Question

If vx = 7.50 units and vy = -6.10 units, determine (a)the magnitude and (b)direction of v⃗ .

Answer 1

Answer:

(a) v= 9.66 units

(b) α = -39.12°

α = 39.12° below the positive axis of the x

Explanation:

Data

vx = 7.50 units

vy = -6.10 units

(a)Calculation of the magnitude of v

$v=\sqrt{(v_{x} )^{2} +(v_{y}) ^{2} }$

$v= \sqrt{(7.5)^{2}+(-6.1)^{2} }$

v= 9.66 units

(b)Calculation of the direction of v

$\alpha = tan^{-1}( \frac{v_{y} }{v_{x} } )$

$\alpha = tan^{-1}( \frac{ -6.1 }{7.5 } )$

α = -39.12°

α = 39.12° below the positive axis

Answer 2

The magnitude of a vector is defined as the square root of its components squared.
 We have then that the magnitude is:
 v = root ((7.50) ^ 2 + (- 6.10) ^ 2)
 v = 9.67 units
 The direction of the vector is:
 x = Atan (vy / vx)
 x = Atan ((- 6.10) / (7.50))
 x = -39.12 degrees (measured from the x axis)
 answer:
 v = 9.67 units
 x = -39.12 degrees (measured from the x axis)