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. 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: (A)

1 m $s^{-2}$

Velocity of car A $=+36km/h=36\times \frac{1000}{3600}=10m/s$

Velocity of car B $=54\times \frac{5}{18}m/s=+15m/s$

Velocity of car C $=-54km/h=-15m/s$

In the frame of reference of car $A:$

velocity of B with respect to A

$V_{BA}=V_B-V_A=15-(+10)$

$V_{BA}=5m/s$

Velocity of C with respect to A

$V_{CA}=V_C-V_A=-15-(+10)$

$V_{CA}=-25m/s$

Therefore time taken by C to reach A

$t=\frac{1000}{25}=40s$(Calculate in the frame of car A)

To avoid accident B will trave same relative distance in 40s

Applying $s=ut+\frac{1}{2}a{t}^{2}1000=5\times 40\times \frac{1}{2}{a}_B\times (40{)}^2{a}_B=1m/s$

Option - A