4. At noon, ship A is 200km east of ship B and ship A is sailing north at 30km/h. Ten minutes later, ship B starts to sail south at 35 km/h.

Question

4. At noon, ship A is 200km east of ship B and ship A is sailing north at 30km/h. Ten minutes later, ship B starts to sail south at 35 km/h.  a. What is the distance between the two ships at 3 pm?  b. How fast (in km/h) are the ships moving apart at 3 pm?

Answer 1

I hope you have a diagram for this already.

 

So at 3 pm, ship A has travelled for 3 hours but ship B has travelled for 2 5/6 hours.

So ship A has travelled 90 km north and ship B has travelled 35(17/6)=99 1/6 km south.

 

You can now draw a big right triangle. The horizontal distance would still be 200 km and the vertical distance would be 189 1/6 km. You can then yse the Pythagorean theorem to fin out how far apart they are. 

 

200² + (189 1/6)² = d² and d = 275.29 km ( rounded to the nearest 100th)

 

b) So I don't know if you're doing this as a related rates problem or as a single function and just a derivative.

 

So the 3 sides of the triangle can be represented as 200, 30x+35(x-1/6), and D.

 

So D² = 40000+ (65x-35/6)²

So D = [40000 + (65x-35/6)² ] ^1/2

and D' = 1/2[ 40000+(65x-35/6)² ]^-1/2 * 2(65x-35/6)(65)

 

now just plug in 3 to find D'(3) = 24591.66666 / 550.58 = 44.67 km/hr

 

if you did this with related rates, you get

a² + b² = c²   so    2a da/dt + 2b db/dt = 2c dc/dt which you can divide by 2 and get a da/dt + b db/dt = c dc/dt

a = 200, b = 189 1/6, c= 275.29

a isn't changing so da/dt = 0

b can be represented by 30t + 35(t-1/6) so db/dt = 30 + 35 = 65.

 

So plug in everything to get

200(0) + (189 1/6)(65) = 275.29 dc/dt solve and get dc/dt = 12295.8333/275.29=44.67 km/hr.  Same amswer!!

 

Hope this helped.