Algebra – Precalculus Vector Questions and Solutions

algebra-precalculus

An airplane is flying on a compass heading (bearing) at 340 degrees at 325 mph. A wind is blowing with the bearing 320 degrees at 40 mph.

Find the actual ground speed and direction of the plane

This is my work:

I know I went wrong someone, I just don't know where. Can someone tell me what I got wrong?

2.A force of 50 lbs acts on a object at an angle of 45 degrees. A second force of 74 lbs acts on the object at an angle of -30 degrees.

Find the direction and magnitude of the resultant force.

This is my work:

The work is correct, except the final step. I looked at the answer and it was the negative angle (-1.226 degrees) I was wondering what the logic was behind this.

Thank you very much

Best Answer

Don't bother converting compass orientations to standard orientations. Just assume that the real world does things like the mathematicians do, even when they don't

$(325\cos 340^\circ, 325\sin 340^\circ) + (40\cos 320, 40\sin 320)$

It should really be minus the wind vector, but I am sure that the person who designed this question doesn't know the actual conventions. But, in my previous statement I just said disregard what is done in the real world, so we will keep doing that.

$\text {ground speed} = \sqrt { (325\cos 340^\circ + 40\cos 320)^2 + (325\sin 340^\circ + 40\sin 320)^2}$

True course.

$\theta = \arctan \frac {325\sin 340^\circ + 40\sin 320}{325\sin 340^\circ + 40\sin 320}$

However the arctan convention is to quote figures in $(-180,180)$ and you will need to add $360^\circ$ to get a number in $(0,360]$

In these sorts of questions it is also possible for the calculation to be off by $180^\circ$ You will have to do a little analysis as to which quadrant you are in and which adjustment you need to make.

Related Solutions

[Math] Trigonometry Airplane question. Finding bearing and distance.

Cosine Law would be easier, but let's do it your way. Consider the small right triangle (call it $\Delta ABC$), which roughly looks like:

A----B
|   /
|  /
| /
|/
C

Note that $\angle ABC =59^\circ$. Thus, $AB=640\cos(59^\circ)$ and $AC=640\sin(59^\circ)$.

Now consider the large right triangle. Its two legs are $AC$ and $AB+480$. Hence, by the Pythagorean Theorem, we have: $$x=\sqrt{(AC)^2 + (AB+480)^2}=\sqrt{(640\cos(59^\circ)+480)^2 + (640\sin(59^\circ))^2} \approx 978 \text{ miles}$$ Finally, for the angle bearing: $$\theta=\tan^{-1}\left(\frac{AC}{AB+480}\right) = \tan^{-1}\left(\frac{640\sin(59^\circ)}{640\cos(59^\circ)+480}\right) \approx 34^\circ \text{ to get the bearing you subtract that value from 90 to get } S56^\circ W$$

[Math] determine the resultant speed of the plane

You have to use vectorial addition - the angle is the difference of direction. Hint: Draw a triangle with

a point (e.g. A) as the current position of the plane
one side is for the velocity of the plane (going east, that is a line/vector going horizontally to the right from a, length: something representing 200, e.g. 5inch/cm)
one side is for the velocity of the wind (going north east, length representing 40km-h, e.g. 1 inch/cm). This line has to be appended at the end of your first line/vector
then draw a line from the starting point A to the end of the second line/vector
that vector represents the resulting velocity vector
in order to calculate it length you could e.g. use trigonometry

By the way, the information about the angle (pi/4) can be retrieved from the directions (east and north east) already and is thus redundant. If you draw your triangle you will see that the angle does have a big influence on the length of the resulting velocity vector.

Best Answer

Related Solutions

[Math] Trigonometry Airplane question. Finding bearing and distance.

[Math] determine the resultant speed of the plane

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