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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 ........$s$
$190$
$50$
$80$
$60$
Solution
Distance covered by First Car in $2$ Minutes
$\mathrm{S}=\mathrm{ut}+\frac 12 \mathrm{at}^{2}$
$\mathrm{u}=0$ from rest $\mathrm{a}=0.4 \mathrm{m} / \mathrm{s}^{2}\;;\; \mathrm{t}=2 \mathrm{mins}=120 \mathrm{sec}$
$S=0+\frac 12(0.4)(120)^{2}$
$\Rightarrow \mathrm{S}=2880 \mathrm{m}$
$\Rightarrow \mathrm{S}=2.88 \mathrm{km}$
distance between the cars $=1.9 \mathrm{km}$
Other car travelled $=2.88-1.9=0.98 \mathrm{km}=980 \mathrm{m}$
Time taken by another car to travel $980 \mathrm{m}$
$980=\frac 12(0.4) t^{2}$
$\Rightarrow t=70 \mathrm{sec}$
Another car started after $120-70=50$ sec $=\frac 56 \mathrm{min}$
time
interval between the departure of the cars is $=\frac 56 \mathrm{min}=50\;sec$
