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(B) Let the time interval between the departures of the two cars be $\Delta t$ seconds.Let the first car travel for $t_1 = 120\, s$ ($2$ minutes).The

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(B) Let the time interval between the departures of the two cars be $\Delta t$ seconds.Let the first car travel for $t_1 = 120\, s$ ($2$ minutes).The distance covered by the first car is $S_1 = \frac{1}{2} a t_1^2 = \frac{1}{2} 	imes 0.4 	imes (120)^2 = 0.2 	imes 14400 = 2880\, m$.The second car starts $\Delta t$ seconds later,so its travel time is $t_2 = (120 - \Delta t)\, s$.The distance covered by the second car is $S_2 = \frac{1}{2} a t_2^2 = \frac{1}{2} 	imes 0.4 	imes (120 - \Delta t)^2 = 0.2(120 - \Delta t)^2$.The distance between the cars is $S_1 - S_2 = 1.9\, km = 1900\, m$.Substituting the values: $2880 - 0.2(120 - \Delta t)^2 = 1900$.$0.2(120 - \Delta t)^2 = 2880 - 1900 = 980$.$(120 - \Delta t)^2 = \frac{980}{0.2} = 4900$.Taking the square root: $120 - \Delta t = 70$.$\Delta t = 120 - 70 = 50\, s$.

Two cars leave one after the other and travel with an acceleration of $0.4\, m/s^2$. Two minutes after the departure of the first,the distance between the cars becomes $1.9\, km$. The time interval between the departures of the cars is ........$s$.

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