2. Suppose that f is the differentiable function shown in the graph on the right and that the position at time t (sec) of a particle moving along a coordinate axis is s = |f(x) dx meters. Use the graph to (4,6) (10,6) answer parts (a) through (g)- (2.3) -3- 2 4 6 8 10 12 14 16 /13 a. What is the particle's velocity at time t= 4? The velocity is (1 ) because v(t) = (2) b. Is the acceleration of the particle at time t = 12 positive or negative? The acceleration is (3) because a(!) = (4) c. What is the particle's position at time t = 4? The particle's position is (5) because s(!) = (6) d. At what time during the first 18 sec does s have the largest value? At time t= (7) because after that time, the region lies (8) the x-axis. e. Approximately when is the acceleration zero? The acceleration is zero at t= (9) because a(t) = (10) (Type a whole number. Use a comma to separate answers as needed.) f. When is the particle moving toward the origin? Away from the origin? The particle is moving toward the origin between t = and t = since (11) (12) on this interval. The particle is moving away from the origin between t= and t= since (13) is (14) on this interval. g. On which side of the origin does the particle lie at time t = 147 At time t= 14, the particle is to (15) - because (16) - is (17) at time t= 14 sec.g. On which side of the origin does the particle lie at time t = 14? At time t= 14, the particle is to (15) because (16) is (17) at time t = 14 sec. (1) O m/sec (2) (3) O negative 0 [ 1(x ) dx. (4) 0 (5) O sec (6) (7) O m/sec (8) O below (9) 0 m (10) O f(t) (11) O v(t) (12) O negative sec positive dt m/sec 0 [ f( x ) dx . O sec O above m/sec O a(t) positive Om O 0 Om 0 Om O sec O s(t ) Of (t ). Of( x ) dx . Of(t). O df Of(t) O df Of(x ) dx. (13) O a(t) (14) O negative (15) O the right or positive side (16) Of(t) (17) O negative O v(t) O positive O the left or negative side O positive O s(t) Of(x ) dx 0 O dt