Should I use arccosine instead?

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Homework Help Overview

The discussion revolves around determining the angle of a resultant velocity vector in a physics problem involving two masses and their respective velocities. The context includes components of motion and trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of angle θ using the tangent function and the implications of negative angles. There is a consideration of how to find the angle based on the x and y components and the potential use of arccosine and arcsine functions.

Discussion Status

Participants are exploring various methods to determine the angle, including the use of arctangent and the limitations of trigonometric functions in identifying angles in different quadrants. Some guidance has been provided regarding the use of atan2 for resolving ambiguities.

Contextual Notes

There is a mention of the original poster's inexperience with math and science, which may influence their understanding of the concepts discussed. The discussion also highlights the limitations of the arccosine and arcsine functions in determining angles based on given ratios.

Uxas
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Homework Statement
On a freeway, a truck of 3500 kg collides with an automobile of 1500 kg that is trying to cut diagonally
across the path of the truck. Just before the collision, the truck was traveling due north at 70 km/h, and the
automobile was traveling at 30° west of north at 100 km/h. After the collision, the vehicles remain joined
together.
(a) What is the velocity (magnitude and direction) of the joined vehicles immediately after the collision?
(b) How much kinetic energy is lost during the collision?

IDK the exact angle or the direction of the wreckage, I am not sure about my answers.
Relevant Equations
m1v1+m2v2=(m1+m2)v'
Xcomp: [-(1500kg)(100km/h)(sin(30))+(3500kg)(70km/h)(cos(90))]/5000kg = -15km/h
ycomp: [(1500kg)(100km/h)(cos(30))+(350(70km/h)(sin(90))]/5000kg = 74.98km/h
 
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Those numbers look right. How do you find the angle? If the answer is speed v at θ W of N, what would the x and y components be?
 
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haruspex said:
Those numbers look right. How do you find the angle? If the answer is speed v at θ W of N, what would the x and y components be?
tan θ = 74.98/-15 right? then to find θ , arctan(74.98/-15)=θ, the angle is somehow negative and does not match my prediction that the wreckage will be heading to the 2nd quadrant.
 
Uxas said:
tan θ = 74.98/-15 right? then to find θ , arctan(74.98/-15)=θ, the angle is somehow negative and does not match my prediction that the wreckage will be heading to the 2nd quadrant.
What does the tangent function give you for angles in the second quadrant?
What does the tangent function give you for angles in the fourth quadrant?

If you know the tangent of an angle and you want the angle, you will need to disambiguate. The arctangent function on your calculator will only return angles in the first and fourth quadrants (-90 degrees to +90 degrees).

[In computer programming we sometimes use the atan2 function. It takes two arguments -- an x and a y component and handles the ambiguity that way. Handy little function, it also deals with the case where x is zero and avoids the divide by zero problem]
 
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jbriggs444 said:
What does the tangent function give you for angles in the second quadrant?
What does the tangent function give you for angles in the fourth quadrant?

If you know the tangent of an angle and you want the angle, you will need to disambiguate. The arctangent function on your calculator will only return angles in the first and fourth quadrants (-90 degrees to +90 degrees).

[In computer programming we sometimes use the atan2 function. It takes two arguments -- an x and a y component and handles the ambiguity that way. Handy little function, it also deals with the case where x is zero and avoids the divide by zero problem]
I did not know that thanks! I am just starting with math and science, not even in college yet. Should I use arccosine instead?
 
I think I got the right answer. Thanks everyone.
 
Uxas said:
I did not know that thanks! I am just starting with math and science, not even in college yet. Should I use arccosine instead?
The arc cosine suffers from the same problem. If all you know is the cosine of an angle (the ratio of the x component to the hypotenuse) then you cannot distinguish between an angle above the x-axis in the second quadrant or the same angle below the x-axis in the third quadrant.

The arc cosine function on your calculator returns angles in the first and second quadrants and ignores the possibility of the third and fourth.

The arc sine suffers from the same problem. If all you know is the sine of an angle (the ratio of the y component to the hypotenuse) then you cannot distinguish between an angle above the x-axis in the first quadrant and the same angle above the x-axis in the second quadrant.

The arc sine function on your calculator returns angles in the first and fourth quadrant and ignores the possibility of the second and third.

As alluded to already, the arc tangent suffers from this problem. If all you know is the tangent of an angle (the ratio of the y component to the x component) then you cannot distinguish between an angle below the x-axis in the fourth quadrant and the same angle above the x-axis in the second quadrant.

The arc tangent function on your calculator returns angles in the first and fourth quadrant and ignores the possibility of the second and third.

But you already know the angle is in the second quadrant.

Yes, the arc cosine function would work in this case. But it is easier to just use the angle you got from the arc tangent function. Instead of a [clockwise] angle below the x-axis in the fourth quadrant, use it as the [clockwise] angle above the x-axis in the second quadrant.
 
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