Respuesta :
Answer:
A) aₓ = -0.717 m / s² , B) a_{y} = -1.024 m / s², C) x = 11.77 m , D) y = -4.6 m ,
E) d = 12.64 m
Explanation:
For this exercise we will use trigonometry to decompose the acceleration, we are told that
a = 1.25 m /s² with an angle tea = 125º in a clockwise direction
A) cos θ = aₓ / a
aₓ = a cos θ
aₓ = 1.25 cos (-125)
aₓ = -0.717 m / s²
B) sin θ = [tex]a_{y}[/tex] / a
a_{y} = a sin θ
a_{y} = 1.25 sin (-125)
a_{y} = -1.024 m / s²
C) The displacement component
x = v₀ₓ t + ½ aₓ t²
let's calculate
x = 5 3 + ½ (-0.717) 3²
x = 11.77 m
D) the y component of the displacement
y = [tex]v_{oy}[/tex] t + ½ [tex]a_{y}[/tex] t²
in this case the initial velocity is zero
y = 0 + ½ (-1,024) 3²
y = -4.6 m
E) Let's use the Pythagorean theorem
d =√ (x² + y²)
d = √ (11.77² + 4.6²)
d = 12.64 m
