Help a highschooler with physics homework?

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Homework Help Overview

The discussion revolves around a physics problem involving vectors, specifically the velocity of a person riding a bike represented in terms of I and J axes. The original poster seeks to determine both the magnitude of the velocity and the total distance traveled over a specified time period.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers drawing a right triangle to visualize the velocity components and questions whether the direction is northeast based on the positive values of the I and J components. They express uncertainty about calculating the total distance traveled after determining the magnitude of the velocity.

Discussion Status

Some participants confirm the approach to finding the magnitude of the velocity and clarify the terminology regarding speed. There is an acknowledgment of the relationship between speed and distance, leading to further calculations by the original poster.

Contextual Notes

The original poster mentions confusion regarding the total distance calculation and expresses intent to seek further assistance in future posts.

phantomcow2
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We just started vectors today and introduced the I and J axes. He insists we use I and J instead of X and Y, that's okay.

So basically here is the question asked:

A person riding a bike has a velocity v=32m/s I + 12m/s J. Determine the magnitude of hte velocity and hte bikes direction. Determine the total distance traveled by the person on the bike after 120 seconds.

SO my first step I would think is to draw quadrant 1 of an XY plane. Is the direction northeast? I would imagine this is the case because there is no mention of any negative velocities, so quadrant 1 is + in both axes.
TO determine the magnitude of velocities...
Do I simply draw a right triangle, with point (0,0) as the vertex with a 90 degree? Move up 32 in the I, and 12 in hte J?
Then use A^2+B^2=C^2 to find the hypotenuse? The hypotenuse being the magnitude?

I THINK that part I have right. BUt it is the total distance that stumps me, can somebody give some insight?
 
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You're correct about how to find the magnitude of the velocity.

Keep in mind that "the magnitude of the velocity" is a fancy way of saying "speed." Once you've found the speed, just multiply it by time to get distance.

- Warren
 
oooh, that threw me off. Speed makes it all come together in my mind :). Well now, I get 34.18m/s as a magnitude of velocity. So x 120 and 4101.6m

Thanks! I will have more questions to come tomorrow I am SURE
 
Excellent homework post. By the way, there's a forum dedicated to those somewhere around here...
 

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