The first thing to do is determine how long it takes the ball to reach the ground. To do this, set [latex]s(t)=0[/latex]. Solving [latex]-16t^2+64=0[/latex], we get [latex]t=2[/latex], so it takes 2 seconds for the ball to reach the ground.

1. The instantaneous velocity of the ball as it strikes the ground is [latex]v(2)[/latex]. Since [latex]v(t)=s^{\prime}(t)=-32t[/latex] m we obtain [latex]v(t)=-64[/latex] ft/s.
2. The average velocity of the ball during its fall is
   [latex]v_{avg}=\frac{s(2)-s(0)}{2-0}=\frac{0-64}{2}=-32[/latex] ft/s.

Begin by finding [latex]v(t)[/latex] and [latex]a(t)[/latex].

[latex]v(t)=s^{\prime}(t)=3t^2-4[/latex] and [latex]a(t)=v^{\prime}(t)=s''(t)=6t[/latex].

Evaluating these functions at [latex]t=1[/latex], we obtain [latex]v(1)=-1[/latex] and [latex]a(1)=6[/latex].

1. Because [latex]v(1)<0[/latex], the particle is moving from right to left.
2. Because [latex]v(1)<0[/latex] and [latex]a(1)>0[/latex], velocity and acceleration are acting in opposite directions. In other words, the particle is being accelerated in the direction opposite the direction in which it is traveling, causing [latex]|v(t)|[/latex] to decrease. The particle is slowing down.

1. The velocity is the derivative of the position function:
   [latex]v(t)=s^{\prime}(t)=3t^2-18t+24[/latex].
2. The particle is at rest when [latex]v(t)=0[/latex], so set [latex]3t^2-18t+24=0[/latex]. Factoring the left-hand side of the equation produces [latex]3(t-2)(t-4)=0[/latex]. Solving, we find that the particle is at rest at [latex]t=2[/latex] and [latex]t=4[/latex].
3. The particle is moving from left to right when [latex]v(t)>0[/latex] and from right to left when [latex]v(t)<0[/latex]. The graph below gives the analysis of the sign of [latex]v(t)[/latex] for [latex]t\ge 0[/latex], but it does not represent the axis along which the particle is moving.
   

   Figure 3. The sign of v(t) determines the direction of the particle.

   Since [latex]3t^2-18t+24>0[/latex] on [latex][0,2)\cup (2,+\infty)[/latex], the particle is moving from left to right on these intervals.
   Since [latex]3t^2-18t+24<0[/latex] on [latex](2,4)[/latex], the particle is moving from right to left on this interval.

4. Before we can sketch the graph of the particle, we need to know its position at the time it starts moving [latex](t=0)[/latex] and at the times that it changes direction [latex](t=2,4)[/latex]. We have [latex]s(0)=4, \, s(2)=24[/latex], and [latex]s(4)=20[/latex]. This means that the particle begins on the coordinate axis at 4 and changes direction at 0 and 20 on the coordinate axis. The path of the particle is shown on a coordinate axis in the graph below.
   

   Figure 4. The path of the particle can be determined by analyzing [latex]v(t)[/latex].

Let [latex]P(t)[/latex] be the population (in thousands) [latex]t[/latex] years from now. Thus, we know that [latex]P(0)=10[/latex] and based on the information, we anticipate [latex]P(5)=30[/latex]. Now estimate [latex]P^{\prime}(0)[/latex], the current growth rate, using

By applying the average rate of change formula to [latex]P(t)[/latex], we can estimate the population 2 years from now by writing

thus, in 2 years the population will be approximately 18,000.

First, find the marginal revenue function: [latex]MR(x)=R^{\prime}(x)=-0.06x+9[/latex].

Next, use [latex]R^{\prime}(100)[/latex] to approximate [latex]R(101)-R(100)[/latex], the revenue obtained from the sale of the 101st dinner. Since [latex]R^{\prime}(100)=3[/latex], the revenue obtained from the sale of the 101st dinner is approximately $3.

The actual revenue obtained from the sale of the 101st dinner is

The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.