Question

A car driving north at 40km/h reaches an intersection at the same time that a second hand car traveling west toward the intersection at 60km/h is 100km directly east of intersection.How long it take for the distance between the cars to be minimum?

Answer 1

at 1.15hr or 1hr09min

Explanation:

Let us denote the north driving car in terms of y and west driving car in terms of x and the intersection as the origin and time, in hours, as t.
so we have the position of car Y as Y = 40t
the position of car X = 100-(60t)
at any time, the distance between the 2 cars is (X^2+Y^2)^0.5

so we have the distance as a function of time as
D = (1600tt + 10000 + 3600tt - 12000t)^0.5
D = (5200tt - 12000t + 10000)^0.5

to minimize the function we have to differentiate it w.r.t t and equate it to 0
dD/dt = (25200t - 12000)/(2(5200tt - 12000t + 10000)^0.5)=0
so 10400*t - 12000 = 0
t = 12000/10400 = 1.15 hr
i.e. 1 hr 09 min