function: = = . Next, we set the velocity equal to
zero. The solutions are: t = ±3. We can ignore the negative solution because t must be positive.
Next, we check the sign of the velocity on either side of t = 3. When 0 < t < 3, the velocity is
positive, so the particle is moving to the right. When t > 3, the velocity is negative, so the
particle is moving to the left. Therefore, the particle is changing direction at t = 3.
- Distance is 2 + sin^2 4 ≈ 2.573
In order to find the distance that the particle travels, we need to look at the position of the
particle at t = 0 and at t = 2. We also need to see if the particle changes direction anywhere on
the interval between the two times. If so, we will need to look at the particle’s position at those
“turning points” as well. The way to find out if the particle is changing direction is to look at
the velocity of the particle, which we find by taking the derivative of the position function. We
get = v(t) = 4 sin (2t) cos (2t). If we set the velocity equal to zero, we get t = and t = .
Actually, there are an infinite number of solutions, but remember that we are restricted to 0 < t
< 2. Next, we check the sign of the velocity on the intervals 0 < t < , < t < , and < t <
- When 0 < t < , the velocity is positive, so the particle is moving to the right. When < t
< , the velocity is negative, so the particle is moving to the left. When < t < 2, the velocity
is positive, so the particle is moving to the right. Now we look at the position of the particle on
the interval. At t = 0, the particle’s position is: x = sin^2 (0) = 0. At t = , the particle’s position
is x = sin^2 = 1. At t = , the particle’s position is: x = sin^2 (π) = 0. And at t = 2, the
particle’s position is x = sin^2 (4). Therefore, the particle travels a distance of 1 on the interval 0
< t < , a distance of 1 in the other direction on the interval < t < , and a distance of
sin^2 (4) on the interval < t < 2. Therefore, the total distance that the particle travels is 2 +
sin^2 4 ≈ 2.573.
