3.2 Instantaneous Velocity and Speed

We have now seen how to calculate the average velocity between two positions. However, since objects in the real world move continuously through space and time, we would like to find the velocity of an object at any single point. We can find the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us better insight into the physics of motion and will be useful in later chapters.

Instantaneous Velocity

The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usually called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two events approaches zero. To illustrate this idea mathematically, we need to express position x as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation is $\overset{–}{v} = \frac{x (t_{2}) - x (t_{1})}{t_{2} - t_{1}}$ . To find the instantaneous velocity at any position, we let $t sub 1 equals t$ and $t_{2} = t + Δ t$ . After inserting these expressions into the equation for the average velocity and taking the limit as $Δ t → 0$ , we find the expression for the instantaneous velocity:

$v (t) = lim_{Δ t \to 0} \frac{x (t + Δ t) - x (t)}{Δ t} = \frac{d x (t)}{d t} = \frac{d}{d t} x (t) .$

Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point $t sub 0$ is the rate of change of the position function, which is the slope of the position function $x (t)$ at $t sub 0$ . Figure 3.6 shows how the average velocity $\overset{–}{v} = \frac{Δ x}{Δ t}$ between two times approaches the instantaneous velocity at $t sub 0.$ The instantaneous velocity is shown at time $t sub 0$ , which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, $t sub 1, t sub 2$ , and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.

Graph shows position plotted versus time. Position increases from t1 to t2 and reaches maximum at t0. It decreases to at and continues to decrease at t4. The slope of the tangent line at t0 is indicated as the instantaneous velocity. — **Figure 3.6** In a graph of position versus time, the instantaneous velocity is the slope of the tangent line at a given point. The average velocities $\overset{–}{v} = \frac{Δ x}{Δ t} = \frac{x_{f} - x_{i}}{t_{f} - t_{i}}$ between times $Δ t = t_{6} - t_{1}, Δ t = t_{5} - t_{2}, and Δ t = t_{4} - t_{3}$ are shown. When $Δ t → 0$ , the average velocity approaches the instantaneous velocity at $t equals t sub 0$ .

Example 1 Finding Velocity from a Position-Versus-Time Graph

Given the position-versus-time graph of Figure 3.7, find the velocity-versus-time graph.

Figure 3.7 The object starts out in the positive direction, stops for a short time, and then reverses direction, heading back toward the origin. Notice that the object comes to rest instantaneously, which would require an infinite force. Thus, the graph is an approximation of motion in the real world. (The concept of force is discussed in Newton’s Laws of Motion.)

Strategy The graph contains three straight lines during three time intervals. We find the velocity during each time interval by taking the slope of the line using the grid.

Solution Time interval 0 s to 0.5 s: $\overset{–}{v} = \frac{Δ x}{Δ t} = \frac{0.5 m - 0.0 m}{0.5 s - 0.0 s} = 1.0 m/s$

Time interval 0.5 s to 1.0 s: $\overset{–}{v} = \frac{Δ x}{Δ t} = \frac{0.5 m - 0.5 m}{1.0 s - 0.5 s} = 0.0 m/s$

Time interval 1.0 s to 2.0 s: $\overset{–}{v} = \frac{Δ x}{Δ t} = \frac{0.0 m - 0.5 m}{2.0 s - 1.0 s} = −0.5 m/s$

The graph of these values of velocity versus time is shown in Figure 3.8.

Figure 3.8 The velocity is positive for the first part of the trip, zero when the object is stopped, and negative when the object reverses direction.

Significance During the time interval between 0 s and 0.5 s, the object’s position is moving away from the origin and the position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can find the instantaneous velocity by taking its slope, which is +1 m/s, as shown in Figure 3.8. In the subsequent time interval, between 0.5 s and 1.0 s, the position doesn’t change and we see the slope is zero. From 1.0 s to 2.0 s, the object is moving back toward the origin and the slope is −0.5 m/s. The object has reversed direction and has a negative velocity.

Speed

In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not have the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.

We can calculate the average speed by finding the total distance traveled divided by the elapsed time:

$Average speed = \overset{–}{s} = \frac{Total distance}{Elapsed time} .$

(3.5)

Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time, then we would be interested in our average speed.

However, we can calculate the instantaneous speed from the magnitude of the instantaneous velocity:

$Instantaneous speed equals | v open parenthesis t close parenthesis |.$

(3.6)

If a particle is moving along the x-axis at +7.0 m/s and another particle is moving along the same axis at −7.0 m/s, they have different velocities, but both have the same speed of 7.0 m/s. Some typical speeds are shown in the following table.

**Table 3.1** Speeds of Various Objects
Speed	m/s	mi/h
Continental drift	$10 to the power −9$	$2 \times 10^{−9}$
Brisk walk	1.7	3.9
Cyclist	4.4	10
Sprint runner	12.2	27
Rural speed limit	24.6	56
Official land speed record	341.1	763
Speed of sound at sea level	343	768
Space shuttle on reentry	7800	17,500
Escape velocity of Earth*	11,200	25,000
Orbital speed of Earth around the Sun	29,783	66,623
Speed of light in a vacuum	299,792,458	670,616,629

Calculating Instantaneous Velocity

When calculating instantaneous velocity, we need to specify the explicit form of the position function $x (t)$ . If each term in the $x (t)$ equation has the form of $A t^{n}$ where $A$ is a constant and $n$ is an integer, this can be differentiated using the power rule to be:

$\frac{d (A t^{n})}{d t} = A n t^{n - 1} .$

(3.7)

Note that if there are additional terms added together, this power rule of differentiation can be done multiple times and the solution is the sum of those terms. The following example illustrates the use of Equation 3.7.

Example 2 Instantaneous Velocity Versus Average Velocity

The position of a particle is given by $x (t) = (3.0 m/s) t + (0.50 {m/s}^{3}) t^{3}$ .

Using Equation 3.4 and Equation 3.7, find the instantaneous velocity at $t equals 2.0$ s.
Calculate the average velocity between 1.0 s and 3.0 s.

StrategyEquation 3.4 gives the instantaneous velocity of the particle as the derivative of the position function. Looking at the form of the position function given, we see that it is a polynomial in t. Therefore, we can use Equation 3.7, the power rule from calculus, to find the solution. We use Equation 3.6 to calculate the average velocity of the particle.

Solution

$v (t) = \frac{d x (t)}{d t} = (3.0 m/s) + (1.5 {m/s}^{3}) t^{2}$ .
Substituting t = 2.0 s into this equation gives $v (2.0 s) = (3.0 m/s) + (1.5 {m/s}^{3}) {(2.0 s)}^{2} = 9.0 m/s$ .
To determine the average velocity of the particle between 1.0 s and 3.0 s, we calculate the values of x(1.0 s) and x(3.0 s):
$x (1.0 s) = [(3.0 m/s) (1.0 s) + (0.50 {m/s}^{3}) {(1.0 s)}^{3}] = 3.5 m$
$x (3.0 s) = [(3.0 m/s) (3.0 s) + (0.50 {m/s}^{3}) {(3.0 s)}^{3}] = 22.5 m$
Then the average velocity is
$\overset{–}{v} = \frac{x (3.0 s) - x (1.0 s)}{(3.0 s) - (1.0 s)} = \frac{(22.5 - 3.5) m}{(3.0 - 1.0) s} = 9.5 m/s .$

Significance In the limit that the time interval used to calculate $v −$ goes to zero, the value obtained for $v −$ converges to the value of v. The value of $9.0 m/s$ that we calculated in part a for the instantaneous value at $t equals 2.0 s$ is close to the average value of $9.5 m/s$ found in b for a short interval that includes $t equals 2.0 s$ . If we were to make the interval shorter, the values in the calculation of the average would become even closer.

Example 3 Instantaneous Velocity Versus Speed

Consider the motion of a particle in which the position is $x (t) = (3.0 m/s) t - (3 {m/s}^{2}) t^{2}$ .

What is the instantaneous velocity at t = 0.25 s, t = 0.50 s, and t = 1.0 s?
What is the speed of the particle at these times?

Strategy The instantaneous velocity is the derivative of the position function and the speed is the magnitude of the instantaneous velocity. We use Equation 3.4 and Equation 3.7 to solve for instantaneous velocity.

Solution

$v (t) = \frac{d x (t)}{d t} = (3.0 m/s - 6.0 {m/s}^{2} t), v (0.25 s) = 1.50 m/s,$
$v (0.5 s) = 0 m/s, v (1.0 s) = −3.0 m/s$
$Speed = | v (t) | = 1.50 m/s, 0.0 m/s, and 3.0 m/s$

Significance The velocity of the particle gives us direction information, indicating the particle is moving to the left (west) or right (east). The speed gives the magnitude of the velocity. By graphing the position, velocity, and speed as functions of time, we can understand these concepts visually Figure 3.9. In (a), the graph shows the particle moving in the positive direction until t = 0.5 s, when it reverses direction. The reversal of direction can also be seen in (b) at 0.5 s where the velocity is zero and then turns negative. At 1.0 s it is back at the origin where it started. The particle’s velocity at 1.0 s in (b) is negative, because it is traveling in the negative direction. But in (c), however, its speed is positive and remains positive throughout the travel time. We can also interpret velocity as the slope of the position-versus-time graph. The slope of x(t) is decreasing toward zero, becoming zero at 0.5 s and increasingly negative thereafter. This analysis of comparing the graphs of position, velocity, and speed helps catch errors in calculations. The graphs must be consistent with each other and help interpret the calculations.

Figure 3.9 (a) Position: x(t) versus time. (b) Velocity: v(t) versus time. The slope of the position graph is the velocity. A rough comparison of the slopes of the tangent lines in (a) at 0.25 s, 0.5 s, and 1.0 s with the values for velocity at the corresponding times indicates they are the same values. (c) Speed:

| v open parenthesis t close parenthesis |

versus time. Speed is always a positive number.

Summary

Instantaneous velocity is a continuous function of time and gives the velocity at any point in time during a particle’s motion. We can calculate the instantaneous velocity at a specific time by taking the derivative of the position function, which gives us the functional form of instantaneous velocity v(t).
Instantaneous velocity is a vector and can be negative.
Instantaneous speed is found by taking the absolute value of instantaneous velocity, and it is always positive.
Average speed is total distance traveled divided by elapsed time.
The slope of a position-versus-time graph at a specific time gives instantaneous velocity at that time.

Conceptual Questions

There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.

Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement/Elapsed time, your average velocity is zero.

Does the speedometer of a car measure speed or velocity?

If you divide the total distance traveled on a car trip (as determined by the odometer) by the elapsed time of the trip, are you calculating average speed or magnitude of average velocity? Under what circumstances are these two quantities the same?

Average speed. They are the same if the car doesn’t reverse direction.

How are instantaneous velocity and instantaneous speed related to one another? How do they differ?

Problems

A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the woodchuck? (b) What is its average speed?

Sketch the velocity-versus-time graph from the following position-versus-time graph.

Graph shows position in meters plotted versus time in seconds. It starts at the origin, reaches 4 meters at 0.4 seconds; decreases to -2 meters at 0.6 seconds, reaches minimum of -6 meters at 1 second, increases to -4 meters at 1.6 seconds, and reaches 2 meters at 2 seconds.

Graph shows velocity in meters per second plotted as a function of time at seconds. Velocity starts as 10 meters per second, decreases to -30 at 0.4 seconds; increases to -10 meters at 0.6 seconds, increases to 5 at 1 second, increases to 15 at 1.6 seconds.

Sketch the velocity-versus-time graph from the following position-versus-time graph.

Graph shows position plotted versus time in seconds. Graph has a sinusoidal shape. It starts with the positive value at zero time, changes to negative, and then starts to increase.

Given the following velocity-versus-time graph, sketch the position-versus-time graph.

Graph shows velocity plotted versus time. It starts with the positive value at zero time, decreases to the negative value and remains constant.

Graph shows position plotted versus time. It starts at the origin, increases reaching maximum, and then decreases close to zero.

An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.

A particle moves along the x-axis according to $x (t) = 10 t - 2 t^{2} m$ . (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between t = 2 s and t = 3 s?

a. $v (t) = (10 - 4 t) m/s$ ; v(2 s) = 2 m/s, v(3 s) = −2 m/s; b. $| v (2 s) | = 2 m/s, | v (3 s) | = 2 m/s$ ; (c) $v – equals 0 m/s$

Unreasonable results. A particle moves along the x-axis according to $x (t) = 3 t^{3} + 5 t $ . At what time is the velocity of the particle equal to zero? Is this reasonable?