Question

time (seconds) 1 3 6 11
distance (meters) 1 9 21 41
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 rate of change of distance over time is true?
a the rate of change of the rate of change is 0 meters per second, and the object is neither speeding up nor slowing down
b the rate of change of the rate 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 rate of change is 4 meters per second, and the object is neither speeding up nor slowing down
d the rate of change of the rate of change is 4 meters per second per second, and the object is speeding up

Explanation:

Step1: Find first - order differences (rate of change of distance, velocity)

The time values are \(t = 1,3,6,11\) and distance values are \(d = 1,9,21,41\).

First, calculate the differences in distance (\(\Delta d\)) and differences in time (\(\Delta t\)):

• Between \(t = 1\) and \(t = 3\): \(\Delta t_1=3 - 1 = 2\), \(\Delta d_1=9 - 1 = 8\). The rate of change (velocity) \(v_1=\frac{\Delta d_1}{\Delta t_1}=\frac{8}{2}=4\) m/s.
• Between \(t = 3\) and \(t = 6\): \(\Delta t_2=6 - 3 = 3\), \(\Delta d_2=21 - 9 = 12\). The rate of change (velocity) \(v_2=\frac{\Delta d_2}{\Delta t_2}=\frac{12}{3}=4\) m/s.
• Between \(t = 6\) and \(t = 11\): \(\Delta t_3=11 - 6 = 5\), \(\Delta d_3=41 - 21 = 20\). The rate of change (velocity) \(v_3=\frac{\Delta d_3}{\Delta t_3}=\frac{20}{5}=4\) m/s.

Step2: Find second - order differences (rate of change of rate of change, acceleration)

Now, calculate the differences in velocity (\(\Delta v\)) and differences in time (\(\Delta t\)) for the velocity values. The velocity values \(v_1 = 4\), \(v_2 = 4\), \(v_3 = 4\).

• Between \(v_1\) and \(v_2\): \(\Delta v_1=4 - 4 = 0\), \(\Delta t_{v1}\) (the time interval between the time intervals of \(v_1\) and \(v_2\)): The mid - points of the time intervals for \(v_1\) (1 - 3) and \(v_2\) (3 - 6) are \(t_{m1}=2\) and \(t_{m2}=4.5\), so \(\Delta t_{v1}=4.5 - 2 = 2.5\). The rate of change of velocity (acceleration) \(a_1=\frac{\Delta v_1}{\Delta t_{v1}}=\frac{0}{2.5}=0\) m/s².
• Between \(v_2\) and \(v_3\): \(\Delta v_2=4 - 4 = 0\), \(\Delta t_{v2}\) (mid - point of 3 - 6 is 4.5, mid - point of 6 - 11 is 8.5), so \(\Delta t_{v2}=8.5 - 4.5 = 4\). The rate of change of velocity (acceleration) \(a_2=\frac{\Delta v_2}{\Delta t_{v2}}=\frac{0}{4}=0\) m/s².

Since the rate of change of the rate of change (acceleration) is 0 meters per second per second, and when acceleration is 0, the object's velocity is constant, so 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