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A ship travels at a constant speed of 25 kilometers per hour (kph) in a
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 - ,

    A(-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.

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