The persons start walking at a steady pace of 3 km/hour from a road 1 intersection along two roads that make an angle of 60 with each other. What will be the (shortest) distance separating them at the end of 20 minutes ?

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Submitted by examrobotsa on

Q: 140 (IAS/1994)
The persons start walking at a steady pace of 3 km/hour from a road 1 intersection along two roads that make an angle of 60° with each other. What will be the (shortest) distance separating them at the end of 20 minutes ?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,10,12,3,3,6,10

keywords: 

{'distance': [0, 3, 3, 3], 'intersection': [0, 1, 0, 1], 'km': [0, 0, 2, 1], 'angle': [0, 0, 1, 0], 'steady pace': [0, 1, 0, 0], 'minutes': [0, 0, 1, 1], 'road': [1, 2, 5, 6], 'roads': [0, 1, 0, 2], 'hour': [5, 5, 11, 12]}

This question is about understanding relative movement in different directions. Here, two persons start walking from the same point, and after 20 minutes, we have to find the shortest distance between them.

Option 1 suggests a 3 km separation. This would be correct if both individuals walked in the same direction, as they`re pacing at 3 km/hour. However, they`re walking in separate directions, making an angle of 60 degrees.

Option 2 proposes a distance of 2 km. This is incorrect for the same reasons as Option 1.

Option 3 puts forth the idea of a 1.5 km distance. Again, this logic only applies if they were walking on the same path or in opposite directions in a straight line, which is not the case here.

Option 4, the correct one, implies a distance of 1 km. In this scenario, the Triangle Law of Vectors applies where the two different directions of the individuals form a triangle. If you calculate, each person walks 1 km in 20 minutes (since 20 minutes is 1/3 of an hour and they walk at 3 km/hour). Using the formula for the separation in the vector law, which is √(a^2+b^