Vectors with a dose of arithmetic :3

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Homework Help Overview

The problem involves two velocities with given conditions regarding their magnitudes and resultant. The context is within the subject area of vectors and arithmetic, specifically focusing on the relationships between the magnitudes and the angle between them.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss transforming the problem statements into equations and explore various algebraic methods to solve for the velocities. There is an attempt to derive relationships between the magnitudes using equations involving sums and products. Some participants suggest using specific equations to find the difference between the magnitudes.

Discussion Status

The discussion includes various approaches to solving for the velocities, with some participants offering algebraic methods and others expressing confusion over the notation used for the results. There is no explicit consensus on the final values of the velocities, and multiple interpretations of the results are being explored.

Contextual Notes

Participants note potential confusion regarding the notation of the results, particularly concerning whether the values represent fractions or whole numbers. There is also mention of the need for clarity in the representation of the velocities.

Lexadis
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Homework Statement


Two velocities acting at a particular point are such that:
  • The sum of their respective magnitudes is 15m/s
  • The product of their respective magnitudes are 56m2/s2
  • The resultant is 13m/s.
    Find the two velocities and the angle between them.

Homework Equations


R^2 = P^2 + Q^2 + 2PQcos\Theta

The Attempt at a Solution



At first I tried to change the statements into equations:
P + Q = 15m/s
P * Q = 56m^2/s^2
R = 13m/s

And then I used the binomial expressions knowledge to try solving it:
(P+Q)^2 = P^2 + 2PQ+Q^2
15^2 = P^2 + 2*56 + Q^2
225 = P^2 + Q^2 + 112
P^2 + Q^2 = 225 - 112
P^2 + Q^2 = 113

I then substituted the value obtained above for the following equation:
R^2 = P^2 + Q^2 + 2PQcos\Theta
13^2 = 113 + 2 X 56 X cos\Theta
169 - 113 = 112cos\Theta
56 = 112cos\Theta
cos\Theta = 56/112
cos\Theta = 1/2
\Theta = 60°

Through this I could find the value of the angle between them. And also, that P^2 + Q^2 = 113. But I can't seem to find the separate values for P and Q. Any ideas? Thank you :3
 
Last edited:
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USE THE EQUATION :-
(P-Q)2 = (P+Q)2 - 4PQ
To find the difference between their magnitudes, after then you can simply use elimination method to solve the two equations you got.
 
(P+Q) P= P2 + PQ

15 P = P2 + 56

Solve that quadratic equation.
 
Oh thank you!
I got the final result as P = 7/8, Q = 7/8. Thank you :3
 
Lexadis said:
Oh thank you!
I got the final result as P = 7/8, Q = 7/8. Thank you :3

I think you mean P = (7, 8) m/s and Q = (7, 8) m/s

Writing P and Q the way you did makes it look like a fraction, which is confusing.
 
Lexadis said:
Oh thank you!
I got the final result as P = 7/8, Q = 7/8. Thank you :3

Don't you mean either P=7 and Q=8 or P=8 and Q=7 ?
 

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