Actually, s is not equal to ut 1 / 2at2.
s = ut + 1/2 at2
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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.
This is the second equation of motion. It has been derived here by the graphical method.
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