4this question has several parts that must be completed sequentially. if you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. tutorial exercisea boat leaves a dock at 8: 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 9: 00 pm how many minutes after 8: 00 pm were the two boats closest together

at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B215km/h>Dock

|

|

B1 at 20km/h

|

|

V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

change tension

f is prop to m1xm2 double m1 double force