The 1-kg collar C is free to slide along the smooth shaft AB. Determine the accleration of collar C if collar A is subjected to an accleration of 1m/s² to the left. Consider the shaft to move in the vertical plane.

1) 3.5 m/s²
2) 9.6 m/s²
3) 10.8 m/s²
4) 7.08 m/s²

Equation of motion:
S F_x^' = ma_x^'   1(9.81)sin45^o = a_C/A + 1(-1cos45^o)
a_C/A = 4.17 m/s²
Relative accleration:
a_C = a_A + a_C/A
a_C = -i + (4.17cos45^o)i - 4.17sin45^oj
a_C = (1.95i - 2.95j) m/s²
Thus, the magnitude of the accleration a_C is

So, the magnitude of accleration is a_C = 3.53 m/s²