Question

Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of 1 m/s and 2 m/s and speeds 3 m/s and 1 m/s respectively. Initially A is 10 m behind B. What is the minimum distance between them?

• A. 4 m
  
• B. 8 m
  
• C. 12 m
  
• D. 16 m
  

Answer 1

The correct option is B 8 m

Acceleration of A w.r.t B $a_{AB}=a_A-A_B=1-2=-1m/s^2$

Initial velocity of A w.r.t B $u_{AB}=u_A-u_B=3-1=2m/s$

$\therefore$ Distance traveled by A w.r.t B, $S_{AB}=u_{AB}t+\frac{1}{2}{a}_{AB}{t}^2$

OR $S_{AB}=2t-\frac{t^2}{2}$

For maximum distance traveled $\frac{d{S}_{AB}}{dt}=0$

$\therefore$ $2-t=0$ $⟹t=2$ s

Thus maximum distance traveled $S_{AB}=2\left(2\right)-\frac{2^2}{2}=2$ m

$⟹$ Minimum distance between the cars $D_{min}=10-2=8$ m