Friday

December 19, 2014

December 19, 2014

Posted by **albert** on Monday, May 9, 2011 at 5:24pm.

straight line from port A located at position (xA, yA) = (−100,−100) to

port B at (xB, yB) = (300, 100).

(a) (i) Find the equation of the line of travel of the ship.

(ii) Find the direction of travel of the ship, as a bearing, with the

angle in degrees correct to one decimal place.

(iii) Calculate the distance between ports A and B, correct to two

decimal places.

(iv) How long does the ship take to cover the distance between ports

A and B? Give the answer in hours and minutes, correct to the

nearest minute.

(v) Find parametric equations for the line of travel of the ship. Your

equations should be in terms of a parameter t, and should be such

that the ship is at port A when t = 0 and at port B when t = 1.

(b) During its journey, the ship passes two lighthouses, L1 and L2, which

are located at positions (0, 0) and (200, 0), respectively.

(i) Write down expressions, in terms of the parameter t of

part (a)(v), for the squares d21

and d22 of distances between the location of the ship at parameter value t and the lighthouses L1

and L2 respectively. Simplify your results.

(ii) Completing the square gives the following result for d21:

d21= 200 000 ((t – 3)^2 + 1/100).

Explain how d1 varies as t increases from 0 to 1. Determine the

shortest distance between the ship and the lighthouse L1, to the

nearest kilometer.

(iii) Complete the square for your result for d22

from part (b)(i), and

determine the shortest distance between the ship and the

lighthouse L2, to the nearest kilometer.

(iv) Lighthouse L1 can be seen from a distance of 50 kilometers.

Calculate the parameter values t1 and t2 when the lighthouse L1

can first and last be seen from the ship. For how many minutes is

the lighthouse visible from the ship?

- maths -
**Henry**, Tuesday, May 10, 2011 at 8:05pmA(-100,-100), B(300,100).

1. m = (100- (-100)) / (300 - (-100)) =

200 / 400 = 1/2.

y = mX + b,

-100/2 + b = -100,

b = -100 + 50 = -50.

Eq: Y = (1/2)X - 50.

2. tanA = m = 1/2 = 0.5.

A = 26.6Deg. CCW = 26.6Deg N. of E.

3. d^2 = (400)^2 + (200)^2 = 200,000,

d = 447.21km.

4. t = d/V = 447.21 / 25 = 17.89h.

**Answer this Question**

**Related Questions**

mayh - A ship travels at a constant speed of 25 kilometres per hour (Kph) in a ...

Maths - How can answer this question ,help please? Information: A ship travels ...

geometry - A ship leaves port and heads due east at a rate of 32 miles per hour...

Math - "A ship leaves port on a bearing of 34.0 degrees and travels 10.4 mi. ...

math- precalculus - I've attempted this problem a few times but I can't get the ...

Physics - Ports A and B are 400 miles apart. One boat starts from port A for ...

Physics - Ports A and B are 400 miles apart. One boat starts from port A for ...

Math - a bus travels 240 kilometers at 60 kph and then returns at 40 kph.. what ...

AP Calculus - Two ships sail from the same port. The first ship leaves port at 1...

mathematics - a ship which started at port p, sailed 15 km due south to port q...