Tutorial Exercise A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15km/h and reaches the same dock at 3:00 PM How many minutes after 2:00 PM were the two boats closest together

Both the boats will closet together at 2:21:36 pm.

Step-by-step explanation:

Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).

Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),

        Formula :   d=v*t

at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)

the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)

      Formula : D=

⇒                    

Now let          

                ∵    

⇒                     t= 450/1250

⇒                     t= .36 hours

⇒                       = 21 min 36 sec

Since F"(t)=0,

∴ This time gives us a minimum.

Thus, The two boats will closet together at 2:21:36 pm.

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

b

step-by-step explanation: