/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Jason drives due west with a spe... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 9

Jason drives due west with a speed of \(35.0 \mathrm{mi} / \mathrm{h}\) for 30.0 min, then continues in the same direction with a speed of $60.0 \mathrm{mi} / \mathrm{h}\( for \)2.00 \mathrm{h},$ then drives farther west at \(25.0 \mathrm{mi} / \mathrm{h}\) for \(10.0 \mathrm{min} .\) What is Jason's average velocity for the entire trip?

Short Answer

Expert verified
Answer: The average velocity of Jason's entire trip is approximately 53.1 mi/h.

Step by step solution

01

Convert time durations to hours

To be consistent, we will work with time in hours for all three parts of the trip. Duration 1: 30.0 min = 30.0/60.0 h = 0.50 h Duration 2: 2.00 h (already in hours) Duration 3: 10.0 min = 10.0/60.0 h = 0.17 h (rounded to two decimal places)
02

Calculate distance traveled in each part

We can use the formula: distance = speed 脳 time Distance 1: \(35.0\,\mathrm{mi}/\mathrm{h} \times 0.50\,\mathrm{h} = 17.5\,\mathrm{mi}\) Distance 2: \(60.0\,\mathrm{mi}/\mathrm{h} \times 2.00\,\mathrm{h} = 120.0\,\mathrm{mi}\) Distance 3: \(25.0\,\mathrm{mi}/\mathrm{h} \times 0.17\,\mathrm{h} = 4.25\,\mathrm{mi}\)
03

Calculate the total distance of the trip

By adding the distances traveled in each part, we get the total distance. Total distance: \(17.5 + 120.0 + 4.25 = 141.75\,\mathrm{mi}\)
04

Calculate the total time duration of the trip

Adding the time durations of each trip part: Total time: \(0.50 + 2.00 + 0.17 = 2.67\,\mathrm{h}\)
05

Calculate the average velocity

To find the average velocity, divide the total distance by the total time. Average velocity: \(\frac{141.75\,\mathrm{mi}}{2.67\,\mathrm{h}} \approx 53.1\,\mathrm{mi}/\mathrm{h}\) (rounded to one decimal place) Jason's average velocity for the entire trip is about \(53.1\, \mathrm{mi}/\mathrm{h}\).

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