Correct Answer - B
(2) `A` to `B: v^(2)=2 alphas_(1)impliess_(1)=v^(2)/(2alpha)`
`B` to `C: 0=v^(2)-2beta`
`s_(2)impliess_(2)=v^(2)/(2beta)`
`s_(1)+s_(2)=v^(2)/2(1/alpha+1/beta)`
`=1/2((alphabetat)/(alpha+beta))^(2)((beta+alpha)/(alphabeta))`
`=(alphabetat^(2))/(2(alpha+beta))`
OR
`alpha=v/t_(1), beta=v/(t-t_(1))`, solving `v=(alphabetat)/(alpha+beta)`
Area of v-t graph gives displacement.
`s=s_(1)+s_(2)=1/2vt=(alphabetat^(2))/(2(alpha+beta))`
