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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
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 time with velocity 6 ms^(−1) and at a constant acceleration of 3 ms^(−2) . Find the time when the: (a) particles are 30 metres apart; (b) particles meet; (c) velocity of P is (3/4) os the velocity of Q.
$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{P}\:\mathrm{and}\:\mathrm{Q}\:\mathrm{move}\:\mathrm{towards}\:\mathrm{each} \\ $$$$\mathrm{other}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:{MN},\:\mathrm{51}\:\mathrm{meters} \\ $$$$\mathrm{long}.\:\mathrm{P}\:\mathrm{starts}\:\mathrm{from}{M}\:\mathrm{with}\:\mathrm{velocity}\:\mathrm{5}\:\mathrm{ms}^{−\mathrm{1}} \\ $$$$\mathrm{and}\:\mathrm{constant}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{1}\:\mathrm{ms}^{−\mathrm{2}} .\:\mathrm{Q}\:\mathrm{starts} \\ $$$$\mathrm{from}\:\mathrm{N}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{time}\:\mathrm{with}\:\mathrm{velocity}\:\mathrm{6}\:\mathrm{ms}^{−\mathrm{1}} \\ $$$$\mathrm{and}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{3}\:\mathrm{ms}^{−\mathrm{2}} . \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{time}\:\mathrm{when}\:\mathrm{the}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{particles}\:\mathrm{are}\:\mathrm{30}\:\mathrm{metres}\:\mathrm{apart}; \\ $$$$\left(\mathrm{b}\right)\:\mathrm{particles}\:\mathrm{meet}; \\ $$$$\left(\mathrm{c}\right)\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{P}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{os}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{Q}. \\ $$
Answered by MJS last updated on 11/May/19
P v_P (t)=t+5 s_P (t)=(1/2)t^2 +5t Q v_Q (t)=3t+6 s_Q (t)=−(3/2)t^2 −6t+51 (a) ∣s_P −s_Q ∣=30 ∣2t^2 +11t−51∣=30 ∧ t≥0 ⇒ t=(3/2)∨t=−((11)/4)+((√(769))/4)≈4.183 but the 2^(nd) value might be after the crash (b) s_P =s_Q (1/2)t^2 +5t=−(3/2)t^2 −6t+51 ∧t≥0 ⇒ t=3 (c) (v_P /v_Q )=(3/4) 4v_P =3v_Q 4t+20=9t+18 ⇒ t=(2/5)
$${P} \\ $$$${v}_{{P}} \left({t}\right)={t}+\mathrm{5} \\ $$$${s}_{{P}} \left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\mathrm{5}{t} \\ $$$${Q} \\ $$$${v}_{{Q}} \left({t}\right)=\mathrm{3}{t}+\mathrm{6} \\ $$$${s}_{{Q}} \left({t}\right)=−\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{51} \\ $$$$ \\ $$$$\left({a}\right) \\ $$$$\mid{s}_{{P}} −{s}_{{Q}} \mid=\mathrm{30} \\ $$$$\mid\mathrm{2}{t}^{\mathrm{2}} +\mathrm{11}{t}−\mathrm{51}\mid=\mathrm{30}\:\wedge\:{t}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:{t}=\frac{\mathrm{3}}{\mathrm{2}}\vee{t}=−\frac{\mathrm{11}}{\mathrm{4}}+\frac{\sqrt{\mathrm{769}}}{\mathrm{4}}\approx\mathrm{4}.\mathrm{183} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{value}\:\mathrm{might}\:\mathrm{be}\:\mathrm{after}\:\mathrm{the}\:\mathrm{crash} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$${s}_{{P}} ={s}_{{Q}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\mathrm{5}{t}=−\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{51}\:\wedge{t}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:{t}=\mathrm{3} \\ $$$$ \\ $$$$\left({c}\right) \\ $$$$\frac{{v}_{{P}} }{{v}_{{Q}} }=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{4}{v}_{{P}} =\mathrm{3}{v}_{{Q}} \\ $$$$\mathrm{4}{t}+\mathrm{20}=\mathrm{9}{t}+\mathrm{18} \\ $$$$\Rightarrow\:{t}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$
Commented by pete last updated on 11/May/19
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by ajfour last updated on 11/May/19
velocity of P w.r.t. Q be v displacement of P w.r.t. Q be s v=11+4t s=11t+2t^2 2t^2 +11t−21=0 ⇒ t=((−11+(√(121+169)))/4) = ((−11+(√(290)))/4) s ≈ 1.5s 2t^2 +11t−51=0 ⇒ t=((−11+(√(121+408)))/4) = 3s 5+t=(3/4)(6+3t) ⇒ 20+4t=18+9t t=(2/5)s .
$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{P}\:\mathrm{w}.\mathrm{r}.\mathrm{t}.\:\mathrm{Q}\:\mathrm{be}\:\mathrm{v} \\ $$$$\mathrm{displacement}\:\mathrm{of}\:\mathrm{P}\:\mathrm{w}.\mathrm{r}.\mathrm{t}.\:\mathrm{Q}\:\mathrm{be}\:\mathrm{s} \\ $$$$\:\:\:\:\:\mathrm{v}=\mathrm{11}+\mathrm{4t} \\ $$$$\:\:\:\:\:\mathrm{s}=\mathrm{11t}+\mathrm{2t}^{\mathrm{2}} \\ $$$$\mathrm{2t}^{\mathrm{2}} +\mathrm{11t}−\mathrm{21}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{t}=\frac{−\mathrm{11}+\sqrt{\mathrm{121}+\mathrm{169}}}{\mathrm{4}}\:=\:\frac{−\mathrm{11}+\sqrt{\mathrm{290}}}{\mathrm{4}}\:\mathrm{s} \\ $$$$\:\:\approx\:\mathrm{1}.\mathrm{5s} \\ $$$$\mathrm{2t}^{\mathrm{2}} +\mathrm{11t}−\mathrm{51}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{t}=\frac{−\mathrm{11}+\sqrt{\mathrm{121}+\mathrm{408}}}{\mathrm{4}}\:=\:\mathrm{3s} \\ $$$$\mathrm{5}+\mathrm{t}=\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{6}+\mathrm{3t}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{20}+\mathrm{4t}=\mathrm{18}+\mathrm{9t} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{t}=\frac{\mathrm{2}}{\mathrm{5}}\mathrm{s}\:. \\ $$
Commented by pete last updated on 11/May/19
Thank you very much sir
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{sir} \\ $$

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