Question

an object is moving in a straight line from a starting point. the distance, in meters, from the starting point at selected times, in seconds, is given in the table. if the pattern is consistent, which of the following statements about the rate of change of the rates of change of distance over time is true? time (seconds) 1 3 6 11 distance (meters) 1 9 21 41 a the rate of change of the rates of change is 0 meters per second, and the object is neither speeding up nor slowing down. b the rate of change of the rates of change is 0 meters per second per second, and the object is neither speeding up nor slowing down. c the rate of change of the rates of change is 4 meters per second, and the object is neither speeding up nor slowing down. d the rate of change of the rates of change is 4 meters per second per second, and the object is speeding up.

Explanation:

Step1: Calculate first - order rate of change (speed)

The rate of change of distance with respect to time (speed) between two points $(t_1,d_1)$ and $(t_2,d_2)$ is given by $v=\frac{d_2 - d_1}{t_2 - t_1}$.
For $(t_1 = 1,d_1 = 1)$ and $(t_2 = 3,d_2 = 9)$: $v_1=\frac{9 - 1}{3 - 1}=\frac{8}{2}=4$ m/s.
For $(t_1 = 3,d_1 = 9)$ and $(t_2 = 6,d_2 = 21)$: $v_2=\frac{21 - 9}{6 - 3}=\frac{12}{3}=4$ m/s.
For $(t_1 = 6,d_1 = 21)$ and $(t_2 = 11,d_2 = 41)$: $v_3=\frac{41 - 21}{11 - 6}=\frac{20}{5}=4$ m/s.

Step2: Calculate second - order rate of change (acceleration)

The rate of change of the rate of change (acceleration) between two speed - values $v_1$ and $v_2$ is given by $a=\frac{v_2 - v_1}{t_{v2}-t_{v1}}$. Since the speed is constant ($v_1 = v_2=v_3 = 4$ m/s), the rate of change of the rates of change (acceleration) $a=\frac{4 - 4}{t_{2}-t_{1}} = 0$ m/s². When the acceleration is 0, the object is neither speeding up nor slowing down.

Answer:

B. The rate of change of the rates of change is 0 meters per second per second, and the object is neither speeding up nor slowing down.