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
