A thief snatched a lady's purse and began running north towards a bus that was 200 meters away. A policeman, who was standing 40 meters south of the lady, noticed this and immediately sent his dog after the thief. The dog caught the thief 2 seconds before reaching the bus. If the dog's speed was 25 m/s, determine the speed of the thief.

Let the speed of the thief be 'x' m/s. Time required by the thief to catch the bus = (200/x) seconds Time taken by the dog to catch the thief = {(200/x) - 2} seconds ATQ; {40/(25 - x)} = {(200/x) - 2} Or, 40x = (200 - 2x) X (25 - x) Or, 20x = (100 - x) X (25 - x) Or, 20x = 2500 - 100x - 25x + x² Or, x² - 145x + 2500 = 0 Or, x² - 125x - 20x + 2500 = 0 Or, x(x - 125) - 20(x - 125) = 0 Or, (x - 125)(x - 20) = 0 So, x = 125 or x = 20 But since the speed of the thief cannot be more than the speed of the dog So, x = 20.