Q. Two cars leave one after the other and travel with and acceleration of $0.4ms^{- 2}$. Two minutes after the departure of the first car, the distance between them becomes 1.90 km. The time interval between their departures is
A) 50 s
B) 60 s
C) 70 s
D) 80 s
Explanation
(a) 1900$= \frac{1}{2} \times 0.4{\lbrack(120)}^{2} - (120 - t^{2})\rbrack$
or $\frac{1900}{0.2} = (240 - t)t$ or 9500=240$t - t^{2}$
or $t^{2} - 240t + 9500 = 0$
or $\ t^{2} - 190t - 50t + 9500 = 0$
or $t(t - 190) - 50(t - 190) = 0$
or $(t - 50)(t - 190) = 0\ \Rightarrow t = 50$s,190s
Rejecting $t = 190$ s, we get $t = 50$ s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.