Word problem: Asking to solve for average velocity and average speed

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Homework Help Overview

The discussion revolves around a word problem involving average speed and average velocity in the context of a harpsichordist driving to a concert. The problem includes calculations related to distance, time, and direction, specifically addressing the impact of traveling at an angle on average speed and velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of average speed and velocity, with some attempting to break down the problem using vector components and the Pythagorean theorem. Questions arise regarding the correct interpretation of the angle and how to apply it to find resultant vectors.

Discussion Status

The discussion is ongoing, with participants seeking clarification on vector addition and the implications of the angle in the problem. Some guidance has been offered regarding the use of triangles and vector components, but confusion remains about specific calculations and the relationships between the vectors.

Contextual Notes

Participants express uncertainty about the calculations involving the angle and the resultant vector, indicating a need for further exploration of vector components and their application in the context of the problem.

slu1986
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1. 1. To get to a concert in time, a harpsichordist has to drive 124 miles in 2.01 hours.

(a) If he drove at an average speed of 53.0 mi/h in a due west direction for the first 1.18 h, what must be his average speed if he is heading 30.0° south of west for the remaining 49.8 min?

(b) What is his average velocity for the entire trip?

2. Equations:

avg velocity = Δr/Δt
avg speed = distance traveled/time of trip


3. I am confused at how to calculate average speed in this problem..do you take the -cos (30.0) 53.0 mi = -45.8 mi for the x component and sin (30.0) 53.0 mi = 26.0 mi for the y component and plug them into the pythagorean theorum = -45.8 mi^2 + 26.0 mi ^2 = 2799.89 mi and take the square root which is = 52.9 mi and divide that by .83 hrs = 63.7 mi/hr b/c 49.8 min is = 0.83 hr Am I doing this wrong? I'm not sure how to calculate the average velocity for this problem. Could someone please explain what I am doing wrong b/c the whole 30 degree angle in the problem throws me off.
 
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Draw a triangle. You have two vectors, one going west and the other going at 30 degrees south of west. You know the first vector and you also know the resultant so with that you can figure out the rest.
 
Jebus_Chris said:
Draw a triangle. You have two vectors, one going west and the other going at 30 degrees south of west. You know the first vector and you also know the resultant so with that you can figure out the rest.

I don't understand what vector value that I'm using the 30 degrees to solve for? Am I using it to get the resultant time or miles/hr? Do I take 53 mi/hr and divide that by .83 hrs and plug that answer in with the sin and cosine of 30.0 degrees to get the x and y components and then plug those into the pythagorean theorum?
 
You don't know the value of the 30 vector, you have to find that out. You know what the resultant vector is. So if you add the components of the first two vectors you get the resultant vector.
 
Jebus_Chris said:
You don't know the value of the 30 vector, you have to find that out. You know what the resultant vector is. So if you add the components of the first two vectors you get the resultant vector.

124 mi/2.01 hrs = 61.7 mi/hr
53.0 mi/0.83 hr =63.8 mi/hr

So would I add 61.7 mi/hr^2 + 63.8 mi/hr^2 = 7877.33 mi/hr and take the square root of this to get = 88.7 mi/hr?

I'm confused..could you explain to me exactly what vectors you are talking about when you say add the two vectors to get the resultant vector
 
Could someone please help me understand how to work this problem?
 

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