Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A canoeist paddled upstream at 10 meters per minute, turned around, and drifted downstream at 15 meters per minute. If the distance traveled in each direction was the same, and the time spent turning the canoe around was negligible, what was the canoeist's average speed over the course of the journey, in meters per minute? A 11.5 B 12 C 12.5 D 13 E 13.5

Short Answer

Expert verified
B 12

Step by step solution

01

Define Variables

Let the distance traveled in each direction be denoted as \( d \) meters and the times spent upstream and downstream be \( t_u \) and \( t_d \) respectively.
02

Set Up Equations for Time

Time can be calculated by dividing distance by speed. So, \( t_u = \frac{d}{10} \) (upstream) and \( t_d = \frac{d}{15} \) (downstream).
03

Find Total Time

The total time for the entire journey is the sum of the times spent upstream and downstream: \( t_{total} = t_u + t_d = \frac{d}{10} + \frac{d}{15} \).
04

Combine Time Fractions

Find a common denominator to combine the fractions: \( t_{total} = \frac{3d}{30} + \frac{2d}{30} = \frac{5d}{30} = \frac{d}{6} \).
05

Calculate Average Speed

Average speed is the total distance divided by total time. The total distance traveled is \( 2d \) (going and returning): \( v_{avg} = \frac{2d}{\frac{d}{6}} = 2d \times \frac{6}{d} = 12 \).
06

Choose the Correct Option

The average speed of the canoeist over the whole journey is 12 meters per minute, which corresponds to option B.

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!

Key Concepts

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

Distance Calculation
In this exercise, the key is to understand the concept of distance calculation. Distance is how far an object has traveled in a straight line. Here, the canoeist paddles upstream and downstream, covering the same distance in both directions. Distance in each direction is denoted by the variable \( d \). This helps simplify calculations using one variable. For total distance, you sum up the distance for both upstream and downstream: total distance = \( d + d = 2d \). It's crucial always to start with a clear understanding of what's being calculated to solve the problem correctly.
Speed and Velocity
Speed and velocity might sound similar but are distinct. Speed is how fast an object moves, while velocity includes direction. In this case, the canoeist's speeds are different for upstream and downstream. Upstream paddle speed = 10 meters per minute and downstream drift speed = 15 meters per minute. These speeds are crucial for calculating time.
Time taken for each leg of the journey is given by:\[ t_u = \frac{d}{\text{speed upstream}} = \frac{d}{10} \]
\[ t_d = \frac{d}{\text{speed downstream}} = \frac{d}{15} \]
Always separate different speeds and times before finding the total or average speed. This organizes the steps clearly and makes solving easier.
Time Management
Understanding how to manage and calculate time is key in this exercise. Total travel time is the sum of the times for upstream and downstream: \( t_{total} = t_u + t_d \). Simplify using a common denominator: \[ t_{total} = \frac{d}{10} + \frac{d}{15} = \frac{3d}{30} + \frac{2d}{30} = \frac{5d}{30} = \frac{d}{6} \]
With total time in hand, you can determine average speed. Average speed is found by dividing total distance by total time: \[ v_{avg} = \frac{2d}{\frac{d}{6}} = 2d \times \frac{6}{d} = 12 \text{ meters per minute} \] Always breaking down the total time calculation helps in efficiently finding the solution.

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

Magnabulk Corp sells boxes holding \(d\) magnets each. The boxes are shipped in crates, each holding \(b\) boxes. What is the price charged per magnet, in cents, if Magnabulk charges \(m\) dollars for each crate? A. $$\frac{100 b d}{m}$$ B. $$\frac{100 m}{b d}$$ C. $$\frac{b d}{100 m}$$ D. $$\frac{m}{b d}$$ E. $$\frac{b d}{m}$$

The moon revolves around the earth at a speed of approximately 1.02 kilometers per second. This approximate speed is how many kilometers per hour? A 60 B 61.2 C 62.5 D 3,600 E 3,672

Truck \(\mathrm{X}\) is 13 miles ahead of Truck \(\mathrm{Y}\), which is traveling the same direction along the same route as Truck X. If Truck \(\mathrm{X}\) is traveling at an average speed of 47 miles per hour and Truck \(Y\) is traveling at an average speed of 53 miles per hour, how long will it take Truck Y to overtake and drive 5 miles ahead of Truck X? A 2 hours B 2 hours 20 minutes C 2 hours 30 minutes D 2 hours 45 minutes E 3 hours

Three hundred students at College \(Q\) study a foreign language. Of these, 110 of those students study French and 170 study Spanish. If at least 90 students who study a foreign language at College \(Q\) study neither French nor Spanish, then the number of students who study Spanish but not French could be any number from A 10 to 40 B 40 to 100 C 60 to 100 D 60 to 110 E 70 to 110

Of the 150 employees at company \(X, 80\) are full-time and 100 have worked at company \(X\) for at least a year. There are 20 employees at company \(X\) who aren't full-time and haven't worked at company \(X\) for at least a year. How many fulltime employees of company \(X\) have worked at the company for at least a year? A 20 B 30 C 50 D 80 E 100

See all solutions

Recommended explanations on English 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