Hi! I was just wondering about how this problem is done. Why isn't density a function of temperature, and therefore be solved by using y=kx+b?

]]>]]>@bulba_bulbasaur said in No one finished Module 6!:

@rz923 said in No one finished Module 6!:

@bulba_bulbasaur said in No one finished Module 6!:

how do you quote?

Here:

E7B11E1E-7DF7-49C0-8EFB-C6BDF897A97C.jpegGreen in the honour of the green shirt! 👕

ok thx!

Ze quote chain

XD

I thought Prof Loh likes anticlockwise ...

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How do you find the answer? \((\)There is only one answer because there is only one triangle with angles \(70^\circ \text{ and } 50^\circ,\) there is only one point on \(\overline{XY}\) so that \(\angle AZY=40^\circ,\) and there is only one point on \(\overline{XZ}\) so that \(\angle BYZ=50^\circ.)\)

]]>I tried it, and I'm sorry that I don't have an elegant way to do it, just a grungy way 🔨 🔨 ⚒

Here is the grungy way:

Angle-chasing (which we learned about in these lessons: M2W2D5-Y-2, M2W4D13-Ch-2) tells us

that \( \angle BYZ = 50^{\circ} \) and \(\angle YAB = 70^{\circ},\) so since \( \angle BZY = 50^{\circ}\) and \(\angle AYZ = 70^{\circ}\) (which is given), we know that triangle \(\bigtriangleup BYZ\) and \(\bigtriangleup AYZ\) are both isosceles.

I'll let \( BZ = BY = 1,\) since all lengths will be relative and only the angles matter, and normalize all other lengths around this length. Then triangle \(\bigtriangleup BYZ\) has lengths of \(1, 1,\) and a third leg, the length of which can be found using the Law of Sines:

$$ \frac{\sin 50^{\circ}}{1} = \frac{\sin 80^{\circ}}{YZ}$$

$$ YZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}} $$

From angle-chasing, we know that \( \overline{AZ} \) and \(\overline{YB}\) intersect at right angles. Let's call this intersection point \(D.\) Then we get four right triangles: \( \bigtriangleup DYZ, \bigtriangleup DAY, \bigtriangleup DBZ, \text{ and } \bigtriangleup DAB.\)

Then \(BD + DY = 1\) and \(AD + DZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}}.\)

From the Pythagorean Theorem on \(\bigtriangleup DBZ,\) we know \(BD^2 + DZ^2 = 1,\) so \(DZ^2 = 1 - DB^2.\)

From the Pythagorean Theorem on \(\bigtriangleup DYZ,\) we know that \(DY^2 + DZ^2 = \left( \frac{\sin 80^{\circ}}{\sin 50^{\circ}} \right)^2 .\)

Substitute for \(DZ^2\) into the first equation to solve for \(DB,\) which equals \( 1 - \frac{1}{2} \left( \frac{\sin 80^{\circ}}{\sin 50^{\circ}} \right)^2.\)

Then solve for \(DZ\) using the Pythagorean Theorem on \(\bigtriangleup DBZ,\) which says \( DZ^2 = 1 - DB^2.\)

Once you have found \(DZ,\) then use it to find \(AD,\) since \(AD + DZ = \frac{\sin 80^{\circ}}{\sin 50^{\circ}}.\)

Once you have found \(AD,\) then using the value for \(DB\) from earlier, we can solve for \(\angle ABY:\)

$$ \tan ABY = \frac{AD}{DB} $$

$$ \angle ABY = \arctan \frac{AD}{DB} $$

Sorry I didn't do the calculations... they were just so grungy! But the logic of the steps is all here.

🙂

]]>I wasn't sure how long the car drove at \(1\) mph for, so I just assumed that it immediately starts accelerating.

We can graph the speed in mph vs. time (in minutes) like this:

20200628-forum-speed-intervals-50-percent.png

The acceleration is constant at \( \pm 1 \text{ } \frac{\text{mph}}{\text{min}}\) for each straight segment of the zigzag. From physics, we learn that

$$ \text{distance} = \text{speed} \times \text{time} $$

(Though actually in physics, they use velocity instead of time, since velocity takes into account the direction of where an object is moving.)

Thus the distance can be interpreted as the area under a graph of speed versus time, like in the following example:

20200628-forum-speed-intervals-averages-area-example-25-percent.png

Here the distance traveled is equal to

$$ \text{ distance } = 4 \frac{\text{miles}}{\cancel{\text{hour}}} \times \frac{1}{12} \text{ } \cancel{\text{hour}} = \textcolor{red}{\frac{1}{3} \text{ miles }}$$

Even though our graph doesn't look as nice and simple as the one above, the distance traveled is also the same: it's the area under the speed versus time curve.

20200628-forum-speed-intervals-area-25-percent.png

The area above is equivalent to the area of the shape below, found by averaging the speed for each interval:

20200628-forum-speed-intervals-averages-area-25-percent.png

Remember that we must convert the time duration of each interval from minutes to hour, to match the units of the speed (miles per hour). Thus we go at

$$\begin{aligned} 2 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 2.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ 3 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 3.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ 4 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 4.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ 5 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 5.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ 6 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 6.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ \dots \\ 59 & \text{ mph for } \frac{1}{30} \text{ hour } \\ 59.5 & \text{ mph for } \frac{1}{60} \text{ hour } \\ \end{aligned} $$In the question, it's not clear how long the car spends driving at \(60 \text{ mph},\) so I'll just assume that once it reaches \(60 \text{ mph} \) it no longer continues driving.

Thus the total distance, in miles, is

$$\begin{aligned} & 2 \times \frac{1}{30} + 2.5 \times \frac{1}{60} + 2 \times \frac{1}{30} + 2.5 \times \frac{1}{60} \\ & + 3 \times \frac{1}{30} + 3.5 \times \frac{1}{60} + 4 \times \frac{1}{30} + 4.5 \times \frac{1}{60} \\ \ldots \end{aligned} $$I'll leave it to you to figure out how to simplify this expression! 🙂

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