Question

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

• A. 9.8 m $s^{-2}$
• B. 10 m $s^{-2}$
• C. 1 m $s^{-2}$
• D. 2.0 m $s^{-2}$

Answer 1

The correct option is C 1 m $s^{-2}$

Speed of $A=36km/hr=36\times \frac{5}{18}m/s=10m/s$

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

Relative speed of $A$ w.r.t $C=10+15=25m/s$

Time taken by $C$ to overtake $A=\frac{1000}{25}$

$=40sec$.

Distance travelled by $A$ all this time $=10\times 40=400m$

So, B have to cover distance of $(1000m+400m)$

i.e. $1400m$ to take over a $A$ before $C$ does. in $40sec$.

Now putting in formula;-

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

$1400=15\times 40+\frac{1}{2}\times a\times (40{)}^2$

or, $800=a\times 800$

$a=1m/s^2$ Ans