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I would like some help with this question.

A plane is flying North East at 600 kph. The wind is blowing to the South at 100 kph. Find the plane's ground speed and true heading.

My work/thinking:

Finding the sum of the plane and wind's vector components would equal the actual ground speed:

<600cos(45) , 600sin(45)> + <0, 100> = <424.264 , 524.264>

Magnitude:

674.4277 kph

The direction:

$\theta$ = 51.018

Have I done this correctly?

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    $\begingroup$ That depends on what "flying northeast at 600 kph" means. Without further, it would be natural to understand that as a claim about the ground velocity itself. $\endgroup$ Jan 13, 2014 at 22:53
  • $\begingroup$ I took it to mean a heading of 45 degrees above the x-axis. $\endgroup$ Jan 13, 2014 at 22:54
  • $\begingroup$ Well done with the results $\endgroup$
    – NasuSama
    Jan 13, 2014 at 22:58

2 Answers 2

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Almost. You have the direction of the wind wrong.

If the wind is blowing south, then this will reduce the north-south component of the velocity, not increase it:

$$\vec{v} = 600 \cos (45^{\circ}) \hat{x} + 600 \sin (45^{\circ}) \hat{y} - 100 \hat{y}.$$

So then $|\vec{v}| = \sqrt{424.264^2 + 324.264^2} \approx 534$ kph and the angle is $\tan^{-1}\frac{324.264}{424.464} \approx 37.39^{\circ}.$

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Using cosine rule, actual ground speed=$\sqrt{600^2+100^2-2\cdot100\cdot 600 \cdot\cos45^o}=533.9917\ldots \approx 534 \text{ kph}$

Using sine rule, $\displaystyle \frac{\sin \theta}{100}=\frac{\sin 45^o}{533.9917\ldots} \implies \theta=\arcsin(\frac{100 \sin 45^o}{533.9917\ldots})=7.609\ldots \approx 7.61^o$

Hence true direction is on a bearing of $045^o+007.61^o=052.61^o$

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