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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




Question Number 17936 by Rishabh#1 last updated on 12/Jul/17
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 s (D)55s
$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{start}\:\mathrm{moving}\:\mathrm{towards} \\ $$$$\mathrm{each}\:\mathrm{other}\:\mathrm{with}\:\mathrm{constant}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\:\mathrm{1}\:{m}/{s}^{\mathrm{2}} .\:\mathrm{If}\:\mathrm{their}\:\mathrm{initial}\:\mathrm{separation}\:\mathrm{is} \\ $$$$\mathrm{100}\:\mathrm{m},\:\mathrm{find}\:\mathrm{after}\:\mathrm{what}\:\mathrm{time}\:\mathrm{they}\:\mathrm{will} \\ $$$$\mathrm{meet}\:\mathrm{each}\:\mathrm{other} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{40}{s}\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{45}{s}\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{50}\:{s}\:\left(\mathrm{D}\right)\mathrm{55}{s} \\ $$
Commented by Tinkutara last updated on 12/Jul/17
It should be 10 s.
$$\mathrm{It}\:\mathrm{should}\:\mathrm{be}\:\mathrm{10}\:\mathrm{s}. \\ $$
Answered by wullyfalcon last updated on 12/Jul/17
(c)50
$$\left({c}\right)\mathrm{50} \\ $$$$ \\ $$
Answered by alex041103 last updated on 12/Jul/17
We know the formulas:  x(t)=x_0  + tv_0  + ((at^2 )/2)  so  100 = 1×t^2   0=(t+10)(t−10)  t_(1;2)  = 10 or −10  but t>0  ⇒t=10s
$${We}\:{know}\:{the}\:{formulas}: \\ $$$${x}\left({t}\right)={x}_{\mathrm{0}} \:+\:{tv}_{\mathrm{0}} \:+\:\frac{{at}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${so} \\ $$$$\mathrm{100}\:=\:\mathrm{1}×{t}^{\mathrm{2}} \\ $$$$\mathrm{0}=\left({t}+\mathrm{10}\right)\left({t}−\mathrm{10}\right) \\ $$$${t}_{\mathrm{1};\mathrm{2}} \:=\:\mathrm{10}\:{or}\:−\mathrm{10} \\ $$$${but}\:{t}>\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{10}{s} \\ $$

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