Question

Car A is moving with a speed of 36 km/h on a two-lane road. Two cars B and C, each moving with a speed of 54 km/h in opposite directions on the other lane are approaching car A. At a certain instant when the distance AB = distance AC = 1 km, the driver of car B decides to overtake A before C does. What must be the minimum acceleration of car B so as to avoid an accident?

• A. $1m{s}^{-2}$
• B. $2m{s}^{-2}$
• C. $3m{s}^{-2}$
• D. $4m{s}^{-2}$

Answer 1

The correct option is A

$1m{s}^{-2}$

Let us suppose that cars A and B are moving in the positive x-direction and car C is moving in the negative x-direction.

Now, $v_A=+36km{h}^{-1}=+10m{s}^{-1},v_B=+54km{h}^{-1}=+15m{s}^{-1}and{v}_C=-54km{h}^{-1}=-15m{s}^{-1}$
The relative velocity B with respect to A is $v_{BA}=v_B-v_A=15-10=5m{s}^{-1}.$
The relative velocity of C with respect to A is $v_{CA}=v_C-v_A=-15-10=-25m{s}^{-1}.$
At time t = 0, the distance between A and B = distance between A and C = 1 km = 1000 m.
The car C will cover a distance AC = 1000 m and just reach car A at a time t given by $t=\frac{AC}{|v_{CA}|}=\frac{1000m}{25m{s}^{-1}}=40s$
Car B will overtake car A just before car C does and avoid an accident, if it acquires a minimum acceleration a such that it covers a distance s = AB = 1000 m in time t = 40 s, travelling at a relative speed $u=v_{BA}=5m{s}^{-1}$. Putting these values in the second equation of motion, $s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow 1000=5\times 40+\frac{1}{2}\times a\times (40{)}^2$
$\Rightarrow a=1m{s}^{-2}$
Hence, the correct choice is (a).