**Actually, s is not equal to ut 1 / 2at ^{2.}**

**s = ut + 1/2 at ^{2}**

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**2. Derive**Velocity–Time graph to derive the equations of motion. Suppose the body travels a distance

*s*=*ut*+ (1/2)*at*^{2}by Graphical Method*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)

*at*

^{2}...... (6) So, Distance travelled,

*s*= Area of rectangle

*OADC*+ Area of triangle

*ABD*or

*s = ut + (1/2) at*^{2}This is the second equation of motion. It has been derived here by the graphical method.

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