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A turbo-jet flies \(50 \mathrm{mph}\) faster than a super-prop plane. If a turbo-jet goes 2000 mi in 3 hr less time than it takes the super-prop to go \(2800 \mathrm{mi}\), find the speed of each plane.

Short Answer

Expert verified
The speed of the super-prop plane is \(192.41 \text{ mph}\), and the speed of the turbo-jet is \(242.41 \text{ mph}\).

Step by step solution

01

- Define Variables

Let the speed of the super-prop plane be denoted as \(s\). Therefore, the speed of the turbo-jet will be \(s + 50 \text{ mph}\).
02

- Set Up Equations for Time

Using the formula \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \], write the time taken for each plane to travel their respective distances. For the turbo-jet: \[ \text{Time}_{\text{turbo-jet}} = \frac{2000}{s + 50} \] and for the super-prop plane: \[ \text{Time}_{\text{super-prop}} = \frac{2800}{s} \]
03

- Establish Relationship Between Times

According to the problem, the turbo-jet takes 3 hours less than the super-prop plane. Therefore, \[ \frac{2000}{s + 50} = \frac{2800}{s} - 3 \]
04

- Solve for \(s\)

Multiply both sides by \(s(s + 50)\) to clear the denominators: \[ 2000s = 2800(s + 50) - 3s(s + 50) \] Simplify to get: \[ 2000s = 2800s + 140000 - 3s^2 - 150s \] Combine like terms and rearrange: \[ 3s^2 - 650s + 140000 = 0 \]
05

- Solve the Quadratic Equation

Use the quadratic formula \[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \(a = 3, b = -650, c = 140000\). Calculate the discriminant: \[ b^2 - 4ac = 650^2 - 4(3)(140000) = 422500 - 168000 = 254500 \] Solve for \(s\): \[ s = \frac{650 \pm \sqrt{254500}}{6} = \frac{650 \pm 504.48}{6} \] Therefore, \[ s = \frac{1154.48}{6} = 192.41 \text{ mph} \text{ (since speed must be positive)}\] \[ s = \frac{145.52}{6} \text{ which does not fit the context since the resulting speed is negative}\]
06

- Calculate Turbo-Jet Speed

Add 50 mph to the speed of the super-prop plane to find the speed of the turbo-jet: \[ s + 50 = 192.41 + 50 = 242.41 \text{ mph} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

speed and distance calculations
Understanding speed and distance calculations is foundational in solving problems related to moving objects. The relationship between speed, distance, and time can be understood using the formula: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] This formula implies that if you know any two of the three variables—distance, speed, and time—you can calculate the third one. In our problem, we used it to find the time taken for the turbo-jet and the super-prop plane to cover their respective distances. Remember:
  • Distance covered is directly proportional to speed when time is constant.
  • For a constant distance, time varies inversely with speed.
In practical terms, this means that faster speeds result in shorter travel times for the same distance.
quadratic equations
Quadratic equations appear frequently in algebra, especially in problems involving squared terms. A quadratic equation typically has the form: \[ ax^2 + bx + c = 0 \] In our exercise, we derived a quadratic equation from the relationship between the times taken by the planes: \[ 3s^2 - 650s + 140000 = 0 \] To solve it, we used the quadratic formula: \[ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, a is 3, b is -650, and c is 140000. By calculating the discriminant \[ b^2 - 4ac \], we found the solutions for s. Key points:
  • The discriminant (b^2 - 4ac) determines the nature of the roots.
  • If it’s positive, we get two real roots; if zero, one real root; if negative, the roots are complex.
For our equation, the positive root gave us the speed of the super-prop plane.
algebraic word problems
Solving algebraic word problems requires translating the given information into mathematical expressions. Here are steps to approach such problems:
1. **Read the problem carefully**: Understand what is being asked. In our case, we need to find the speeds of two planes.
2. **Define the variables**: Decide what each variable represents. We let s be the speed of the super-prop plane, making the turbo-jet's speed s + 50.
3. **Set up equations**: Translate the conditions in the problem into algebraic equations. We used the given distances and times to form our initial equations.
4. **Solve the equations**: Use algebraic methods to solve the equations. We solved a quadratic equation using the quadratic formula.
5. **Interpret the solution**: Make sure your solution makes sense in the context of the problem. We confirmed our speeds were reasonable for planes.
These steps help break down complex problems into manageable parts, making it easier to find a solution.

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Most popular questions from this chapter

Solve. During the first part of a trip, Tara drove \(120 \mathrm{mi}\) at a certain speed. Tara then drove another \(100 \mathrm{mi}\) at a speed that was \(10 \mathrm{mph}\) slower. If the total time of Tara's trip was 4 hr, what was her speed on each part of the trip?

Solve. Canoeing. During the first part of a canoe trip, Ken covered \(60 \mathrm{km}\) at a certain speed. He then traveled \(24 \mathrm{km}\) at a speed that was \(4 \mathrm{km} / \mathrm{h}\) slower. If the total time for the trip was 8 hr, what was the speed on each part of the trip?

Find an equation in slope–intercept form of a line with the given characteristics. Slope: \(-\frac{1}{3} ; y\) -intercept: \((0,16)\)

Solve. $$ \sqrt{2 x-8}=15 $$

Hydrology. The following drawing shows the cross section of a river. Typically rivers are deepest in the middle, with the depth decreasing to 0 at the edges. A hydrologist measures the depths \(D,\) in feet, of a river at distances \(x,\) in feet, from one bank. The results are listed in the following table. $$ \begin{array}{|c|c|} \hline \text { Distance x, } & {\text { Depth } D \text { of }} \\ {\text { from the Left }} & {\text { the River }} \\ {\text { Bank (in feet) }} & {\text { (in feet) }} \\ \hline 0 & {0} \\ {15} & {10.2} \\ {25} & {17} \\ {50} & {20} \\ {90} & {7.2} \\ {100} & {0} \\ \hline \end{array} $$ a) Use regression to find a quadratic function that fits the data. b) Use the function to estimate the depth of the river 70 ft from the left bank.

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