Question

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

• A. $1 \, m / s ^{2}$
• B. $1.5\, m / s ^{2}$
• C. $2 \,m / s ^{2}$
• D. $3\, m / s ^{2}$

Answer 1

Answer: (B)

$v_{A}=36\, kmh ^{-1}=36 \times \frac{5}{18} ms ^{-1}=10\, ms ^{-1}$
$v_{B}=v_{C}=54\, km\, h ^{-1}=\frac{54 \times 5}{18} ms ^{-1}=15\, ms ^{-1}$
$s_{A B}=s_{A C}=1000 \,m$
$a_{A B}=a_{A}-a_{B}=0-a$
$s_{A B}=u_{A B} t+\frac{1}{2} a_{A B} t^{2}$
$1000\, m =(10-15) \frac{ m }{ s } \cdot t+\frac{1}{2}(0-a) t^{2}$
$\Rightarrow 1000=-5 t-\frac{1}{2} a t^{2}$
$\Rightarrow a=-\frac{2(1000+5 t)}{t^{2}}$
For car $C, s_{A C}=u_{A C} \cdot t$
$1000 \,m =\left(\frac{25\, m }{ s }\right) \cdot t,$
$ t=\left(\frac{1000}{25}\right)=40 s$
$\Rightarrow a=-\frac{2(1000+5(40))}{40^{2}}$
$=-\frac{2400}{1600}=-1.5 m / s ^{2}$