a. The mail carrier's position at
.
b. The position remains at
seconds.
c. At t = 6 seconds, the mail carrier returns to the initial position at
.
d. The position stays at
seconds.
From the velocity vs. time graph, we can determine the position vs. time graph by integrating the velocity function.
Given the velocity graph:
a. For the first 2 seconds (from 0 to 2 seconds), the velocity is constant at +2 m/s.
![\[ \text{Position} = \text{Initial Position} + \text{Area under the velocity-time graph}\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/70lx0vd32tz6cpqehme4euivnde9q79w7b.png)
The area under the velocity-time graph during this period (a rectangle) is the velocity (+2 m/s) multiplied by the time interval (2 seconds).
![\[ \text{Area} = \text{Velocity} * \text{Time} = 2 \, \text{m/s} * 2 \, \text{s} = 4 \, \text{m}\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/b9k5zdol9frqxk7pyn3vmg22nnml8uycn5.png)
Therefore, the position at t = 2 seconds is

b. For the next 2 seconds (from 2 to 4 seconds), the velocity is

During this interval, the mail carrier remains stationary; hence, the position does not change from
.
c. For the next 2 seconds (from 4 to 6 seconds), the velocity is
.
The area under the velocity-time graph during this period (a rectangle below the time axis) is the velocity
multiplied by the time interval (2 seconds).
![\[ \text{Area} = \text{Velocity} * \text{Time} = -2 \, \text{m/s} * 2 \, \text{s} = -4 \, \text{m}\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/9dcude4nn8eeccvnbl7qw7fplobyh76ugc.png)
Since the velocity is negative, it means the mail carrier is moving in the negative direction, which implies a decrease in position.
Therefore, at t = 6 seconds, the position is
.
d. For the next 4 seconds (from 6 to 10 seconds), the velocity remains 0 m/s.
During this interval, the mail carrier remains stationary; hence, the position stays at 0 m.
Therefore, based on the position vs. time graph, the position of the mailbox is at 0 m (Option A).