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 $1km,B$ decides to overtake $A$ before $C$ does. What minimum acceleration of car $B$ is required to avoid an accident?

Answer 1

Suppose car A, B, and C are moving with velocity

$V_a,V_b,and{V}_c,$

let the posotive direction of X axis is taken as motion of car A Car B. now

$V_a=36km/hr{V}_b=54km/hrand{V}_c=-54km/hr$

relative velocity of car B with car A

$V_{ba}=V_b-V_a=54-36=18km/hr$

relative velocity of car with car A

$V_{ca}=V_c-V_a=-54-36=-90km/hr$

negative sign shows Vca is in -ve x direction

now suppose car C takes time t to cover distance BA of 1 km in slightly less then (1/90)h at a speed of $V_{ba}=18km/hr$

Substituting u=18km/hr t= 1/90h s = 1 km

$s=ut+\frac{1}{2}a{t}^2$

$1000=5\times 40+\frac{1}{2}a{\left(40\right)}^2$

800a=1000 - 200

a = 800/800

$a=1m/s^2$ (Ans.)