answer this please i need it by tomorrow i tried but i keep getting it wrong(physics)

Question

A Coast Guard cutter detects an unidentified ship at a distance of 16.8 km in the direction 14.2° east of north. The ship is traveling at 22.5 km/h on a course at 38.9° east of north. The Coast Guard wishes to send a speedboat to intercept the vessel and investigate it. If the speedboat travels 41.7 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.

Answer 1

Hi Porcia!

 

What I would do with this one, following the pattern of some of your other questions, is break it down into a question of components.

 

First, let's look at the initial position of the unknown ship, in terms of i-hat and j-hat.  The unknown's original location is:

 

16.8 km at 14.2^o east of north; which we can rewrite as 16.8 km at 75.8^o north of east

 

This leads to an initial position for the unknown ship (with respect to the Coast Guard station) we can express in unit vector notation as:

 

x_ou = (16.8 km)cos(75.8^o) = 4.12 km

y_ou = (16.8 km)sin(75.8^o) = 16.3 km

 

In unit vector notation, the original position of the unknown ship is thus:

 

r_ou = 4.12i+16.3j

 

The initial position for the Coast Guard ship is at the Coast Guard station, so (0,0)

 

The velocity of the unknown ship can also be expressed in unit vector notation, since the ship moves at 38.9^o east of north (51.1^o north of east).

 

v_xu = (22.5 km/h)cos(51.1^o)= 14.1 km/h

v_yu = (22.5 km/h)sin(51.1°) = 17.5 km/h

 

So v_u = 14.1i + 17.5j

 

And since displacement is just (velocity*time) for constant velocity motion

 

r_u(t) = r_ou + (14.1t)i+(17.5t)j  (where t is measured in hours)

 

where r_ou is the initial position of the unknown (with respect to the Coast Guard station) calculated above, so:

 

r_u(t) = 4.12i + 16.3j + (14.1t)i + (17.5t)j

 

Grouping the i and j terms:

 

r_u(t) = (4.12 + 14.1t)i + (16.3+17.5t)j

 

 

Now, for the Coast Guard ship has an unknown angle bearing, but we can still write:

 

v_CG = (41.7cosθ)i + (41.7sinθ)j   where θ is the angle measured counterclockwise from the +x axis

 

Then r_CG(t) = [(41.7cosθ)t]i + [(41.7sinθ)t]j   (no initial position is needed, since it is (0,0) for the Coast Guard ship]

 

The condition for the two ships meeting is that they are at the same position at the same time.  Thus, their x and y (i-hat and j-hat) positions need to be identical at the same time t.  So, for the condition of meeting, we can write

 

r_u(t_m) = r_CG(t_m)   (where t_m denotes the time to intercept)

 

Hence, we need:

 

x (i-hat) direction:  4.12 + 14t_m = (41.7cosθ)t_m

y (j-hat) direction:  16.3 + 17.5t_m = (41.7sinθ)t_m

 

This represents two equations with two unknowns, and you want to solve for θ.  This is not the easiest thing ever, but it feasible.  It may help to remember the trig identity that (sin/cos) = tan.

 

Actually, it may be easier to solve for t_m first, by squaring both sides of both equations (squaring the left sides will be a little work, but feasible).  Then, if you add the two right-hand sides, and factor out (41.7t_m)², you are left with (41.7t_m)²(sin²θ + cos²θ) = (41.7t_m)²(1).  You can then rearrange all the terms and simplify into a quadratic equation for t_m, which you can solve with the quadratic formula (I think you should get only one positive solution, which will be the one you want).  Then plug the solution for t_m back into either of the last two equations and solve for θ.  

 

Once you solve for θ as I have defined it, you will have the angle measured clockwise with respect to the +x axis.  To get the angle with respect to true north, you will want to take the complement of θ, and that should be east of north.

 

I am sorry the math on this is a little involved, but I do not immediately see a more straightforward way to calculate it.  Hope that helps a little!  Try it out, and, if you get stuck, we can look at it further.