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

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

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

Answered by mrW1 last updated on 17/Nov/17

$$velocityofparticle1:$$$$v_1\left(t\right)=\frac{{dx}_1}{dt}=2\times 6t+3=12t+3$$$$velocityofparticle2:$$$$v_2\left(t\right)=\int a_2\left(t\right)dt=\int (-8t)dt=-4t^2+C$$$$since{v}_2=20att=0,\Rightarrow C=20$$$$\Rightarrow v_2\left(t\right)=-4t^2+20$$$$$$$$when{v}_1=v_2$$$$\Rightarrow 12t+3=-4t^2+20$$$$\Rightarrow 4t^2+12t-17=0$$$$\Rightarrow t=\frac{-12+\sqrt{{12}^2+4\times 4\times 17}}{2\times 4}\approx 1.05s$$$$i.e.att=1.05stheirvelocitiesmatch.$$$$v_1=v_2\approx 12\times 1.05+3=15.6m/s$$

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

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

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