How to Find the Rate of Change in Distance for a Ship's Journey?

[wolfsprint]

Ship started its journey from a Location A with velocity 18km/h to the north . after 2 hours the ship sailed in the direction 30 north of east find the rate of change in the distance between the location A and the position of the ship after 4 hours from the start.

i have no idea how to solve this problem and would really like how to?

 

[Dustinsfl]

Re: Related time rate , ship problem

Ship started its journey from a Location A with velocity 18km/h to the north . after 2 hours the ship sailed in the direction 30 north of east find the rate of change in the distance between the location A and the position of the ship after 4 hours from the start.

i have no idea how to solve this problem and would really like how to?

You have a related rates problem. Have you drawn the triangle the problem forms? Have you labeled the triangle with what you know?

If not, that is step one. Then what are the equations related to a triangle we could use?

 

[wolfsprint]

Re: Related time rate , ship problem

I understand i have to draw the x-axis and the y-axis and plot and join the dots to form a triangle but the problem is I am having trouble understanding the positions of the ships

 

[Dustinsfl]

Re: Related time rate , ship problem

I understand i have to draw the x-axis and the y-axis and plot and join the dots to form a triangle but the problem is I am having trouble understanding the positions of the ships

The 30 degrees north of east part?

If we let the east-west axis be our x-axis, then 30 degrees north of east is $\frac{\pi}{6}$.

 

[wolfsprint]

Re: Related time rate , ship problem

so A is the origin and when the ship sails north i put a dot a bit further up and for the 30 north of east i draw an imaginary x-axis on the dot on the north direction and put a point in the 30 degrees direction? what's next?

 

[Dustinsfl]

Re: Related time rate , ship problem

so A is the origin and when the ship sails north i put a dot a bit further up and for the 30 north of east i draw an imaginary x-axis on the dot on the north direction and put a point in the 30 degrees direction? what's next?

That is what I would do, let A be the origin. Then the ship sailed 2 hours at 18km/h so the from A to B we have 36km.
Then the ship sailed at 30 degrees from point b for 2 hours to point C. What you want to find is the rate of change of the AC

 

[wolfsprint]

Re: Related time rate , ship problem

ok so this is how i solved the problem :
After t hours i presumed the following
AB = 36+2t
BC = 36+2t
we need to find AC so
AC^2 = (36+2t)^2 + (36+2t)^2
AC = 12(Root)26
is this correct?

 

[wolfsprint]

Re: Related time rate , ship problem

Thanks for understanding , and i would like to know if my method of solving is correct please?

 

[Dustinsfl]

Re: Related time rate , ship problem

Thanks for understanding , and i would like to know if my method of solving is correct please?

Look at the this example page 192