Vectors with a dose of arithmetic :3

[Lexadis]

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

 

[paras02]

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.

 

[dauto]

(P+Q) P= P² + PQ

15 P = P² + 56

Solve that quadratic equation.

 

[Lexadis]

Oh thank you!
I got the final result as P = 7/8, Q = 7/8. Thank you :3

 

[SteamKing]

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.

 

[dauto]

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 ?