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Question Number 24387 by chernoaguero@gmail.com last updated on 17/Nov/17

Two particle move along an x−axis. The position of particle 1 is given; by x=6.00t^2 +3.00t+2.00((m/s)) the acceleration of particle 2 is given by a=−8.00t((m/s^2 )) and,at t=0,its velocity is 20 ((m/s)).when the velocities of the particles match, what is their velocity? plzz help

$$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{particle}}\:\boldsymbol{\mathrm{move}}\:\boldsymbol{\mathrm{along}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{axis}}. \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{position}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{particle}}\:\mathrm{1}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}}; \\ $$$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{x}}=\mathrm{6}.\mathrm{00}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{3}.\mathrm{00}\boldsymbol{\mathrm{t}}+\mathrm{2}.\mathrm{00}\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}}\right) \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{acceleration}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{particle}}\:\mathrm{2}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}} \\ $$$$\boldsymbol{\mathrm{by}}\:\:\boldsymbol{\mathrm{a}}=−\mathrm{8}.\mathrm{00}\boldsymbol{\mathrm{t}}\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}^{\mathrm{2}} }\right)\:\boldsymbol{\mathrm{and}},\mathrm{at}\:\mathrm{t}=\mathrm{0},\boldsymbol{\mathrm{its}} \\ $$$$\boldsymbol{\mathrm{velocity}}\:\boldsymbol{\mathrm{is}}\:\mathrm{20}\:\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}}\right).\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{velocities}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{particles}}\:\boldsymbol{\mathrm{match}}, \\ $$$$\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{their}}\:\boldsymbol{\mathrm{velocity}}? \\ $$$$\boldsymbol{\mathrm{plzz}}\:\boldsymbol{\mathrm{help}} \\ $$

Answered by mrW1 last updated on 17/Nov/17

velocity of particle 1: v_1 (t)=(dx_1 /dt)=2×6t+3=12t+3 velocity of particle 2: v_2 (t)=∫a_2 (t)dt=∫(−8t)dt=−4t^2 +C since v_2 =20 at t=0, ⇒C=20 ⇒v_2 (t)=−4t^2 +20 when v_1 =v_2 ⇒12t+3=−4t^2 +20 ⇒4t^2 +12t−17=0 ⇒t=((−12+(√(12^2 +4×4×17)))/(2×4))≈1.05 s i.e. at t=1.05 s their velocities match. v_1 =v_2 ≈12×1.05+3=15.6 m/s

$${velocity}\:{of}\:{particle}\:\mathrm{1}: \\ $$$${v}_{\mathrm{1}} \left({t}\right)=\frac{{dx}_{\mathrm{1}} }{{dt}}=\mathrm{2}×\mathrm{6}{t}+\mathrm{3}=\mathrm{12}{t}+\mathrm{3} \\ $$$${velocity}\:{of}\:{particle}\:\mathrm{2}: \\ $$$${v}_{\mathrm{2}} \left({t}\right)=\int{a}_{\mathrm{2}} \left({t}\right){dt}=\int\left(−\mathrm{8}{t}\right){dt}=−\mathrm{4}{t}^{\mathrm{2}} +{C} \\ $$$${since}\:{v}_{\mathrm{2}} =\mathrm{20}\:{at}\:{t}=\mathrm{0},\:\Rightarrow{C}=\mathrm{20} \\ $$$$\Rightarrow{v}_{\mathrm{2}} \left({t}\right)=−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{20} \\ $$$$ \\ $$$${when}\:{v}_{\mathrm{1}} ={v}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{12}{t}+\mathrm{3}=−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{20} \\ $$$$\Rightarrow\mathrm{4}{t}^{\mathrm{2}} +\mathrm{12}{t}−\mathrm{17}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{−\mathrm{12}+\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{4}×\mathrm{4}×\mathrm{17}}}{\mathrm{2}×\mathrm{4}}\approx\mathrm{1}.\mathrm{05}\:{s} \\ $$$${i}.{e}.\:{at}\:{t}=\mathrm{1}.\mathrm{05}\:{s}\:{their}\:{velocities}\:{match}. \\ $$$${v}_{\mathrm{1}} ={v}_{\mathrm{2}} \approx\mathrm{12}×\mathrm{1}.\mathrm{05}+\mathrm{3}=\mathrm{15}.\mathrm{6}\:{m}/{s} \\ $$

Commented by chernoaguero@gmail.com last updated on 17/Nov/17

Thank you sir vry much

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{vry}\:\mathrm{much} \\ $$

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