Question

Sven starts walking due south at 7 feet per second from a point 190 feet north of an intersection. At the same time Rudyard starts walking due east at 4 feet per second from a point 130 feet west of the intersection.
A. Write an expression for the distance d between Sven and Rudyard t seconds after they start walking.
B. What is the minimum distance between them?
C. When are Sven and Rudyard closest?

Answer 1

Answer: A. $d=\sqrt{(190-7t)^2+(130-4t)^2}$

B. Minimum distance between them  = 18.61 feet.

C. After 28.76 seconds Sven and Rudyard are closest.

Step-by-step explanation:

A) Let (0,0) be the intersection point.

Then, Initial Location of Sven (0,190).

Speed of Sven = 7 feet per second

Then, position of Sven after t seconds = (0,190-7t)  [speed = distance x time]

Similarly, Initial position of Rudyard= (130,0)

Speed of Rudyard = 4 feet per second

Position after t seconds = (130-4t,0)

Distance d between Sven and Rudyard t seconds after they start walking:

$d=\sqrt{(190-7t)^2+(130-4t)^2}$

B) Let $d(t)=\sqrt{(190-7t)^2+(130-4t)^2}$

$d'(t)=2(190-7t)(-7)+(2)(130-4t)(-4)\\\\=130t-3700$

Put d'(t)=0

$130t-3700=0\\\\\Rightarrow\ t=\dfrac{3700}{130}\approx28.46\ sec$

Minimum distance :

$d(28.46)=\sqrt{(190-7(28.46))^2+(130-4(28.46))^2}\\\\=\sqrt{346.154}\approx18.61\ feet$

Hence, the minimum distance between them  = 18.61 feet.

c) After 28.76 seconds Sven and Rudyard are closest.