You are a lifeguard and spot a drowning child 30 meters along the shore and 60 meters from the shore to the child. You run along the shore and for a while and then jump into the water and swim from there directly to child. You can run at a rate of 5 meters per second and swim at a rate of 1 meter per second. How far along the shore should you run before jumping into the water in order to save the child? Round your answer to three decimal places.

The lifeguard should run approximately 17.752 meters along the shore, before, jumping in the water

Explanation:

The given parameters are;

The rate at which the lifeguard runs = 5 m/s

The rate at which the lifeguard swims = 1 m/s

The horizontal distance of the child from the lifeguard = 30 meters along the shore

The vertical distance of the child from the lifeguard = 60 meters along the shore

Let x represent the distance the lifeguard runs

We have;

The distance the lifeguard swims = √((30 - x)² + 60²)

Time = Distance/Speed  

The time the lifeguard runs = x/5

The time the lifeguard swims = √((30 - x)² + 60²)/1

The total time = √((30 - x)² + 60²) + x/5

The minimum time is given by finding the derivative and equating the result to zero, as follows;

Using an online application, we have;

d(√((30 - x)² + 60²) + x/5)/dx = 1/5 - (30 - x)/(√((30 - x)² + 60²)) = 0

Which gives;

1/5 - (30 - x)/(√(x² - 60·x + 4500) = 0

(30 - x)/(√(x² - 60·x + 4500)) = 1/5

5×(30 - x) = √(x² - 60·x + 4500)

We square both sides to get;

(5×(30 - x))² = (x² - 60·x + 4500)

(5×(30 - x))² - (x² - 60·x + 4500) = 0

25·x² - 1500·x + 22500 - x² + 60·x - 4500 = 0

24·x² - 1440·x + 18000 = 0

Dividing n=by 24 gives;

24/24·x² - 1440/24·x + 18000/24 = 0

x² - 60·x + 750 = 0

By the quadratic formula, we have;

x = (60 ± √((-60)² - 4×1×750))/(2 × 1) =

Using an online application, we have;

x = (60 ± 10·√6)/(2)

x = 30 + 5·√6 or x = 30 - 5·√6

x ≈ 42.25 m and x ≈ 17.752 m

At x = 42.25

Time = √((30 - 42.247)² + 60²) + 42.247/5 ≈ 69.69 seconds

At x = 17.75

Time = √((30 - 17.752)² + 60²) + 17.752/5 ≈ 64.79 seconds

Therefore, the route with the shortest time is when the lifeguard runs approximately 17.752 meters (rounded to three decimal places) along the shore, before, diving in the water

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14