Question

Sven starts walking due south at 7 feet per second from a point 140 feet north of an intersection. At the same time Rudyard starts walking due east at5 feet per second from a point 160 feet west of the intersection.(a) Write an expression for the distance d between Sven and Rudyard t seconds after they start walking. (b) When are Sven and Rudyard closest? (Round your answer to two decimal places.)(c) What is the minimum distance between them? (Round your answer to two decimal places.)

Answer 1

a) d = sqrt24t^2 - 360t + 27560

b) 20 seconds

c) 40ft

Step-by-step explanation:Sven walks 7ft/s . His distance is decreasing ,

Sven= 140 - 7t

Rudyard's distance is 160 - 5t

Using Pythagorean theorem to find d

d^2 = (160 - 5t)^2 + (140 - 7t)^2

d^2 = 25600 - 1600t - 25t^2 + 19600+ 1960t + 49t^2

d^2 = 24t^2 - 360t + 27560

d = sqrt 24t^2 - 360t + 27560

b) 140 - 5t = 0

t = 140/7 = 20seconds

c) 160 - 5t = Rudyard's distance

160 - 5(20) = 160- 100 = 40feet