#11.91

x = (t+1)²        y = 4(t+2)^-2
Vx = x' = 2(t+1)           Vy = y' = -8(t+1)^-3
a_x = v'x = 2                a_y = v'y = 24(t+1)^-4

Now solve for (t+1)² from the the expression for x:
(t+1)² = x     
Substitute this into the expression for y:
y = 4/x
so xy = 4
This is the equation of a rectangular hyperbola, so

A) at t = 0
Vx = 2 m/s         Vy = - 8m/s
√[2² + (-8)²] = 8.25 m/s
θ = tan^-1 (-8/2) = -76^o
So v = 8.25 m/s < 76^o
a_x = 2 m/s²      a_y = 24 m/s²
a = √[(2²) + (24²)] = 24.1 m/s²
θ = tan^-1(24/2) = 85.2^o
so a = 24.1 m/s² < 85.2^o

 

B) at t = 1/2 s
Vx = 3 m/s       Vy = -2.37 m/s
v = √3² + 2.37² = 3.82 m/s
θ = tan^-1(-2.37/3)
 v = 3.82 m/s > 38.3^o
a_x = 2 m/s²     a_y = 4.74 m/s²
a = √2² + 4.74² = 5.14 m/s²
θ = tan^-1(4.74/2) = 67.1^o
so a = 5.14 m/s² > 67.1^o
Note: > and < represent reference angles and not inequalities