Calculus—invented simultaneously in England by Newton and in Germany by Leibniz—is designed to deal with instantaneous quantities. In other words, it provides us with the tools for evaluating limits such as the one in Equation 2.4.

The notation $ds/dt$ is called the derivative of s with respect to t, and Equation 2.4 defines it as the limiting value of a ratio. As FIGURE 2.8 showed, $ds/dt$ can be interpreted graphically as the slope of the line that is tangent to the position graph.

The most common functions we will use in Parts I and II of this book are powers and polynomials. Consider the function $u\left(t\right)=c{t}^n$, where $c$ and $n$ are constants. The symbol $u$ is a “dummy name” to represent any function of time, such as $x\left(t\right)$ or $y\left(t\right)$. The following result is proven in calculus:

For example, suppose the position of a particle as a function of time is $s\left(t\right)=2t^2\text{m,}$ where $t$ is in s. We can find the particle’s velocity $v_s=ds/dt$ by using Equation 2.6 with $c=2$ and $n=2$ to calculate

This is an expression for the particle’s velocity as a function of time.

FIGURE 2.11 shows the particle’s position and velocity graphs. It is critically important to understand the relationship between these two graphs. The value of the velocity graph at any instant of time, which we can read directly off the vertical axis, is the slope of the position graph at that same time. This is illustrated at $t=3\text{s}$.

FIGURE 2.11 Position-versus-time graph and the corresponding velocity-versus-time graph.

A value that doesn’t change with time, such as the position of an object at rest, can be represented by the function $u=c=\text{constant}$. That is, the exponent of $t^n$ is $n=0$. You can see from Equation 2.6 that the derivative of a constant is zero. That is,

This makes sense. The graph of the function $u=c$ is simply a horizontal line. The slope of a horizontal line—which is what the derivative $du/dt$ measures—is zero.

The only other information we need about derivatives for now is how to evaluate the derivative of the sum of two functions. Let $u$ and $w$ be two separate functions of time. You will learn in calculus that

That is, the derivative of a sum is the sum of the derivatives.

EXAMPLE 2.4 Using calculus to find the velocity

A particle’s position is given by the function $x\left(t\right)=(-t^3+3t)$ m, where $t$ is in s.

1. What are the particle’s position and velocity at $t=2$ s?

2. Draw graphs of $x$ and $v_x$ during the interval $-3\text{s}\le t\le 3\text{s}$.

3. Draw a motion diagram to illustrate this motion.

SOLVE a. We can compute the position directly from the function $x$:

$$x(\text{at }t=2\text{s})=-{\left(2\right)}^3+\left(3\right)\left(2\right)=-8+6=-2\text{m}$$

The velocity is $v_x=dx\text{/}dt$. The function for $x$ is the sum of two polynomials, so

$$v_x=\frac{dx}{dt}=\frac{d}{dt}(-t^3+3t)=\frac{d}{dt}(-t^3)+\frac{d}{dt}\left(3t\right)$$

The first derivative is a power with $c=-1$ and $n=3;$ the second has $c=3$ and $n=1$. Using Equation 2.6, we have

$$v_x=\left(-3t^2+3\right)\text{ m}\text{/}\text{s}$$

where $t$ is in s. Evaluating the velocity at $t=2\text{s}$ gives

$$v_x(\text{at }t=2\text{ s})=-3{\left(2\right)}^2+3=-9\text{ m}\text{/}\text{s}$$

The negative sign indicates that the particle, at this instant of time, is moving to the left at a speed of 9 $\text{m}/\text{s}$.

b. FIGURE 2.12 shows the position graph and the velocity graph. You can make graphs like these with a graphing calculator or graphing software. The slope of the position-versus-time graph at $t=2\text{s}$ is $-9\text{m}/\text{s};$ this becomes the value that is graphed for the velocity at $t=2\text{s}$.

FIGURE 2.12 Position and velocity graphs.

FIGURE 2.12 Full Alternative Text

c. Finally, we can interpret the graphs in FIGURE 2.12 to draw the motion diagram shown in Figure 2.13.

FIGURE 2.13 Motion diagram for Example 2.4.

FIGURE 2.13 Full Alternative Text

• The particle is initially to the right of the origin $(x>0$ at $t=-3\text{s})$ but moving to the left $(v_x<0)$. Its speed is slowing $(v=|v_x|$ is decreasing$)$, so the velocity vector arrows are getting shorter.

• The particle passes the origin $x=0\text{m}$ at $t\approx -1.5\text{s,}$ but it is still moving to the left.

• The position reaches a minimum at $t=-1\text{s};$ the particle is as far left as it is going. The velocity is instantaneously $v_x=0\text{m}\text{/}s$ as the particle reverses direction.

• The particle moves back to the right between $t=-1\text{s}$ and $t=1\text{s }(v_x>0)$.

• The particle turns around again at $t=1\text{s}$ and begins moving back to the left $(v_x<0)$. It keeps speeding up, then disappears off to the left.

A point in the motion where a particle reverses direction is called a turning point. It is a point where the velocity is instantaneously zero while the position is a maximum or minimum. This particle has two turning points, at $t=-1\text{s}$ and again at $t=+1\text{s}$. We will see many other examples of turning points.