x = (t+1)2 y = 4(t+2)-2
Vx = x' = 2(t+1) Vy = y' = -8(t+1)-3
ax = v'x = 2 ay = 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) = -76o
So v = 8.25 m/s < 76o
ax = 2 m/s2 ay = 24 m/s2
a = √[(22) + (242)] = 24.1 m/s2
θ = tan-1(24/2) = 85.2o
so a = 24.1 m/s2 < 85.2o
B) at t = 1/2 s
Vx = 3 m/s Vy = -2.37 m/s
v = √3
2 + 2.37
2 = 3.82 m/s
θ = tan
-1(-2.37/3)
v = 3.82 m/s > 38.3
o
a
x = 2 m/s
2 a
y = 4.74 m/s
2
a = √2
2 + 4.74
2 = 5.14 m/s
2
θ = tan
-1(4.74/2) = 67.1
o
so a = 5.14 m/s
2 > 67.1
o
Note: > and < represent reference angles and not inequalities