The distance coverd by an object between times ` t_(1) and t_(2)` is given by the area under the ` v-t` gaph between ` t_(1) and t_(2)` Prove this staement for an object moving with negative acceleration and gaving and having a positive velocity at time ` t_(1) and and negative velocity at time `t_(2)`.

Explanation

The velocity-time graph for the motion of object as per question is shown in Fig. 2 (b) . 28 . Let ` v_(1) be the velocity of object at time ` T_(1) and -v_(2) ` be the velocity of object at time ` t_(2)`.
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Let (S) be the distance travelled in between timings ` t_(1) and t_(2)`. Let `C` be a point on the straight line ` AB`, where the graph cuts the time axis and (t') be the time corresponding to point ` C`.
Area under the ` u-t`grapg b between timings `t_(1)` and `t_(2)` is
A =area ` AA_1 C + area CBB_1`
`= 1/2 A_1 C xx AA_1 + 1/2 CB_1 xx BB_1`
`= 1/2 (t' -t') xx v-1 + 1/2 (_2 -t') xx (-v_2)`
`= 1/2 v_1 (t' -t_1) - 1/2 v-2 (t _2 - t')` ...(i)
We know that the acceleration (a) of an object is equal to the slope of ` v-t` graph. Taking slope the ` v-t` graph betweem the points ` A and C` , we have
` a= (velocity at t'-velocity at t_1)/(t'-t_1)`
`= (o-v_1)/(t'-t') =- v_1/(t'-t_1) or t' -t =- v_1/a`
Slope of ` v-t` graph between the points ` B and C`, we have
` a= (velocity at t'-velocity at t_')/(t_2-t')`
` = (-v_2)-0)/9t_2 -t' ) =- v_2/(t_2 -t') or t_2 -t' =- v_2/a`
Pittomg tje va,ies in (i), we get
` A= 1/2 v_1 (- v_1/a) - 1/2 v_2 (- v_2/a)`
`=- v_1^2/(2 a) + v_2^2/(2 a) = (v_2^2-v_1^2)/92 a) = (2 a S) /(2 a) =S`
Hence area under ` v-t` graph between timings ` t_1 and t_2` is equal to the distance coverd by object between timings ` t_1 and t-2`.