Question

Two ships are 60 km apart on North-South vertical at an instant as shown in the figure. The one farther North is streaming South at 30√3 km/hr and the other is streaming East at 30 km/hr. What is the time taken by the ships to reach the closest approach?

A
60 min
B
52 min
C
30 min
D
45 min

Solution

The correct option is D 52 minAs per the question taking everything in B frame we have −−→vAB=−→vA−−→vB So, we have from above equation |−−→vAB|=√(30√3)2+(−30)2=60 km/hr and, angle θ is given by θ=tan−1(3030√3)=tan−1(1√3)=30∘ Thus, the relative motion can be shown as So, from the above figure we can say that the relative distance moved by ship A to raech the closest approach is AC, which is given by AC=60cosθ=60cos30∘≈52 km Hence, time taken by the ships is given by t=52|−−→vAB|=5260=52 min

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