You are flying on a commercial airline on your way from Houston, Texas, to Oklahoma City, Oklahoma. Your pilot announces that the plane is directly over Austin, Texas, traveling at a constant speed of \(245 \mathrm{mph}\), and will be flying directly over Dallas, Texas, \(362 \mathrm{~km}\) away. How long will it be before you are directly over Dallas, Texas?

Short Answer

Expert verified
Answer: __________ hours (or __________ minutes)

Step by step solution

01

Convert distance from kilometers to miles

To make the units consistent, let's convert the distance from kilometers to miles. We can use the conversion factor: 1 mile = 1.60934 kilometers. So, let's convert 362 km to miles.$$ \text{Distance in miles} = \frac{362}{1.60934} $$
02

Calculate the time taken to travel the distance

Now that we have distance in miles, we can use the time-distance-speed formula:$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$Plug in the values we have:\(\text{Time} = \frac{\text{Distance in miles}}{245 \text{ mph}}\)
03

Simplify the expression and find the time

Simplify the expression for time:$$\text{Time} = \frac{\frac{362}{1.60934}}{245}$$Calculate the time:$$\text{Time} = \frac{362}{1.60934\times245}$$After calculating the time, we get the result in hours.
04

Convert the time to minutes (optional)

It might be useful to convert the time to minutes. To do this, multiply the time in hours by 60:$$\text{Time}(\text{in minutes}) = \text{Time}(\text{in hours}) \times 60$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A car travels \(22.0 \mathrm{~m} / \mathrm{s}\) north for \(30.0 \mathrm{~min}\) and then reverses direction and travels \(28.0 \mathrm{~m} / \mathrm{s}\) for \(15.0 \mathrm{~min}\). What is the car's total displacement? a) \(1.44 \cdot 10^{4} \mathrm{~m}\) b) \(6.48 \cdot 10^{4} \mathrm{~m}\) c) \(3.96 \cdot 10^{4} \mathrm{~m}\) d) \(9.98 \cdot 10^{4} \mathrm{~m}\)

A girl is standing at the edge of a cliff \(100 . \mathrm{m}\) above the ground. She reaches out over the edge of the cliff and throws a rock straight upward with a speed \(8.00 \mathrm{~m} / \mathrm{s}\). a) How long does it take the rock to hit the ground? b) What is the speed of the rock the instant before it hits the ground?

The Bellagio Hotel in Las Vegas, Nevada, is well known for its Musical Fountains, which use 192 HyperShooters to fire water hundreds of feet into the air to the rhythm of music. One of the HyperShooters fires water straight upward to a height of \(240 \mathrm{ft}\). a) What is the initial speed of the water? b) What is the speed of the water when it is at half this height on its way down? c) How long will it take for the water to fall back to its original height from half its maximum height?

A car is traveling due west at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the velocity of the car after \(3.00 \mathrm{~s}\) if its acceleration is \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) due west. Assume the acceleration remains constant. a) \(17.0 \mathrm{~m} / \mathrm{s}\) west b) \(17.0 \mathrm{~m} / \mathrm{s}\) east c) \(23.0 \mathrm{~m} / \mathrm{s}\) west d) \(23.0 \mathrm{~m} / \mathrm{s}\) east e) \(11.0 \mathrm{~m} / \mathrm{s}\) south

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). What maximum height above the projection point does it reach?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free