Blocks-P-and-R-starts-from-rest-and-moves-to-the-right-with-acceleration-a-P-12t-m-s-2-and-a-R-3-m-s-2-Here-t-is-in-seconds-The-time-when-block-Q-again-comes-to-rest-is-

Question Number 19509 by Tinkutara last updated on 12/Aug/17

$$\mathrm{Blocks}\:{P}\:\mathrm{and}\:{R}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{and} \\ $$$$\mathrm{moves}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{with}\:\mathrm{acceleration} \\ $$$${a}_{{P}} \:=\:\mathrm{12}{t}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{and}\:{a}_{{R}} \:=\:\mathrm{3}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{Here}\:{t} \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{seconds}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{when}\:\mathrm{block}\:{Q} \\ $$$$\mathrm{again}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{is} \\ $$

Commented by Tinkutara last updated on 12/Aug/17

Answered by ajfour last updated on 12/Aug/17

$$\mathrm{a}_{\mathrm{P}} −\mathrm{a}_{\mathrm{R}} =\mathrm{2a}_{\mathrm{Q}} \\ $$$$\mathrm{a}_{\mathrm{Q}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{12t}−\mathrm{3}\right)=\mathrm{6t}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\int_{\mathrm{0}} ^{\:\:\mathrm{v}_{\mathrm{Q}} } \mathrm{dv}_{\mathrm{Q}} \:=\:\int_{\mathrm{0}} ^{\:\:\mathrm{t}} \left(\mathrm{6t}−\frac{\mathrm{3}}{\mathrm{2}}\right)\mathrm{dt} \\ $$$$\:\:\:\:\mathrm{v}_{\mathrm{Q}} =\:\mathrm{3t}^{\mathrm{2}} −\frac{\mathrm{3t}}{\mathrm{2}}\: \\ $$$$\Rightarrow\:\:\:\mathrm{v}_{\mathrm{Q}} =\mathrm{0}\:\:\:\mathrm{at}\:\mathrm{t}=\mathrm{0},\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{s}. \\ $$$$\mathrm{Q}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{again}\:\mathrm{at}\:\boldsymbol{\mathrm{t}}=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{s}. \\ $$

Commented by Tinkutara last updated on 12/Aug/17

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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