Question

On a two-lane road, car A is travelling with a speed of $36km{h}^{-1}$. Two cars B and C approach car A in opposite directions with a speed of $54km{h}^{-1}$ each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Answer 1

Velocity of car A, $v_A$ = 36 km/h = 10 m/s
Velocity of car B, $v_B$ = 54 km/h = 15 m/s
Velocity of car C, $v_C$ = 54 km/h = 15 m/s
Relative velocity of car B with respect to car A, $v_{BA}=v_B–v_A=15–10=5m/s$
Relative velocity of car C with respect to car A, $v_{CA}=v_C–(–v_A)=15+10=25m/s$
At a certain instance, both cars B and C are at the same distance from car A i.e.,
s = 1 km = 1000 m
Time taken (t) by car C to cover $1000m=\frac{1000}{25}=40s$
Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.
From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
$s=ut+\frac{1}{2}a{t}^{2}1000=5\times 40+\frac{1}{2}\times a\times (40{)}^{2}a=\frac{1600}{1600}=1m/s^2$