Two-particles-P-and-Q-move-towards-each-other-along-a-straight-line-MN-51-meters-long-P-starts-fromM-with-velocity-5-ms-1-and-constant-acceleration-of-1-ms-2-Q-starts-from-N-at-the-same-tim

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Question Number 59505 by pete last updated on 11/May/19

$$\mathrm{Two}\mathrm{particles}\mathrm{P}\mathrm{and}\mathrm{Q}\mathrm{move}\mathrm{towards}\mathrm{each}$$$$\mathrm{other}\mathrm{along}\mathrm{a}\mathrm{straight}\mathrm{line}MN,51\mathrm{meters}$$$$\mathrm{long}.\mathrm{P}\mathrm{starts}\mathrm{from}M\mathrm{with}\mathrm{velocity}5{\mathrm{ms}}^{-1}$$$$\mathrm{and}\mathrm{constant}\mathrm{acceleration}\mathrm{of}1{\mathrm{ms}}^{-2}.\mathrm{Q}\mathrm{starts}$$$$\mathrm{from}\mathrm{N}\mathrm{at}\mathrm{the}\mathrm{same}\mathrm{time}\mathrm{with}\mathrm{velocity}6{\mathrm{ms}}^{-1}$$$$\mathrm{and}\mathrm{at}\mathrm{a}\mathrm{constant}\mathrm{acceleration}\mathrm{of}3{\mathrm{ms}}^{-2}.$$$$\mathrm{Find}\mathrm{the}\mathrm{time}\mathrm{when}\mathrm{the}:$$$$\left(\mathrm{a}\right)\mathrm{particles}\mathrm{are}30\mathrm{metres}\mathrm{apart};$$$$\left(\mathrm{b}\right)\mathrm{particles}\mathrm{meet};$$$$\left(\mathrm{c}\right)\mathrm{velocity}\mathrm{of}\mathrm{P}\mathrm{is}\frac{3}{4}\mathrm{os}\mathrm{the}\mathrm{velocity}\mathrm{of}\mathrm{Q}.$$

Answered by MJS last updated on 11/May/19

$$P$$$$v_P\left(t\right)=t+5$$$$s_P\left(t\right)=\frac{1}{2}{t}^2+5t$$$$Q$$$$v_Q\left(t\right)=3t+6$$$$s_Q\left(t\right)=-\frac{3}{2}{t}^2-6t+51$$$$$$$$\left(a\right)$$$$\mid s_P-s_Q\mid =30$$$$\mid 2t^2+11t-51\mid =30\wedge t⩾0$$$$\Rightarrow t=\frac{3}{2}\vee t=-\frac{11}{4}+\frac{\sqrt{769}}{4}\approx 4.183$$$$\mathrm{but}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{value}\mathrm{might}\mathrm{be}\mathrm{after}\mathrm{the}\mathrm{crash}$$$$$$$$\left(b\right)$$$$s_P=s_Q$$$$\frac{1}{2}{t}^2+5t=-\frac{3}{2}{t}^2-6t+51\wedge t⩾0$$$$\Rightarrow t=3$$$$$$$$\left(c\right)$$$$\frac{v_P}{v_Q}=\frac{3}{4}$$$$4v_P=3v_Q$$$$4t+20=9t+18$$$$\Rightarrow t=\frac{2}{5}$$

Commented by pete last updated on 11/May/19

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by ajfour last updated on 11/May/19

$$\mathrm{velocity}\mathrm{of}\mathrm{P}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{Q}\mathrm{be}\mathrm{v}$$$$\mathrm{displacement}\mathrm{of}\mathrm{P}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{Q}\mathrm{be}\mathrm{s}$$$$\mathrm{v}=11+4\mathrm{t}$$$$\mathrm{s}=11\mathrm{t}+{2\mathrm{t}}^2$$$${2\mathrm{t}}^2+11\mathrm{t}-21=0$$$$\Rightarrow \mathrm{t}=\frac{-11+\sqrt{121+169}}{4}=\frac{-11+\sqrt{290}}{4}\mathrm{s}$$$$\approx 1.5\mathrm{s}$$$${2\mathrm{t}}^2+11\mathrm{t}-51=0$$$$\Rightarrow \mathrm{t}=\frac{-11+\sqrt{121+408}}{4}=3\mathrm{s}$$$$5+\mathrm{t}=\frac{3}{4}(6+3\mathrm{t})$$$$\Rightarrow 20+4\mathrm{t}=18+9\mathrm{t}$$$$\mathrm{t}=\frac{2}{5}\mathrm{s}.$$

Commented by pete last updated on 11/May/19

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}$$

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