Resultant velocity from easterly and northerly components

[rhxoehwhfh]

The easterly and northerly components of a car’s velocity are 24 m/s and 30 m/s, respectively. In what direction and with what speed is the car moving? In other words, what is car’s velocity. Hint: this question requires input from Trigonometry for the right angle triangle.
Thus, the car was moving with the velocity of 38 m/s [E510N] (or, to the north of due east).


I have no clue with this one ... can someone please help me?..

 

[Feldoh]

Velocity is a vector and we're given the components of the velocity vector. How do we find the magnitude of a vector?

 

[rhxoehwhfh]

I don't know T.T

 

[Feldoh]

Ok, forget about velocity for a second. Say we move 24m east, then move 30m north what's the total distance we have gone? On a coordinate system say we started at (0,0) now we're at (24,30) right? How do we find the total distance?

 

[Antineutron]

what about thinking it as pathagorean theorem. Hypotenuse?

 

[rhxoehwhfh]

Is there like a formula or something...? because its really confusing..

 

[EugP]

The magnitude of a vector is the square root of the sum of its components squared. I'm sure you're confused by now, I know I'd be, so here's a usefull formula you should remember:

\vec v = \sqrt{v_x^2 + v_y^2}

 

[Feldoh]

We can use the distance formula which (kinda what antineutron is saying) will give us the distance.

Magnitude: |\vec v| = \sqrt{v_x^2+v_y^2}

Direction: \theta = \arctan{\frac{v_y}{v_x}}

 

[rhxoehwhfh]

ok i thinkk i get it ... Thank youvery much everyone. ^^