Respuesta :
Answer:
a) a = [tex]- \frac{v_1}{T}[/tex] i ^ +[tex]\frac{v_3 - v_2}{T}[/tex] j^, b) r = 2 v₃ T j ^, c) v = -v₁ i ^ + (2 v₃ - v₂) j ^
Explanation:
This is a two-dimensional kinematics problem
a) Let's find the acceleration of the body, for this let's use a Cartesian coordinate system
X axis
initial velocity v₀ₓ = v₁ for t = 0, velocity reaches vₓ = 0 for t = T, let's use
vₓ = v₀ₓ + aₓ t
we substitute
for t = T
0 = v₁ + aₓ T
aₓ = - v₁ / T
y axis
the initial velocity is [tex]v_{oy}[/tex] = v₂ at t = 0 s, for time t = T s the velocity is v_{y} = v₃
v₃ = v₂ + a_{y} T
a_{y} = [tex]\frac{v_3 - v_2}{T}[/tex]
therefore the acceleration vector is
a = [tex]- \frac{v_1}{T}[/tex] i ^ +[tex]\frac{v_3 - v_2}{T}[/tex] j^
b) the position vector at t = 2T, we work on each axis
X axis
x = v₀ₓ t + ½ aₓ t²
we substitute
x = v₁ 2T + ½ (-v₁ / T) (2T)²
x = 2v₁ T - 2 v₁ T
x = 0
Y axis
y = [tex]v_{oy}[/tex] t + ½ a_{y} t²
y = v₂ 2T + ½ [tex]\frac{v_3 - v_2}{T}[/tex] 4T²
y = 2 v₂ T + 2 (v₃ -v₂) T
y = 2 v₃ T
the position vector is
r = 2 v₃ T j ^
c) the velocity vector for t = 2T
X axis
vₓ = v₀ₓ + aₓ t
we substitute
vₓ = v₁ - [tex]\frac{v_1}{T}[/tex] 2T = v₁ - 2 v₁
vₓ = -v₁
Y axis
[tex]v_{y}[/tex] = v_{oy} + a_{y} t
v_{y} = v₂ + [tex]\frac{ v_3 - v_2}{T}[/tex] 2T
v_{y} = v₂ + 2 v₃ - 2v₂
v_{y} = 2 v₃ - v₂
the velocity vector is
v = -v₁ i ^ + (2 v₃ - v₂) j ^
