Two-particles-start-moving-towards-each-other-with-constant-acceleration-of-1-m-s-2-If-their-initial-separation-is-100-m-find-after-what-time-they-will-meet-each-other-A-40s-B-45s-C-50

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Question Number 17936 by Rishabh#1 last updated on 12/Jul/17

$$\mathrm{Two}\mathrm{particles}\mathrm{start}\mathrm{moving}\mathrm{towards}$$$$\mathrm{each}\mathrm{other}\mathrm{with}\mathrm{constant}\mathrm{acceleration}$$$$\mathrm{of}1m/s^2.\mathrm{If}\mathrm{their}\mathrm{initial}\mathrm{separation}\mathrm{is}$$$$100\mathrm{m},\mathrm{find}\mathrm{after}\mathrm{what}\mathrm{time}\mathrm{they}\mathrm{will}$$$$\mathrm{meet}\mathrm{each}\mathrm{other}$$$$\left(\mathrm{A}\right)40s\left(\mathrm{B}\right)45s\left(\mathrm{C}\right)50s\left(\mathrm{D}\right)55s$$

Commented by Tinkutara last updated on 12/Jul/17

$$\mathrm{It}\mathrm{should}\mathrm{be}10\mathrm{s}.$$

Answered by wullyfalcon last updated on 12/Jul/17

$$\left(c\right)50$$$$$$

Answered by alex041103 last updated on 12/Jul/17

$$Weknowtheformulas:$$$$x\left(t\right)=x_0+{tv}_0+\frac{{at}^2}{2}$$$$so$$$$100=1\times t^2$$$$0=(t+10)(t-10)$$$$t_{1;2}=10or-10$$$$butt>0$$$$\Rightarrow t=10s$$

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