On a two lane road, car A is travelling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both 1km, B decided to overtake A before C does. What minimum acceleration (in m/s²) of car B is required so that B overtake A before C does?

At the instant when car B decides to overtake car A, the velocities of cars are;
$\begin{array}{l}{\mathrm{v}}_{\mathrm{A}}=36\times \frac{5}{18}=10\mathrm{m}/\mathrm{s}\\ {\mathrm{v}}_{\mathrm{B}}=54\times \frac{5}{18}=15\mathrm{m}/\mathrm{s}\\ \text{ And }{\mathrm{v}}_{\mathrm{C}}=-54\times \frac{5}{18}=-15\mathrm{m}/\mathrm{s}\end{array}$

Velocity of car C relative to A, v_CA = v_C - v_A = -15-10 = -25 m/s
Velocity of car B relative to A, v_BA = v_B - v_A = 15-10 = 5 m/s
Time the car C requires to just cross $\mathrm{A}=\frac{1000}{{\mathrm{v}}_{\mathrm{CA}}}=\frac{1000}{25}=40\mathrm{s}$
In order to avoid accident, car B must overtake A in this time, so
$\begin{array}{l}1000={\mathrm{u}}_{\mathrm{BA}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{BA}}{\mathrm{t}}^2\text{ or }1000=5\times 40+\frac{1}{2}{\mathrm{a}}_{\mathrm{BA}}\times {40}^2\\ \therefore {\mathrm{a}}_{\mathrm{BA}}=1\mathrm{m}/{\mathrm{s}}^2\end{array}$
Thus the minimum acceleration that car B requires to avoid an accident is 1m/s²