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?

Options

A. 1 min
B. 2 min
C. `1/2` min
D. `3/2` min

Answer

a

Explanation

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