Question

Person $A$ walking along a road at $3m{s}^{-1}$ sees another person $B$ walking on another road at right angle to his road. Velocity of B is $4m{s}^{-1}$ when he is $10m$ off. They are nearest to each other when person A has covered a distance of

• A. $3.6$
  
• B. $3.7$
  
• C. $6.6$
  
• D. $8$
  

Answer 1

The correct option is A $3.6$

when $A$ is at origin with velocity $3m/s$ towards positive X.

and $B$ has velocity $4m/s$ in positive Y direction and initial position $(0,-10)$.

so position at any time $t$ for body $A$ is given by $(3t,0)$ and for body $B$ given by $(0,-10+4t)$

distance between them at any time $t$ is given by $S=\sqrt{(3t{)}^2+(-10+4t{)}^2}$

minimum value of $S$ Will when $S^2$ is minimum,

$S^2=(3t{)}^2+(-10+4t{)}^2$

so minimizing $S^2$ we get $\frac{d{S}^2}{dt}=6t+8(-10+4t)=0$

$=>t=\frac{40}{19}$

also $\frac{d(S^2{)}^2}{d{t}^2}=+ve$ so this value is minimum.

at this $t$, $S=3.6m$

so the answer is option A.