If we take due east to be the positive direction, we can write the equation for the position (x) of the particle at time (t) as

x = 9t - (1/2)(2)t^2 = t(9-t)

The 5th second is the one ending at t=5. A graph of the travel in that time is seen here.

When we put t=4 and t=5 into the equation above, we find that the value

x(4) = x(5) = 20 meters

Thus, there is no net travel by the particle in that time period. However, we know the particle is not standing still. The farthest east it will travel is at time t=4.5, at which point

x(4.5) = 20.25 meters

In the 5th second, the particle travels from 20 m east to 20.25 m east and back again, for a

x = 9t - (1/2)(2)t^2 = t(9-t)

The 5th second is the one ending at t=5. A graph of the travel in that time is seen here.

When we put t=4 and t=5 into the equation above, we find that the value

x(4) = x(5) = 20 meters

Thus, there is no net travel by the particle in that time period. However, we know the particle is not standing still. The farthest east it will travel is at time t=4.5, at which point

x(4.5) = 20.25 meters

In the 5th second, the particle travels from 20 m east to 20.25 m east and back again, for a

**total distance traveled**of**0.50 meters**. As stated above, its net displacement in that period is zero.