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lara [203]
3 years ago
6

[need math help] Study buddies part #2

Mathematics
1 answer:
rosijanka [135]3 years ago
6 0

Answer:

Part A) The distance between Whitney's house and Anh's house is 6.5 miles.

Part B) The total distance traveled by Whitney during the trip was 9.5 miles.

Step-by-step explanation:

Part A)

Notice that during Whitney's journey to Anh's house, her velocity was negative for some time. So, Whitney was travelling backwards. Hence, the total distance Whitney traveled will be greater than the actual distance from Whitney's house to Anh's.

So, instead, we can calculate Whitney's total displacement. Total displacement is given by:

\displaystyle \text{Displacement}=\int_a^bv(t)\, dt

Where <em>v(t)</em> is the velocity function.

So, the displacement of Whitney when she traveled from her house to Anh's house will be:

\displaystyle \text{Displacement}=\int_0^{24}v(t)\, dt

To calculate the integral, we can split it off at <em>t</em> = 11, <em>t</em> = 16, and <em>t</em> = 24.

The three regions represent three geometric figures, which we can use to calculate the area and then find the integral. Rewriting:

\displaystyle \text{Displacement}=\int_0^{11}v(t)\, dt+\int_{11}^{16}v(t)\, dt+\int_{16}^{24}v(t)\, dt

Now, we can calculate the area of each figure.

The first figure is a trapezoid with a height of 0.8 and two bases of 11 and 3. So, its area is:

\displaystyle A_1=\frac{1}{2}(0.8)(11+3)=5.6\text{ miles}

The second figure is a triangle with a height of 0.6 and a base of 5. So, its area is:

\displaystyle A_2=\frac{1}{2}(0.6)(5)=1.5\text{ miles}

Lastly, the third figure is another triangle will a height of 0.6 and a base of 8. So, its area is:

\displaystyle A_3=\frac{1}{2}(0.6)(8)=2.4\text{ miles}

However, notice that the second figure is below the <em>x</em>-axis. During that interval, Whitney was traveling backwards. Hence, we will subtract the second area from the rest of the areas.

Then the displacement will be:

\displaystyle \text{Displacement}=5.6-1.5+2.4=6.5\text{ miles}

So, the distance between Whitney's house and Anh's house is 6.5 miles.

Part B)

The total distance differs from displacement in that it includes all the distances in both directions. It is given by:

\displaystyle \text{Distance}=\int_a^b|v(t)|\, dt

In other words, all of our values are now positive. We will utilize the same strategy:

\displaystyle \text{Distance}=\int_0^{11}|v(t)|\, dt+\int_{11}^{16}|v(t)|\, dt+\int_{16}^{24}|v(t)|\, dt

Substitute. The only difference is that we will add 1.5 instead of subtracting it. So:

\displaystyle \text{Distance} =5.6+1.5+2.4=9.5

Therefore, in this instance, Whitney traveled a total of 9.5 miles to get to Anh's house.

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Part (a)

Focus on triangle PSQ. We have

angle P = 52

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Use of the law of sines to determine angle S

sin(S)/PQ = sin(P)/SQ

sin(S)/(6.8) = sin(52)/(5.4)

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S = arcsin(0.99230983787513)

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Which is approximate

------------

Use this to find angle Q. Again we're only focusing on triangle PSQ.

P+S+Q = 180

Q = 180-P-S

Q = 180-52-82.889762826274

Q = 45.110237173726

Which is also approximate.

A more specific name for this angle is angle PQS, which will be useful later in part (b).

------------

Now find the area of triangle PSQ

area of triangle = 0.5*(side1)*(side2)*sin(included angle)

area of triangle PSQ = 0.5*(PQ)*(SQ)*sin(angle Q)

area of triangle PSQ = 0.5*(6.8)*(5.4)*sin(45.110237173726)

area of triangle PSQ = 13.0074347717966

------------

Next we'll use the fact that RS:SP is 2:1.

This means RS is twice as long as SP. Consequently, this means the area of triangle RSQ is twice that of the area of triangle PSQ. It might help to rotate the diagram so that line PSR is horizontal and Q is above this horizontal line.

We found

area of triangle PSQ = 13.0074347717966

So,

area of triangle RSQ = 2*(area of triangle PSQ)

area of triangle RSQ = 2*13.0074347717966

area of triangle RSQ = 26.0148695435932

------------

We're onto the last step. Add up the smaller triangular areas we found

area of triangle PQR = (area of triangle PSQ)+(area of triangle RSQ)

area of triangle PQR = (13.0074347717966)+(26.0148695435932)

area of triangle PQR = 39.0223043153899

------------

<h3>Answer: 39.0223043153899</h3>

This value is approximate. Round however you need to.

===========================================

Part (b)

Focus on triangle PSQ. Let's find the length of PS.

We'll use the value of angle Q to determine this length.

We'll use the law of sines

sin(Q)/(PS) = sin(P)/(SQ)

sin(45.110237173726)/(PS) = sin(52)/(5.4)

5.4*sin(45.110237173726) = PS*sin(52)

PS = 5.4*sin(45.110237173726)/sin(52)

PS = 4.8549034284642

Because RS is twice as long as PS, we know that

RS = 2*PS = 2*4.8549034284642 = 9.7098068569284

So,

PR = RS+PS

PR = 9.7098068569284 + 4.8549034284642

PR = 14.5647102853927

-------------

Next we use the law of cosines to find RQ

Focus on triangle PQR

c^2 = a^2 + b^2 - 2ab*cos(C)

(RQ)^2 = (PR)^2 + (PQ)^2 - 2(PR)*(PQ)*cos(P)

(RQ)^2 = (14.5647102853927)^2 + (6.8)^2 - 2(14.5647102853927)*(6.8)*cos(52)

(RQ)^2 = 136.420523798282

RQ = sqrt(136.420523798282)

RQ = 11.6799196828694

--------------

We'll use the law of sines to find angle R of triangle PQR

sin(R)/PQ = sin(P)/RQ

sin(R)/6.8 = sin(52)/11.6799196828694

sin(R) = 6.8*sin(52)/11.6799196828694

sin(R) = 0.4587765387107

R = arcsin(0.4587765387107)

R = 27.3081879220073

--------------

This leads to

P+Q+R = 180

Q = 180-P-R

Q = 180-52-27.3081879220073

Q = 100.691812077992

This is the measure of angle PQR

subtract off angle PQS found back in part (a)

angle SQR = (anglePQR) - (anglePQS)

angle SQR = (100.691812077992) - (45.110237173726)

angle SQR = 55.581574904266

--------------

<h3>Answer: 55.581574904266</h3>

This value is approximate. Round however you need to.

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It also means that -4 is the y intercept.

<u>-4 is the y intercept.</u>

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