Homework SourseA ship leaving port at 10am travels north with a speed of 10
A ship leaving port at 10am travels north with a speed of 10 mi/hr. Another ship, leaving the same port at 12 noon travels east with a speed of 20 mi/hr. How fast is the distance between the two ships is increasing at 4:00PM?
Thank you!
Thank you!
Solution
let the distance travelled by first ship =y and second ship = x after time t hours, y = 10t, at4PM, y = 60mi x = 20(t-2), at 4PM, x = 80mi distance between them = s x^2 + y^2 = s^2 2xdx/dt + 2ydy/dt = 2sds/dt 2*80*20 + 2*60*10 = 2*100*ds/dt ds/dt = 16+6 = 22mi/hr
