Answer :

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) *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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