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

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

Commented by Tinkutara last updated on 12/Aug/17

Answered by ajfour last updated on 12/Aug/17

$${\mathrm{a}}_{\mathrm{P}}-{\mathrm{a}}_{\mathrm{R}}={2\mathrm{a}}_{\mathrm{Q}}$$$${\mathrm{a}}_{\mathrm{Q}}=\frac{1}{2}(12\mathrm{t}-3)=6\mathrm{t}-\frac{3}{2}$$$$\Rightarrow {\int }_0^{{\mathrm{v}}_{\mathrm{Q}}}{\mathrm{dv}}_{\mathrm{Q}}={\int }_0^{\mathrm{t}}(6\mathrm{t}-\frac{3}{2})\mathrm{dt}$$$${\mathrm{v}}_{\mathrm{Q}}={3\mathrm{t}}^2-\frac{3\mathrm{t}}{2}$$$$\Rightarrow {\mathrm{v}}_{\mathrm{Q}}=0\mathrm{at}\mathrm{t}=0,\frac{1}{2}\mathrm{s}.$$$$\mathrm{Q}\mathrm{comes}\mathrm{to}\mathrm{rest}\mathrm{again}\mathrm{at}\mathbf{t}=\frac{1}{2}\mathrm{s}.$$

Commented by Tinkutara last updated on 12/Aug/17

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

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