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ivanzaharov [21]
3 years ago
5

Two trucks leave a warehouse at the same time. One travels due west at an average speed of 58 miles per hour, and the other trav

els due east at an average speed of 46 miles per hour. After how many hours will the two trucks be 208 miles apart?
Mathematics
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

17.33 hours or 17 hours 20 minutes.

Step-by-step explanation:

The distance travelled per hour for truck "A" = 58 miles.

The distance travelled per hour for truck "B" = 46 miles.

Now, it is obvious that the first truck (Truck A) was travelling with a fairly greater speed than the second truck (Truck B) so it is safe to say that the distance covered by truck A in a given number of hours will certainly be greater than the distance that is covered by truck B over the same period. In this situation, let us use "y" to represent that number of hours that it will take for the distance between the two trucks to now become 208 miles assuming the two trucks (A and B) maintained their respective average speed.

This means that, for that number of hours(y), the distance covered by the first truck (truck A) will be 58 × y = 58ymiles and the distance that the second truck would cover over the same period will be 46 × y = 46ymiles.

This ultimately implies that the distance between them or the difference between the distances that the trucks have covered after this unknown number of hours is 208miles. Since they are travelling due west and due east respectively and the first truck is travelling at a greater speed, then:

58y miles - 46y miles = 208 miles.

12y = 208

y = 208/12

y = 17.33 hours.

Since 1 hour = 60 minutes, then 0.33 hours is:

1 hour --------- 60 mins

0.33 hour -------- ? mins

= 0.33/1 × 60/1 = 20 minutes

If the drivers of both trucks maintained their average speeds, then the time it will take for them to become 208 miles apart is 17 hours 20 minutes or 17.33 hours

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For the following integral, find the approximate value of the integral with 4 subdivisions using midpoint, trapezoid, and Simpso
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\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}

\textsf{Trapezium rule}: \quad \pi

\textsf{Simpson's rule}: \quad \dfrac{4 \pi}{3}

Step-by-step explanation:

<u>Midpoint rule</u>

\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}

<u>Trapezium rule</u>

\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}

<u>Simpson's rule</u>

\displaystyle \int_{a}^{b} y \:\:\text{d}x \approx \dfrac{1}{3}h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{n-2}+4y_{n-1}+y_n\right)\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}

<u>Given definite integral</u>:

\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x

Therefore:

  • a = 0
  • b = 2π

Calculate the subdivisions:

\implies h=\dfrac{2 \pi - 0}{4}=\dfrac{1}{2}\pi

<u>Midpoint rule</u>

Sub-intervals are:

\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]

The midpoints of these sub-intervals are:

\dfrac{1}{4} \pi, \dfrac{3}{4} \pi, \dfrac{5}{4} \pi, \dfrac{7}{4} \pi

Therefore:

\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}

<u>Trapezium rule</u>

\begin{array}{| c | c | c | c | c | c |}\cline{1-6} &&&&&\\ x & 0 & \dfrac{1}{2}\pi & \pi & \dfrac{3}{2} \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}

\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x &  \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}

<u>Simpson's rule</u>

<u />

<u />\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}

6 0
2 years ago
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