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. The minimum required acceleration of car $B$ to avoid an accident is

• A. $1ms{}^{-2}$
• B. $1.5ms{}^{-2}$
• C. $2ms{}^{-2}$
• D. $3ms{}^{-2}$

Answer 1

Answer: (A)

Velocity of car $A$ ,
$v_A=36km{h}^{-1}=36\times \frac{5}{18}m{s}^{-1}=10m{s}^{-1}$
Velocity of car $B$ ,
$v_B=54km{h}^{-1}=54\times \frac{5}{18}m{s}^{-1}=15m{s}^{-1}$
Velocity of car $C$ ,
$v_C=-54km{h}^{-1}$
$=-54\times \frac{5}{18}m{s}^{-1}=-15m{s}^{-1}$
Relative velocity of car $B$ w.r.t. car $A$
$v_{BA}=v_B-v_A=15m{s}^{-1}-10m{s}^{-1}=5m{s}^{-1}$
Relative velocity of car $C$ w.r.t. car $A$ is
$v_{CA}=v_C-v_A=-15m{s}^{-1}-10m{s}^{-1}=-25m{s}^{-1}$
At a certain instant, both cars $B$ and $C$ are at the same distance from car $A$
i.e. $AB-BC=1km=1000m$
Time taken by car $C$ to cover $1km$ to reach car $A$
$=\frac{1000m}{25m{s}^{-1}}=40s$
In order to avoid an accident, the car $B$ accelerates such that it overtakes car $A$ in less than $40s$ . Let the minimum required acceleration be a. Then,
$u=5m{s}^{-1}$ , $t=40s$ , $S=1000m$ , $a=?$
As $S=ut+\frac{1}{2}a{t}^2$
$\therefore 1000=5\times 40+\frac{1}{2}\times a\times 40^2$
$800a=1000-200=800$ or $a=1m{s}^{-2}$