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 $54km{h}^{-1}$. At a certain instant, when the distance of $AB$ 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. $1m/s^2$
• B. $1.5m/s^2$
• C. $2m/s^2$
• D. $3m/s^2$

Answer 1

Answer: (B)

$v_A=36km{h}^{-1}=36\times \frac{5}{18}m{s}^{-1}=10m{s}^{-1}$
$v_B=v_C=54km{h}^{-1}=\frac{54\times 5}{18}m{s}^{-1}=15m{s}^{-1}$
$s_{AB}=s_{AC}=1000m$
$a_{AB}=a_A-a_B=0-a$
$s_{AB}=u_{AB}t+\frac{1}{2}{a}_{AB}{t}^2$
$1000m=(10-15)\frac{m}{s}\cdot t+\frac{1}{2}(0-a)t^2$
$\Rightarrow 1000=-5t-\frac{1}{2}a{t}^2$
$\Rightarrow a=-\frac{2(1000+5t)}{t^2}$
For car $C,s_{AC}=u_{AC}\cdot t$
$1000m=\left(\frac{25m}{s}\right)\cdot t,$
$t=\left(\frac{1000}{25}\right)=40s$
$\Rightarrow a=-\frac{2(1000+5(40))}{40^2}$
$=-\frac{2400}{1600}=-1.5m/s^2$