`132`

Ta có:

`v_B^2-v_C^2=2a_[BC]s_[BC]`

`-> 0^2-v_C^2=2a_[BC]*s_[AB]/2`

`-> v_C=sqrt[-a_[BC]s_[AB]] (1)`

`v_B^2-v_A^2=2a_[AB]s_[AB]`

`-> 0^2-v_A^2=2a_[AB]s_[AB]`

`-> v_A=sqrt[-2a_[AB]s_[AB]] (2)`

Lại có `a_[AB]=a_[BC]`

`-> (1)`, `(2) -> v_C=[v_A]/[sqrt2]`

`v_B=v_A+a_[AB]t_[AB]`

`-> a_[AB]=a_[AC]=[-v_A]/[t_[AB]]`

Lại có:

`v_C=v_A+a_[AC]t`

`-> [v_A]/[sqrt2]=v_A+a_[AC]t`

`-> [v_A]/[sqrt2]=v_A-[v_A]/[t_[AB]]t`

`-> 1/[sqrt2]=1-1/[t_[AB]]t`

`-> t_[AB]=sqrt2(sqrt2+1)t`

`t_[BC]=t_[AB]-t=sqrt2(sqrt2+1)t-t=t+sqrt2t`

Vì mặt nghiêng nhẵn `-> t'=2t_[BC]=2(t+sqrt2t)=2(sqrt2+1)t`

`-> bbB`

`133`

Gọi `C` là nơi Bình giảm tốc

Ta có:

`v_B^2-v_C^2=2a_[BC]*s_[BC]`

`-> 0^2-v_C^2=2(-0.4)s_[BC]`

`-> v_C=sqrt[0.8s_[BC]]`

`v_C^2-v_A^2=2a_[AC]*s_[AC]`

`-> (sqrt[0.8s_[BC]])^2-0^2=2*0.4s_[AC]`

`-> 0.8s_[BC]=0.8s_[AC]`

`-> s_[BC]=s_[AC]=[s_[AB]]/2=90/2=45m`

`v_C=sqrt[0.8s_[BC]]=sqrt[0.8*45]=6m//s`

`t_[BC]=[v_B-v_C]/[a_[BC]]=[0-6]/[-0.4]=15s`

`t_[AC]=[v_C-v_A]/[a_[AC]]=[6-0]/[0.4]=15s`

`-> t=t_[BC]+t_[AC]=15+15=30s`

`-> bbB`