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

Answer 1

Step 1: Given data

Speed of $A$ $=36km{h}^{-1}=36\times \frac{5}{18}m{s}^{-1}=10m{s}^{-1}$

Speed of $B$= Speed of $C$ $=54\times \frac{5}{18}m{s}^{-1}=15m{s}^{-1}$

Step 2: Find the time taken by $C$ to overtake $A$.

With respect to $C$, the relative speed of $A$ $=10+15=25m{s}^{-1}$

So, the time taken by $C$ to overtake $A$$=\frac{1000}{25}=40sec$.

Distance traveled by $A$ during the time C overtook $A=10\times 40=400m$

So, the distance $B$ has to cover $=(1000m+400m)$ i.e. $1400m$, to take over $A$ before $C$ does in 40 sec.

Step 3: Find the minimum acceleration required by $B$.

Using the $2^{nd}$equation of motion, $s=ut+\frac{1}{2}a{t}^2$

Putting the values, we get

$$\begin{gathered} 1400=15\times 40+\frac{1}{2}\times a\times {\left(40\right)}^2 \\ \Rightarrow 800=a\times 800 \\ \Rightarrow a=1m{s}^{-2} \end{gathered}$$

Final Answer: The minimum acceleration required by car $B$ to avoid an accident is $1m{s}^{-2}$.