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Two buses leave with a minute gap and move with acceleration of 0.2 ms - How long after the departure of the second bus do the distance hew them become equal to three times its initial value? |
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Answer» Solution :Initial distance between two BUSES is `s_1 =(1)/(2)at^(2)=(1)/(2)0.2xx(60)^(2)=360 m`. Let after t SECOND from the departure of second bus the disatance between two buses becomes `s_1-s_2=3xx360=1080` m `s_1=(1)/(2)a(t+60)^(2)=(1)/(2)xx0.2(t+60)^(2)` `s_2 =(1)/(2)(0.2)(t)^(2)` On solving, we GET t=60s =1 min |
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