ut+ 1/2 at2

s=1/2 (u+v)t

v2 = u2 + 2as

s= vt- 1/2 at2

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derive s=ut+1/2at2
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s=ut+1/2at2
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2. Derive s = ut + (1/2) at2 by Graphical Method Velocity–Time graph to derive the equations of motion. Suppose the body travels a distance s in time t. In the above Figure, the distance travelled by the body is given by the area of the space between the velocity – time graph AB and the time axis OC,which is equal to the area of the figure OABC. Thus: Distance travelled = Area of figure OABC   = Area of rectangle OADC + Area of triangle ABD We will now find out the area of the rectangle OADC and the area of the triangle ABD. (i) Area of rectangle OADC = OA × OC   = u × t   = ut ...... (5) (ii) Area of triangle ABD = (1/2) × Area of rectangle AEBD   = (1/2) × AD × BD   = (1/2) × t × at (because AD = t and BD = at)   = (1/2) at2...... (6) So, Distance travelled, s = Area of rectangle OADC + Area of triangle ABD or s = ut + (1/2) at2
This is the second equation of motion. It has been derived here by the graphical method.
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I'm Sorry but I'm not able to put the figure in the message....Still hope that my derivation helps...
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How it solve
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derive s = ut +half at 2

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ut +
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ut+half at square derive it kkk
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s= speed x time
=velocity x time
=average velocity x   time
=( v+u)  x  t
(2)
=from the first equation of motion we know that v= u+at
=(u+at + u) x t
(2)
= (2u+at)t
(2)
=2ut + a(tsquare)
(2)
= 2ut  +  at(square)
​     2            2
= ut + 1atsquare
2

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S=(u+u+at)/2*t
s=(2u+at)/2*t
S=(2ut+at^2)/2
s=ut+1/2at^2
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distance travelled= area of figure OABC
s= area of rectangle OAD+ area of triangle ABD