Sven starts walking due south at 5 feet per second from a po

Sven starts walking due south at 5 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 170 feet west of the intersection. (a) Write an expression for the distance d between Sven and Rudyard t seconds after they start walking. d = (b) When are Sven and Rudyard closest? t = s What is the minimum distance between them? d = ft

Solution

(a) we need to get distance in terms of t. Speed is distance over time, so distance is equal to speed multiplied by time, or d = v*t.

For Sven, d = 190 - 5t
For Rudyard, d = 170 - 4t

On our drawing, you see that Sven\'s path is simply the distance for the y-component, and Rudyard\'s the distance for the x-component. Because this is a right triangle, we can use the Pythagorean theorem. We see that the distance we are trying to calculate (between Sven and Rudyard) is the hypotenuse of the triangle.

The theorem states that a^2 + b^2 = c^2. Let\'s denote the distance for Sven as dy and the distance for Rudyard as dx. These are equal to a and b. The distance we want, let\'s set as c, since it is the hypotenuse.

So we get d^2 = (190 - 5t)^2 + (170 - 4t)^2
Thus, d = sqrt[(190 - 5t)^2 + (170 - 4t)^2]
d = sqrt[(41t^2 - 3260t + 65000)]

(b) Normally, we would set the expression found in part (a) equal to zero and solve for t. But Sven and Rudyard will clearly not meet at the intersection, so the distance will never be zero. But we know the local minimum or maximum for d will be when the derivative of d is equal to zero. Then solve for t.

derivative of d=
[(41t - 1630] / [sqrt(41t^2 - 3260t + 65000)]
derivative of d=0

therefore 41t-1630=0

41t=1630

t=1630/41

t=39.76s

(c) To get the distance, you take the time from part (b) and plug it into the equation in part (a).
This gave me d = 14.056 ft