Multiple choice

A car, 'A' is moving towards south with a speed of 20 m/s. Another car 'B' is moving towards east with a speed of 15 m/s. At a certain instant, the car 'A' is due south of the car 'B' and is at distance of 50 m from car B. The shortest distance between the car 'A' & 'B'.

  1. 30 m

  2. 10 m

  3. 40 m

  4. 35 m
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A Correct answer
Explanation

Assuming the cars are moving towards each other along perpendicular paths, their relative distance squared at time t is given by D^2 = (15t)^2 + (50 - 20t)^2 = 625t^2 - 2000t + 2500. Minimizing this quadratic expression by setting its derivative to zero gives t = 1.6 seconds. Substituting t = 1.6 back into the distance formula yields a minimum distance of sqrt(900) = 30 meters.

AI explanation

Let car B be at the origin (0,0); then car A starts at (0, -50). After time t, car B is at (15t, 0) and car A is at (0, -50 + 20t). The distance between them is the square root of the sum of the squares of the differences in their coordinates. Squaring the distance gives S^2 = (15t)^2 + (-50 + 20t)^2, which we differentiate to find the minimum: 2(15t)(15) + 2(-50 + 20t)(20) = 0. Solving this gives t = 2 seconds, and substituting t back yields a distance of 30 m.

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