Question

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.

Answer 1

Answer:

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