The initial position is given by plugging $t=0$ into the position function: $$s(0) = 0^3 -3(0)^2 + 2 = 2 \text { feet}.$$ Assuming that the motion is taking place along a horizontal line, with the positive direction to the right and the negative direction to the left, then the initial position of the particle is 2 feet to the right of the origin. At time $t=2$ seconds, the particle is at position $$s(2) = 2^3 -3(2)^2 + 2 = 8 - 12 + 2 = -2 \text { feet},$$ so that at time $t = 2$, the particle is 2 feet to the left of the origin.

From this information, one might be tempted to conclude that the particle has traveled a distance of 4 feet during the time interval [0,2]. However, we cannot be certain that the particle did not change direction during this time interval and so that conclusion might be false. One thing that we can conclude is that the displacement from time $t = 0$ to time $t = 2$ is $$s(2) - s(0) = -2 - 2 = -4 \text { feet},$$ meaning that the particle has been displaced 4 feet to the left.

Since $$v(t) = s'(t),$$ we have $$v(t) = 3t^2 - 6t.$$ From this we can note that $$v(0) = 3(0)^2 - 6(0) = 0.$$ This means that the initial velocity of the particle was 0 ft/sec, i.e. the particle was initially at rest.

To find all of the times that the particle is at rest, we need to solve the equation $$v(t) = 0$$ for $t$. We have $$3t^2 - 6t = 0$$ $$3t(t-2) = 0$$ $$3t = 0 \ \ \text {or} \ \ t-2 = 0$$ $$t = 0 \ \ \text {or} \ \ t = 2.$$

Thus the particle was at rest at times $t = 0$ seconds and $t = 2$ seconds.

Using the fact that these are the only times that the particle can change direction, we can now find distance traveled. The total distance traveled by the particle from time $t=0$ to time $t=2$ is

$$|s(2) - s(0)| = |-2 - 2| = |-4| = 4 \text { feet.}$$

For the time interval, $[1,3]$ the situation is a little bit more complicated since the particle changes direction.

The particle is moving to the left when the velocity function is negative, and the particle is moving to the right when the velocity is positive. Since $v(t)$ can be written as $$v(t) = 3t(t-2)$$ we can conclude that the velocity is negative when $0<t<2$ and positive when $t>2$. Thus the particle begins at rest, then moves to the left between $t = 0$ and $t = 2$, comes to rest again at $t=2$ and moves to the right for $t>2$.

To find the total distance the particle traveled from time $t = 1$ to time $t = 3$, we would need to take into consideration the fact that it changes direction at $t = 2$: $$\text {distance} \ = |s(2) - s(1)| + |s(3) -s(2)|$$ $$= |-2-0|+|2-(-2)| = 2 + 4 = 6 \text { feet}.$$ Notice that the absolute value is used in computing distance but not when computing displacement.

To find the initial height, we plug $t = 0$ into $s(t)$: $$s(0) = 128(0) - 16(0)^2 = 0.$$ Since the initial height is zero feet, the object was launched from the ground.

To compute the initial velocity, we first find the velocity function and then we plug in $t=0$. We have $$v(t) = s'(t) = 128 - 32t$$ and $$v(0) = 128 - 32(0) = 128.$$ Hence the initial velocity of the projectile was 128 ft/sec.

The total distance traveled is the distance up plus the distance down. The distance down is the maximum height of the object, $68$ feet. The distance up is the maximum height minus the initial height, $68 - 32$ feet. Thus the total distance traveled is $$68 + (68 - 32) = 68 + 36 = 104 \ \text {feet}.$$

To find the time in the air we need to solve $s(t) = 0$ for $t$. We get: $$80 + 64t-16t^2 = 0$$ $$5 + 4t-t^2 = 0$$ $$t^2 -4t -5 = 0$$ $$(t-5)(t+1) = 0$$ and so, $t = 5$ or $-1$ seconds. Since $-1$ seconds does not make sense in this situation, we conclude that the total time that the object was in the air was 5 seconds. Next, to determine the instantaneous velocity when the object was 128 ft. up, we need to first figure out the times when this occurred and then plug them into the velocity function, $$v(t) = s'(t) = 64 - 32t.$$ We have $$s(t) = 128$$ $$80 + 64t -16t^2 = 128$$ $$16t^2 - 64t + 48 = 0$$ $$t^2 - 4t + 3 = 0$$ $$(t-1)(t-3) = 0$$ and so $t= 1$ or $3$ seconds. In words, the object is at a height of 128 feet after 1 second (while it is on the way up) and after 3 seconds (on the way down). The instantaneous velocities at these times are $$v(1) = 64 - 32(1) = 32 \ \text {ft/sec}$$

$$\ v(3) = 64 - 32(3) = -32 \text {ft/sec.}$$ The positive velocity at $t= 1$ second indicates that the object was on the way up and the negative velocity at time $t=3$ seconds indicates that the object was on its way down. The absolute value of the velocity tells us the objects’ speed. Hence the object is going at the same speed of 32 ft/sec at the height of 128 feet when it is on the way up as it is on the way down. This is a general phenomenon of projectiles. The object loses speed on the way up and regains it at exactly the same rate on its way down so that at a given height it is going at the same speed on the way up as it is on the way down.