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Chapter 2 One-Dimensional Kinematics
2.9 New – Motion with a non-constant acceleration – Derivatives
Learning Objectives
By the end of this section, you will be able to:
Explain the difference between average velocity and instantaneous velocity.
Describe the difference between velocity and speed.
Calculate the instantaneous velocity given the mathematical equation for the velocity.
Calculate the speed given the instantaneous velocity.
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.
The quantity that tells us how fast an object is moving anywhere along its path is theinstantaneous velocity, usually called simplyvelocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points 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
Instantaneous Velocity
The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t:
Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point t0is the rate of change of the position function, which is the slope of the position function x(t) at t0. Figure 3.6 shows how the average velocity”>v¯=ΔxΔtv¯=ΔxΔtbetween two times approaches the instantaneous velocity at t0. The instantaneous velocity is shown at time t0, 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, t1, t2, 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.
FigureIn a graph of position versus time, the instantaneous velocity is the slope of the tangent line at a given point. The average velocities. Δt → 0, the average velocity approaches the instantaneous velocity at t = t0.
Example 3.2: Finding Velocity from a Position-Versus-Time Graph
Given the position-versus-time graph of Figure 3.7, find the velocity-versus-time graph.
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
The graph of these values of velocity versus time is shown in Figure 3.8.
FigureThe 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:
Table 3.1 Speeds of various objects in metres per second
Continental drift = 10 -7 m/s
Brisk walk = 1.7 m/s
Cyclist = 4. 4 m/s
Sprint runner = 12.2 m/s
Official land speed record = 341.4 m/s
Speed of sound = 343 m/s
Escape velocity = Escape velocity is the velocity at which an object must be launched so that it overcomes Earth’s gravity and is not pulled back toward Earth = 11,200 m/s
ISS = International Space Station =. 7,600 m/s
Orbital speed of the Earth =. 29,700 m/s
Speed of light in.a vacuum = 3.00 x 108 m/s
Calculating Instantaneous Velocity
When calculating instantaneous velocity, we need to specify the explicit form of the position function x(t). For the moment, let’s use polynomials x(t) = Atn, because they are easily differentiated using the power rule of calculus:
The following example illustrates the use of Equation 3.7.
Example 3.3: Instantaneous Velocity Versus Average Velocity
The position of a particle is given by x(t) = 3.0t + 0.5t3m.
Using Equation 3.4 and Equation 3.7, find the instantaneous velocity at t = 2.0 s.
Calculate the average velocity between 1.0 s and 3.0 s.
Strategy
Equation 3.4 give 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
dx/dt = v(t) = 3.0 + 1.5t2m/s
Substituting t = 2.0 s into this equation gives v(2.0 s) = [3.0 + 1.5(2.0)2] m/s = 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):
Example 3.4: Instantaneous Velocity Versus Speed
Consider the motion of a particle in which the position is x(t) = 3.0t − 3t2m.
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.
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(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(t)| versus time. Speed is always a positive number.
Check Your Understanding 3.2
The position of an object as a function of time is x(t) = −3t2m. (a) What is the velocity of the object as a function of time? (b) Is the velocity ever positive? (c) What are the velocity and speed at t = 1.0 s?
Contributors
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
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.
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Given the following velocity-versus-time graph, sketch the position-versus-time graph.
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. 35. A particle moves along the x-axis according to x(t) = 10t − 2t2 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?
Unreasonable results. A particle moves along the x-axis according to x(t) = 3t3 + 5t . At what time is the velocity of the particle equal to zero? Is this reasonable?
The acceleration of a particle varies with time according to the equation a(t) = pt2 − qt3. Initially, the velocity and position are zero. (a) What is the velocity as a function of time? (b) What is the position as a function of time?
Between t = 0 and t = t0, a rocket moves straight upward with an acceleration given by a(t) = A − Bt1/2, where A and B are constants. (a) If x is in meters and t is in seconds, what are the units of A and B? (b) If the rocket starts from rest, how does the velocity vary between t = 0 and t = t0? (c) If its initial position is zero, what is the rocket’s position as a function of time during this same time interval?
The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1, where A = 2 m/s, B = 0.25 m, and 1.0 s ≤ t ≤ 8.0 s. Determine the acceleration and position of the particle at t = 2.0 s and t = 5.0 s. Assume that x(t = 1 s) = 0.
A particle at rest leaves the origin with its velocity increasing with time according to v(t) = 3.2t m/s. At 5.0 s, the particle’s velocity starts decreasing according to [16.0 – 1.5(t – 5.0)] m/s. This decrease continues until t = 11.0 s, after which the particle’s velocity remains constant at 7.0 m/s. (a) What is the acceleration of the particle as a function of time? (b) What is the position of the particle at t = 2.0 s, t = 7.0 s, and t = 12.0 s?
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