Equation 2.4 allows us to find the instantaneous velocity $v_s$ if we know the position $s$ as a function of time. But what about the reverse problem? Can we use the object’s velocity to calculate its position at some future time t? Equation 2.3, $s_{\text{f}}=s_{\text{i}}+v_s\text{Δ}t$, does this for the case of uniform motion with a constant velocity. We need to find a more general expression that is valid when $v_s$ is not constant.

FIGURE 2.14a is a velocity-versus-time graph for an object whose velocity varies with time. Suppose we know the object’s position to be $s_{\text{i}}$ at an initial time $t_{\text{i}}$. Our goal is to find its final position $s_{\text{f}}$ at a later time $t_{\text{f}}$.

FIGURE 2.14 Approximating a velocity-versus-time graph with a series of constant-velocity steps.

Because we know how to handle constant velocities, using Equation 2.3, let’s approximate the velocity function of Figure 2.14a as a series of constant-velocity steps of width $\text{Δ}t$. This is illustrated in FIGURE 2.14b. During the first step, from time $t_{\text{i}}$ to time $t_{\text{i}}+\text{Δ}t$, the velocity has the constant value ${\left(v_s\right)}_1$. The velocity during step $k$ has the constant value ${\left(v_s\right)}_k$. Although the approximation shown in the figure is rather rough, with only 11 steps, we can easily imagine that it could be made as accurate as desired by having more and more ever-narrower steps.

The velocity during each step is constant (uniform motion), so we can apply Equation 2.3 to each step. The object’s displacement $\text{Δ}{s}_1$ during the first step is simply $\text{Δ}{s}_1={\left(v_s\right)}_1\text{Δ}t$. The displacement during the second step $\text{Δ}{s}_2={\left(v_s\right)}_2\text{Δ}t$, and during step $k$ the displacement is $\text{Δ}{s}_k={\left(v_s\right)}_k\text{Δ}t$.

The total displacement of the object between $t_{\text{i}}$ and $t_{\text{f}}$ can be approximated as the sum of all the individual displacements during each of the $N$ constant-velocity steps. That is,

where $\sum$ (Greek sigma) is the symbol for summation. With a simple rearrangement, the particle’s final position is

Our goal was to use the object’s velocity to find its final position $s_{\text{f}}$. Equation 2.10 nearly reaches that goal, but Equation 2.10 is only approximate because the constant-velocity steps are only an approximation of the true velocity graph. But if we now let $\text{Δ}t\to 0$, each step’s width approaches zero while the total number of steps $N$ approaches infinity. In this limit, the series of steps becomes a perfect replica of the velocity-versus-time graph and Equation 2.10 becomes exact. Thus

The expression on the right is read, “the integral of $v_{s}dt$ from $t_{\text{i}}$ to $t_{\text{f}}$.” Equation 2.11 is the result that we were seeking. It allows us to predict an object’s position $s_{\text{f}}$ at a future time $t_{\text{f}}$.

We can give Equation 2.11 an important geometric interpretation. FIGURE 2.15 shows step $k$ in the approximation of the velocity graph as a tall, thin rectangle of height ${\left(v_s\right)}_k$ and width $\text{Δ}t$. The product $\text{Δ}{s}_k={\left(v_s\right)}_k\text{Δ}t$ is the area $(\text{base}\times \text{height})$ of this small rectangle. The sum in Equation 2.11 adds up all of these rectangular areas to give the total area enclosed between the t-axis and the tops of the steps. The limit of this sum as $\text{Δ}t\to 0$ is the total area enclosed between the t-axis and the velocity curve. This is called the “area under the curve.” Thus a graphical interpretation of Equation 2.11 is

FIGURE 2.15 The total displacement $\text{Δ}s$ is the “area under the curve.”

EXAMPLE 2.5 The displacement during a drag race

FIGURE 2.16 shows the velocity-versus-time graph of a drag racer. How far does the racer move during the first 3.0 s?

FIGURE 2.16 Velocity-versus-time graph for Example 2.5.

FIGURE 2.16 Full Alternative Text

MODEL Model the drag racer as a particle with a well-defined position at all times.

VISUALIZE FIGURE 2.16 is the graphical representation.

SOLVE The question “How far?” indicates that we need to find a displacement $\text{Δ}x$ rather than a position $x$. According to Equation 2.12, the car’s displacement $\text{Δ}x=x_{\text{f}}-x_{\text{i}}$ between $t=0\text{s}$ and $t=3\text{s}$ is the area under the curve from $t=0\text{s}$ to $t=3\text{s}$. The curve in this case is an angled line, so the area is that of a triangle:

$$\begin{array}{lll}\text{Δ}x\hfill & =\hfill & \text{area of triangle between }t=0\text{ s and }t=3\text{ s}\hfill \\ \hfill & =\hfill & \frac{1}{2}\times \text{base}\times \text{height}\hfill \\ \hfill & =\hfill & \frac{1}{2}\times 3\text{ s}\times 12\text{ m}\text{/}\text{s}=18\text{ m}\hfill \end{array}$$

The drag racer moves 18 m during the first 3 seconds.

REVIEW The “area” is a product of s with m/s, so $\text{Δ}x$ has the proper units of m.

EXAMPLE 2.6 Finding the turning point

FIGURE 2.17 is the velocity graph for a particle that starts at $x_{\text{i}}=30\text{m}$ at time $t_{\text{i}}=0\text{s}$.

1. Draw a motion diagram for the particle.

2. Where is the particle’s turning point?

3. At what time does the particle reach the origin?

FIGURE 2.17 Velocity-versus-time graph for the particle of Example 2.6.

FIGURE 2.17 Full Alternative Text

VISUALIZE The particle is initially 30 m to the right of the origin and moving to the right $(v_x>0)$ with a speed of 10 m/s. But $v_x$ is decreasing, so the particle is slowing down. At $t=2\text{s}$ the velocity, just for an instant, is zero before becoming negative. This is the turning point. The velocity is negative for $t>2\text{s,}$ so the particle has reversed direction and moves back toward the origin. At some later time, which we want to find, the particle will pass $x=0\text{m}$.

SOLVE a. FIGURE 2.18 shows the motion diagram. The distance scale will be established in parts b and c but is shown here for convenience.

FIGURE 2.18 Motion diagram for the particle whose velocity graph was shown in Figure 2.17.

FIGURE 2.18 Full Alternative Text

b. The particle reaches the turning point at $t=2\text{s}$. To learn where it is at that time we need to find the displacement during the first two seconds. We can do this by finding the area under the curve between $t=0\text{s}$ and $t=2\text{s}\text{:}$

$$\begin{array}{ll}\hfill x\left(\text{at }t=2\text{s}\right)& =x_{\text{i}}+\text{area under the curve between 0 s and 2 s}\hfill \\ \hfill & =30\text{ m +}\frac{1}{2}\left(2\text{s}-0\text{s}\right)\left(10\text{m}\text{/}\text{s}-0\text{m}\text{/}\text{s}\right)\hfill \\ \hfill & =40\text{m}\hfill \end{array}$$

The turning point is at $x=40\text{m}$.

c. The particle needs to move $\text{Δ}x=-40$ m to get from the turning point to the origin. That is, the area under the curve from $t=2\text{s}$ to the desired time $t$ needs to be $-40\text{m}$. Because the curve is below the axis, with negative values of $v_x$, the area to the right of $t=2\text{s}$ is a negative area. With a bit of geometry, you will find that the triangle with a base extending from $t=2\text{s}$ to $t=6\text{s}$ has an area of $-40\text{m}$. Thus the particle reaches the origin at $t=6\text{s}$.