What is the Velocity of the Meteor?

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Homework Help Overview

The problem involves calculating the average velocity of a meteor based on its coordinates at two points in time. The coordinates provided are (5.20 km, 1.20 km) at point A and (6.24 km, 0.915 km) at point B, with a time interval of 1.18 seconds. The discussion focuses on finding both the magnitude and direction of the average velocity.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants describe methods for calculating the distance between the two points using the Pythagorean theorem and discuss the components of the coordinates. Some suggest drawing a figure to visualize the problem, while others question the calculations of magnitude and direction.

Discussion Status

There is ongoing exploration of different methods to find the magnitude and direction of the average velocity. Some participants have provided guidance on using inverse tangent for direction, while others are questioning the accuracy of their calculations and the assumptions made in their approaches.

Contextual Notes

Participants express uncertainty about their calculations and the proper use of coordinate systems. There are indications of confusion regarding the correct application of trigonometric functions and the interpretation of the results.

teababy
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Homework Statement


A meteor streaking through the night sky is located with radar. At point A its coordinates are (5.20km, 1.20km), and 1.18 s later its has moved to point B with coordinates (6.24, 0.915).


1. Find the magnitude of its average velocity between these two points.

2. Find the direction of its average velocity between these two points. ( degree is counterwise from +x axis.)

Homework Equations


No equations were given

The Attempt at a Solution



First i added all the x components and squared it. Then I added all the y components and squared it. Took the square root and got 8.2616

I took the 8.2616 and divided by the time of 1.18seconds

But the answer is wrong. :( !


Help please~
Teababy
 
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teababy said:
First i added all the x components and squared it. Then I added all the y components and squared it. Took the square root and got 8.2616

I would try drawing a figure, perhaps on some graph paper, and locate the two given points on the figure. Draw the right triangle that is appropriate for figuring out the distance between the points, and take it from there.
 
to find the distance it covers, you subtract the x coordinates and square the result. do the same for the y coordinates. add the result and then find its square root.
 
Leong said:
to find the distance it covers, you subtract the x coordinates and square the result. do the same for the y coordinates. add the result and then find its square root.

I subtracted x2-x1 and added it with y2-y1 and took the square root

my magnitude came out to be 1.0703 but it was wrong
 
i get 1.0783. then divide it by 1.18 s, then the answer is ... km/s. if you need to express the answer in other unit, then convert the unit.
 
You have to realize that the problem you are dealing with is with a coordinate system. So to find the distance it covers in the x direction you would do as you did and subtract the two. Same for y.

Knowing that, use a triangle and Pythagorean's theorem to find the magnitude.

-Sorry guess you already did all that, I didn't check your answer I thought you forgot to square it.
 
First of all thank you Redbelly98, Leong, and Sheneron for all your help. It's very much appreciated. I got the right answer which came out to be 0.914 =D

now on to the 2nd part:

2. Find the direction of its average velocity between these two points. ( degree is counterwise from +x axis. )

what i did was take the inverse tangent of -0.242/0.881 = 12.41

I also tried -12.41 degree and it was also wrong

:(
 
I would try sketching a figure and locate the two given points on the figure. Draw the right triangle that is appropriate for figuring out the angle.
 
find the inverse tangent is right.
the two values are wrong and irrelevant.
 
  • #10
I sketched a right triangle, and I for the height i added up the y-components and for the x i added up all the x-components. the hypot i put the magnitude.

So to find the angle, does it mean i can use either cos, sin, and tangent?
 
  • #11
Leong said:
find the inverse tangent is right.
the two values are wrong and irrelevant.

oh wait! i posted before i saw your responds. I'll take a look again at my numbers.
 
  • #12
figure attached.
 

Attachments

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  • #13
I did inverse tangent of sum of y's over the sum of x's. Thus: tangent-1 (2.125/11.24) = 10.70 .. am i doing this right?
 
  • #14
I'm sorry I'm new here. How do i approve your pending graph?
 
  • #15
you can't. only the moderators can.
 
  • #16
i think u will just have to wait.
 
  • #17
no it is wrong too.
 
  • #18
teababy said:
I did inverse tangent of sum of y's over the sum of x's. Thus: tangent-1 (2.125/11.24) = 10.70 .. am i doing this right?

It's a mistake to take the sums here. Look at your figure again ... the side along x is no where close to 11 units long. Think more carefully about how to calculate that side length. Then do the same for the y-direction side.
 

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