Applied maths 1999 Q2

finality

a) An aeroplane has a speed of 160m/s in still air. When the wind blows from the east, the velocity of the aeroplane as observed from the ground is 120m/s towards the north-east. Find the speed of the wind correct to two decimal places.


b) Two ships A and B move with constant speeds 48km/h and 60km/h respectively. At a certain instant ship B is 30km west of A and travelling due south. Find
i) the direction ship A should steer in order to get as close as possible to ship B
ii) the shortest distance between the ships.



Help? I can't do either part, I'm so horribly bad at relative motion. :L


The answers are
a) 50.8 m/s
b)i) 53.13° S of W
ii) 18km

Stalin and rugby

You gotta draw out your triangle for starters that's prob the hardest bit I guess.
I don't know how to actually show you the drawing so I'll do my best

You should get an Alpha of 135 degrees then use the cosine rule where

(160) squared = (120) ²+ velocity wind ² - 2(120)(velocity wind)cos135

Should get a v of 50.79 correct to 2 decimal places.

Part 2

Closest approach question hasn't come up since that year but it's kinda hard.

What you gotta do is draw out a diagream. The closest it will get is when the angle is at a max to the actual required angle. Sorry it's kinda hard to explain when there's no diagram. What you wanna do is say

SinA/60 = SinB/48

Max B will occur for max A. max angle sin can be is 90 = 1

So you get 1/60 * 48 = Sin B

B = 53.13

Second part is a standard question.

Cheers brah!