A particle at rest leaves the origin with its velocity increasing with time according to v(f) = 3.4r m/s. At 5.0 s, the particle's velocity starts decreasing according to [17.0 - 1.4(t - 5.0)] m/s. This decrease continues until t = 11.0 s, after which the particle's velocity remains constant at 8.6 m/s. (a) What is the acceleration of the particle as a function of time? (Use the following as necessary: t. Do not use other variables, substitute numeric values. Assume that a is in m/s and t is in seconds. Do not include units in your answer.) 3.4 f 110s (b) What is the position of the particle (in m) att = 2.0 s, t = 7.0 s, and t = 12.0 s? x(2.0 s) = 6.8 m x(7.0 5) = 87.7 X m x(12.0 s) = 244.8 X m