Calculating Average Velocity: Solving a Train Motion Problem in Physics

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Homework Help Overview

The problem involves calculating the average velocity of a train moving in different directions over specified time intervals. The subject area is kinematics, focusing on vector addition and average velocity calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate total displacement and time, with some questioning the distinction between average speed and average velocity. There are attempts to clarify the vector components involved in the displacement calculations.

Discussion Status

Some participants have provided guidance on ensuring the correct use of angles and components in vector addition. There is an ongoing exploration of the calculations and potential misunderstandings regarding average velocity.

Contextual Notes

Participants are addressing the original poster's calculations and assumptions, noting the importance of correctly interpreting vector components and the resultant displacement. There is an acknowledgment of the need for clarity in distinguishing between average speed and average velocity.

Dan Feerst
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Homework Statement



A train at a constant 42.0 km/h moves east for 33 min, then in a direction 64.0° east of due north for 25.0 min, and then west for 69.0 min. What are the (a) magnitude (in km/h) and (b) angle (relative to north, with east of north positive and west of north negative) of its average velocity during this trip?

Homework Equations


The way I inturpreted the problem I needed to first find the time in hours and the displacement in kilometers for each vector then I added the displacements over the times. here is the work I did.


The Attempt at a Solution


First I calculated the total time. Since time isn't a vector I added everything.
33'*/60'=.550h
25'*/60'=.417h
69'/60'=1.15h
Sum = 2.12h

Next I found the total distance
42km/h*.550h=23.1km
42km/h*.417h=17.5km
42km/h*1.15h=48.3km

then I added the magnitudes of the three vectors taking the magnitude of the x coordinate for the second
r=(23.1km)+(17.5km *Cos(26))-(48.3km)= -9.47

Finally I divided displacement over time to get an unconvincing and completely wrong answer of -4.47

What did I do wrong.
 
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For average velocity, it should be sufficient to consider the total displacement (resultant vector of the sum of all displacement vectors). Correct me if I'm wrong.
 
You need to sort out the difference between average speed and average velocity.
 
.
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Dan Feerst said:

The Attempt at a Solution


First I calculated the total time. Since time isn't a vector I added everything.
33'*/60'=.550h
25'*/60'=.417h
69'/60'=1.15h
Sum = 2.12h
Looks good so far.

Next I found the total distance
42km/h*.550h=23.1km
42km/h*.417h=17.5km
42km/h*1.15h=48.3km

then I added the magnitudes of the three vectors taking the magnitude of the x coordinate for the second
r=(23.1km)+(17.5km *Cos(26))-(48.3km)= -9.47
This is just the x-coordinate of r. You need to also:

1. Use the angle w.r.t. the x-axis when you take the cosine to get the x-component.
2. Find the y-coordinate of r, so that you get the complete displacement vector r.
 

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