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 15 km/h and reaches the same dock at 3:00 pm. at what time were the two boats closest together

Answer 1

make a diagram.
It took the eastbound ship 1 hr to reach the dock, so when the southbound ship was at the dock, the eastbound ship was 15 km from the dock

So after t hrs, the southbound ship went 20t
and the eastbound ship went 15t
I see a right-angled triangle with a vertical of 20t and a horizontal of 15-15t
let d be the distance between them
d^2 = (20t)^2 + (15-15t)^2
2d dd/dt = 2(20t)(20) + 2(15-15t)(-15)
at a minimum of d, dd/dt = 0
800t - 30(15-15t) = 0
800t - 450 + 450t = 0
1250t = 450
t = .36 hrs or 21.6 minutes
so they were closest at 2:21:36 pm

check:
when t = .36 , d^2 = 51.84 + 92.16 = 144, d = 12
take a value slightly higher and smaller
t = .37 , d^2 = 54.76 + 89.3025 = 144.0625 , d = 12.0026 , slightly farther
t = .35 , d^2 = 49 + 95.0625 = 144.0625 , d = 12.0026 , again slightly farther

my answer is correct

Answer 2

first, find the integral and divide by the derivative. Then, find the time component of the acceleration vector on the complex plane.

Answer 3

(-5)|36|

Answer 4

The answer is 14 protons.