What is the average velocity of the train during this run?

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Homework Help Overview

The discussion revolves around calculating the average velocity of a train that travels in multiple directions over a specified time period. The problem involves vector addition and understanding the relationship between displacement and average velocity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the average velocity using displacement vectors and the total time of travel. There are questions about the correct interpretation of velocity in terms of direction and magnitude, as well as unit conversions.

Discussion Status

Participants are actively engaging with the problem, clarifying the definition of average velocity as a vector and discussing the necessary calculations for both magnitude and direction. There is no explicit consensus yet, as different interpretations of the calculations are being explored.

Contextual Notes

There are mentions of unit conversions from miles per hour to miles per minute, and the need to consider both the magnitude and direction of the average velocity vector. Some participants question the necessity of calculating the angle with respect to the x-axis.

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A train moving at an essentially constant speed of 60 miles/hr moves eastward for 40 min, then in a direction 45 degrees east of north for 20 min, and finally westward for 50 min. What is the average velocity of the train during this run?

So average velocity is [itex]\frac{\Delta r}{\Delta t}[/itex]. I am starting at the origin, and have three vectors a , b , and c .

[itex]a = 40i[/itex]
[itex]b = (10\sqrt{2}) i + (10\sqrt{2}) j[/itex]
[itex]c = -50i[/itex]

So the resultant vector would be [itex](10\sqrt{2} - 10)i + (10\sqrt{2})j[/itex]. So to find the average velocity would I just do [itex]\frac{(10\sqrt{2} - 10)^{2}) + (10\sqrt{2})^{2})}{110}[/itex]

Thanks
 
Last edited:
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The magnitude of the average velocity is the magnitude of that resultant vector. (I think you mistyped your answer.)

Careful about units. Since you have the time in minutes, you are measuring velocity in miles/minute, not the given miles per hour.

The average velocity is a vector: it has both direction (figure out what angle it makes with the x-axis) and magnitude.
 
Yeah I converted 60 miles/hr to 1 mile/min. So your saying that the average velocity is just [itex](10\sqrt{2} - 10)^{2}) + (10\sqrt{2})^{2})[/itex]? The direction would be [itex]\tan^{-1}(\frac{10\sqrt{2}-10}{10\sqrt{2}})[/itex] so [itex]\theta = 16.3[/itex] degrees north of east?

Thanks
 
Last edited:
No, he is saying that the average velocity is a vector- the displacement vector you calaculated divided by the 110 minutes (in miles per minute of course).

Although Doc Al said "it has both direction (figure out what angle it makes with the x-axis) and magnitude." I don't think you really need to do that. One way to give a vector is to give direction (angle) and magnitude, but component form should be perfectly acceptable.
 
As Halls said, describing the vector in component form is perfectly acceptable. But if you need the magnitude, calculate it properly. If the resultant displacement vector is [itex]R_x i + R_y j[/itex], then its magnitude is [itex]\sqrt(R_x^2 + R_y^2)[/itex]. To find the magnitude of the average velocity, divide by the time.
 

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