While the bridge is closing with a constant rotation of .5 rad/s, a man runs along the roadway such that when d=10ft, he is running outward from the center at 5 ft/s with an acceleration of 2 ft/s^2, both measured relative to the roadway. Determine his velocity and acceleration at this instant.
V(a) = V(o) + Omega X r(a/o) + v(a/o)
A(a) = A(o) + Omega' X r(a/o) + Omega X (Omega X r(a/o)) + 2Omega X V(a/o) + A(a/o)
The Attempt at a Solution
V(o) = 0, Omega = .5k, Omega' = 0, r(a/o) = 10j, V(a/o) = -5j, A(a/o) = -2j
Lets do velocity first
Omega X r(a/o) = -5i
V(a) = 0 + -5i + -5j = -5i - 5j ft/s
Omega' X r(a/o) = 0
Omega X (Omega X r(a/o)) = 2.5j
2Omega X v(a/o) = 5j
A(a) = 0 + 0 + 2.5j + 5j + -2j = 5.5j ft/s^2
Shouldn't the velocity and acceleration be in the same direction.