Bunuel
Andrew starts moving from A to B, a distance of 200 km, at a speed of 40 km/hr. At the same time, Bob starts moving from B at a speed of 20 Km/hr along a road which is perpendicular to AB. In how many hours the distance between Andrew and Bob will be the least?
A. 1.5 hr
B. 2 hr
C. 3 hr
D. 4 hr
E. 5 hr
Let t be the number of hours each has traveled. So the distance between Andrew and point B is 200 - 40t km (say, west of point B) and the distance between Bob and point B is 20t km (say, north of point B). Point B and the positions of Andrew and Bob form the vertices of a right triangle, where B is the right angle. Letting d = the distance between Andrew and Bob t hours after they started, by the Pythagorean theorem we have:
d^2 = (200 - 40t)^2 + (20t)^2
d^2 = 40,000 - 16,000t + 1600t^2 + 400t^2
d^2 = 2000t^2 - 16,000t + 40,000
We see that the right hand side of the equation is a quadratic of at^2 + bt + c, which is a parabola. The value of a is positive, which means the parabola is up-opening and thus must attain a minimum at its vertex. Since the x-value of the vertex (in this case, the t-value) has the formula x = -b/(2a) (in this case, t = -b/(2a)), the distance between Andrew and Bob will be the least when t = -(-16,000)/(2(2000)) = 16,000/4000 = 4.
Answer: D