Question

On a two-lane road, car A is travelling at a speed of 36 km/hr. Two cars B and C approach car A in opposite directions with a speed of 54 km/hr 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 is a minimum acceleration of car B required to avoid an accident?

Answer 1

Velocity of car A, $V_A$ = 36 km/h = 10m/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 – 5 m/s
Relative velocity of car C with respect to car A,
$V_{CA}=V_C-(-V_A)$ = 15 + 10 = 25 m/s
At a certain instance, both cars B and C are at the same distance from car A i.e., 5 – 1 km = 1000m
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 40s.
From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:
$s=ut+\frac{1}{2}a{t}^2$
$1000=5\times 40+\frac{1}{2}\times a\times (40{)}^2$
$a=\frac{1600}{1600}=1m/s^2$