Two cars leave one after the other and travel with an acceleration of 0.4 ms. Two minutes after the departure of the first car, the distance between them becomes 1.90 km. The time interval between their departures is :

Options

A. 50 s
B. 60 s
C. 70 s
D. 80 s

Answer

a

Explanation

Here distance between the two =9 km =1900m
Time =120 sec.
Then `1900=(1)/(2)xx0.4[(120)^(2)-(120-t)^(2)]`
or `(1900)/(0.2)=(240-t) t` or 9500=`240t-t^(2)`
or `t^(2)-240t+9500=0`
or `t^(2)-240t+9500=0`
or `t^(2)-190t-50t+9500=0`
or t(t-190)-50(t-190)=0
or (t-50)(t-190)=0
t=50 s 190 s
Rejecting t=190 s, we get t=50 s