Two ships are 80km apart on North-South vertical at an instant as shown in the figure. The one farther North is streaming South at 40km/hr and the other is streaming East at 30km/hr. What is their distance of closest approach?
A
48km
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B
32km
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C
60km
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D
52km
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Solution
The correct option is A48km As per the question taking everything in B frame we have −−→vAB=−→vA−−→vB
So, we have from above equation |−−→vAB|=√(40)2+(−30)2=50km/hr
and, angle θ is given by θ=tan−1(3040)=tan−1(34)=37∘
Thus, the relative motion can be shown as
So, from the above figure we can say that the distance of closest approach is dshort=80sinθ=80sin37∘=48km