Two straight tracks AOB and COD intersect at right angles at O. A person walking at 5 km h⁻¹ along AOB is at O at 12:00 noon. Another person walking at 5 km h⁻¹ along COD reaches O at 1:30 PM. Which option gives the time of closest approach and minimum distance?
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Question

Two straight tracks AOB and COD intersect at right angles at O. A person walking at 5 km h⁻¹ along AOB is at O at 12:00 noon. Another person walking at 5 km h⁻¹ along COD reaches O at 1:30 PM. Which option gives the time of closest approach and minimum distance?

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G1 Physics · Accepted Answer

At 12:00, the second person is 1.5 h away from O: OC =5×1.5=7.5 km =15/2 km. Relative velocity of one person with respect to the other has equal perpendicular components 5 km h⁻¹, so it is along a line at 45°. The minimum separation is the perpendicular distance from O to the relative-motion line: dₘᵢₙ = OC sin45° = (15/2)(1/√2) =15/(2√2) km. Relative speed =5√2 km h⁻¹. Time to closest approach: t = [OC cos45°]/(5√2) = 3 4 \frac{3}{4} 4 3 ​ h. Thus the closest approach occurs at 12:45 PM.