Vectors anyone?

Question

A supertanker is crossing a channel heading N25 degrees 45 minutes 00 seconds W at 28km/h. The tanker encounters a strong currect moving 6km/h towards the south west. Determine the effective speed and direction of the tanker

Answer 1

Ugh, those units. Do mariners still think that way? Get the ship's direction in radains or degrees, and draw a vector of length 28 in that direction from the origin. From the head of that vector, draw a length 6 vector going southwest (direction 5pi/4). The vector from the origin to the head of the second vector gives the direction & speed on the boat. (If you don't like drawing, just convert into (x,y) coordinates and add.)

Answer 2

Method 1. Use components.
Tanker north 28*cos(25deg 45 min); west 28*sin(25deg 45min)

Current north -6*cos 45 deg; west 6*sin 45 deg.

Resultant: north 28*cos(25deg 45min) - 6*cos(45 deg)
........................... = 20.9769 approx
.................. west 28*sin(25deg 45min) + 6*sin(45deg)
............................= 16.4071 approx

Resultant magnitude: Square, add, then square root:
gives 26.6313 approx,
so the effective speed is 26.63 km/h

The effective direction is North A° West, where
tan A° = 16.4071/20.9769

i.e. North 38°1min51sec West


Method 2. The tanker velocity, the current velocity, and the resultant velocity form a triangle, in which the angle between the tanker velocity and the current velocity is 45° + 25°45min
which is 70° 45min

The cosine rule applied to this triangle gives the magnitude of the effective speed
x^2 = 28^2 + 6^2 - 2*28*6*cos(70.75°)
since 45 minutes = 0.75°

This gives x = 26.63125876


To find the direction, use the sine rule to find the angle T between the tanker velocity and the effective velocity:

(sin T)/6 = sin (70.75°)/26.63125876

Hence T = 12° 16min 51sec,
add 25° 45min to this to get the bearing
North 38° 1min 51sec

Answer 3

If the two velocities and resultant velocity are drawn , a velocity diagram of sides 6 and 28 with included angle 70.75° is obtained.
Let resultant velocity = v
v² = 28² + 6² - [2 x 28 x 6 cos(70.75°)]
v² = 784 + 36 - 110.8 = 709.2
v = 26.6 km/h

Now use sine rule:-
26.6 / sin 70.75° = 6 / sin X where X is angle between tanker velocity vector and resultant velocity vector
26.6.sinX = 6.sin 70.75°
sin X = 6.sin 70.75° / 26.6
X = 12.3°

Bearing of tanker = 25.75° + 12.3° W of N
= 38.05° W of N