JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Kinematics

Ship A is sailing towards north - east with velocity

v = 30 \hat{i} + 50 \hat{j} km / hr

where

\hat{i}

points east and

\hat{j}

, north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at

10 km / hr

. A will be at minimum distance from B ins:

Answer: (C)

Solution

Work in A's frame. The position of B relative to A is

r = 80 \hat{i} + 150 \hat{j}

km, and the relative velocity is

v = v_{B} - v_{A} = (- 10 \hat{i}) - (30 \hat{i} + 50 \hat{j}) = - 40 \hat{i} - 50 \hat{j}

km/hr. The separation is least when the relative position is perpendicular to the relative velocity, at

t = - \frac{r \cdot v}{∣ v ∣ ^{2}} = - \frac{( 80 ) ( - 40 ) + ( 150 ) ( - 50 )}{4 0 ^{2} + 5 0 ^{2}} = \frac{3200 + 7500}{4100} = \frac{10700}{4100} \approx 2.6

hr.

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option