When a car drives along a road, its position, velocity and acceleration will all change at various points in the trip. These three properties are all inherently linked: if the speed of the car changes, it must have accelerated or decelerated, and this will then cause the position of the car to change too.
Calculus is the study of change. For this reason, the use of calculus is an integral part of the study of motion, pun intended.
Integrals of Motion Definition
Integration of motion is a method of studying how objects move in space, through the use of integration. Before you cover this, it is important to recap some ideas to do with integration first.
Integration and Area Under a Curve
Integration is a method that can be used to find the area under a graph.
Fig. 1,. The integral of this function between \(a\) and \(b\) gives the value of the area shaded.
The definite integral between \(a\) and \(b\) of the function above gives the area under the graph. If the function is \(f(t),\) this is written
\[ \int_a^b f(t) \, \textrm{d}t. \]
For more information on integrals, as well as how to find the integral of a certain function, see Integrals.
Displacement, Velocity and Acceleration
Displacement, velocity and acceleration are the most important factors when working with any sort of moving object. All of these definitions are assuming an object is moving in one direction, along the real line.
Displacement \(s(t)\) is how far away an object is from a certain position, taken as the origin.
Displacement is directional, meaning it can be positive or negative depending on which direction the object is from the origin.
Velocity \(v(t)\) is the rate of change of displacement.
- If the velocity is \(0,\) the object is stationary.
- If the velocity is positive, the object is traveling forward (towards infinity on the real axis).
- If the velocity is negative, the object is traveling backward (towards negative infinity on the real axis).
Acceleration \(a(t)\) is the rate of change in velocity.
If the acceleration is 0, the object will travel at the same speed.
If the acceleration is positive, the object will be
In the above definitions, \(t\) represents time. Since these definitions use rate of change, you can be sure that differentiation will be used when working with them. The following formulas hold:
\[ \begin{align} v(t) & = \frac{\mathrm{d}s}{\mathrm{d}t} \\ a(t) & = \frac{\mathrm{d}v}{\mathrm{d}t}. \end{align} \]
First Integral of Motion
Given the functions \(s(t), v(t) \) and \( a(t),\) denoting displacement, velocity and acceleration, the following indefinite integral formulas exist:
\[ \begin{align} \int v(t) \, \mathrm{d}t & = s(t) + c \\ \int a(t) \, \mathrm{d}t & = v(t) + c. \end{align} \]
Don't forget to include your constant of integration when taking indefinite integrals.
This makes sense given the derivative relations in the previous section, as these are simply the inverse formulas. Let's look at some examples of using these formulas.
A particle is traveling with velocity \( v(t) = 6t^2, \) and has displacement \(100\) after 5 seconds. Find the velocity function of this particle.
Solution:
The first step is to take the indefinite integral of the particle.
\[ \begin{align} s(t) & = \int v(t) \, \mathrm{d}t \\ & = \int 6 t^2 \, \mathrm{d}t \\ & = 2 t^3 + c. \end{align} \]
Now, you can substitute \(t=5\) into the displacement vector to find the value of the constant of integration.
\[ \begin{align} v(5) = 100 & = 2 \cdot 5^3 + c \\ & = 2 \cdot 125 + c \\ & = 250 + c. \end{align} \]
Hence, you have that
\[ \begin{align} 250 + c & = 100 \\ \implies c & = -150. \end{align} \]
Hence, the displacement function must be
\[ s(t) = 2 t^3 - 150. \]
Now let's look at one where you must find the velocity function from the acceleration vector.
A particle is traveling with acceleration \( a(t) = \cos{t} \) and is stationary at \(t=0.\) Find the velocity function for this particle.
Solution:
First, integrate the acceleration function.
\[ \begin{align} v(t) & = \int a(t) \, \mathrm{d}t \\ & = \int \cos{t} \, \mathrm{d}t \\ & = \sin{t} + c. \end{align} \]
Now, since the particle is stationary at \(t=0,\) the velocity must be \(0.\) Substitute this in to find the value of \(c.\)
\[ v(0) = 0 = \sin(0) + c = c \]
Hence, \(c\) must be \(0.\) This means the final velocity function must be
\[ v(t) = \sin{t}. \]
Integrals of Motion Formula
Definite integrals can be used to calculate the total displacement between two times \(t=a\) and \(t=b\). The total displacement is the distance between the position of the particle at \(t=a\) and \(t=b\). This can be found by taking the definite integral of the particle with bounds \(a\) and \(b\):
\[ \text{total displacement} = \int_a^b v(t) \, \mathrm{d}t. \]
You can also find the total distance that the particle has traveled during this time, by taking the absolute value of the velocity before integrating it.
\[ \text{total distance} = \int_a^b |v(t)| \, \mathrm{d}t. \]
The easiest way to see the difference between total displacement and total distance is an example.
If you throw a ball \(2\) meters into the air and catch it, the total displacement is \(0\) because it has returned to its original spot. The total distance, however, is \(4\) meters, since it traveled \(2\) meters into the air and \(2\) meters back down. If the object is always traveling with positive velocity, the total displacement and total distance traveled will always be the same.
Let's look at an example of finding the displacement and distance of a particle.
A particle travels with velocity \(v(t) = 8 -4t.\) Find the total displacement and the total distance of the particle between times \(t=1\) and \(t=3.\)
Solution:
First, find the total displacement. Take the definite integral of the velocity function between \(t=1\) and \(t=3.\)
\[ \begin{align} \int_1^3 8 - 4t \, \mathrm{d}t & = [8t - 2t^2]_1^3 \\ & = (8 \cdot 3 - 2 \cdot 3^2) - (8 \cdot 1 - 2 \cdot 1^2) \\ & = (24 - 18) - (8 - 2) \\ & = 6 - 6 \\ & = 0. \end{align} \]
So the total displacement is \(0.\)
To find the total distance travelled, you must find the integral of the absolute value of the function.
\[ \int_1^3 | 8 - 4t | \, \mathrm{d}t. \]
The function will go negative at \(t=2,\) hence the integral can be split into two at this point.
\[ \begin{align} \int_1^3 | 8 - 4t | \, \mathrm{d}t & = \int_1^2 8 - 4t \, \mathrm{d}t + \int_1^2 -(8-4t) \, \mathrm{d}t \\ & = \int_2^3 8 - 4t \, \mathrm{d}t + \int_1^2 -8+4t \, \mathrm{d}t. \end{align} \]
Now you can evaluate the integrals on the left to deduce the total distance travelled.
\[ \begin{align} \int_1^3 | 8 - 4t | \, \mathrm{d}t & = [8t - 2t^2]_1^2 + [-8t + 2t^2]_2^3 \\ & = [(8\cdot 2 - 2\cdot 2^2) - (8 \cdot 1 - 2 \cdot 1^2)] + [(-8\cdot 3 + 2 \cdot 3^2) - (-8 \cdot 2 - 2 \cdot 2^2)] \\ & = [(16 - 8) + (8 - 2)] + [(-24 + 18) - (-16 + 8)] \\ & = [8 - 6] + [-6 + 8] \\ & = 4. \end{align} \]
So the total distance travelled is \(4.\)
Additive Integrals of Motion
If a particle is travelling with velocity \(v_1(t)\) for the first \(a\) seconds and then travels at velocity \(v_2(t)\) for the following \(b\) seconds, the displacement can be found by adding these two integrals together.
\[ \text{total displacement} \int_0^a v_1(t) \, \mathrm{d}t + \int_a^b v_2(t) \, \mathrm{d}t. \]
This same technique can be used for any finite number of separate velocity functions too, and can be used to find the total distance travelled by integrating the absolute values of the velocities instead.
A particle travels with velocity \(v_1(t) = t^2 \) for 3 seconds, and then has velocity \(v_2(t) = 3t\) for a further 6 seconds. Calculate the total distance travelled by the particle in this time.
Solution:
First, since the question asks for the total distance travelled, you must take the absolute values of the velocities. But since both of the velocities are always positive, the total distance travelled will be equal to the displacement, and the absolute values are irrelevant. Hence, the total distance travelled is
\[ \begin{align} \int_0^3 3t^2 \, \textrm{d}t + \int_3^9 3t \, \textrm{d}t & = [t^3]_0^3 + \left[\frac{3}{2} t^2 \right]_3^9 \\ & = (3^3 - 0^3) + \left(\frac{3}{2} \cdot 9^2 - \frac{3}{2} \cdot 3^2 \right) \\ & = 27 + (\frac{243}{2} - \frac{27}{2}) \\ & = 27 + 116 \\ & = 143 \end{align} \]
So the total displacement and total distance travelled are 143 units.
Integrals of Motion Sample Problems
Let's look at some more complicated examples of questions using integration and motion.
The acceleration of a particle is given by \[a(t) = 6t\] and the position of the particle is \(10\) at \(t=0\) and \(14\) at \(t=2.\) calculate the displacement function of the particle.
Solution:
This time, you will have to integrate twice, since the function given is acceleration and not velocity. Integrating the function once will give you:
\[ \begin{align} v(t) & = \int a(t) \, \textrm{d}t \\ & = \int 6t \, \textrm{d}t \\ & = 3t^2 + c. \end{align}\]
Now, you can integrate the velocity function to find the displacement function.
\[ \begin{align} s(t) & = \int v(t) \, \textrm{d}t \\ & = \int 3 t^2 + c \, \textrm{d}t \\ & = t^2 + ct + d. \end{align} \]
\(d\) is another constant of integration. Now, you can substitute in the initial conditions to find the exact displacement function. If \(t=0:\)
\[ \begin{align} s(0) = 10 & = 0^2 + c \cdot 0 + d \\ \implies d & = 10. \end{align} \]
Hence, \(d\) must be \(0.\) If \(t = 2:\)
\[ \begin{align} s(2) = 14 & = 2^2 + c \cdot 10 + 10 \\ & = 14 + 10c \\ \implies 0 & = 10c \end{align} \]
so \(c = 0.\) Hence, the final formula for displacement is
\[ s(t) = t^2 + 10. \]
Now let's look at another example of a question that may be a little more difficult.
Particle \(A\) is travelling at velocity \(v_A(t) = 2t^2 \) and particle \(B\) is travelling at velocity \(v_B(t) = 4t.\) Both particles are in the same position at time \(t=0.\)
- Find the distance between the particles at time \(t=5.\)
- At what time are the particles in the same position again?
Solution:
1) Since the particles are only moving in one dimension, the distance between them will simply be the absolute value of the difference between their distances from the starting points.
\[ \begin{align} \text{Distance between A and B at time 5} & = \int_0^5 v_A(t) \, \textrm{d}t - \int v_B(t) \, \textrm{d}t \\ & = \int v_A(t) - v_B(t) \, \textrm{d}t. \end{align} \]
Now, you can plug the formulas for \(v_A(t)\) and \(v_B(t)\) into the formula above to get the distance between them.
\[ \begin{align} \int_0^5 v_A(t) - v_B(t) \, \textrm{d}t & = \int_0^5 2t^2 - 4t \, \textrm{d}t \\ & = \left[ \frac{2}{3} t^3 - 2 t^2 \right]_0^5 \\ & = \frac{2}{3} \cdot 5^3 - 2 \cdot 5^2 \\ & = \frac{250}{3} - 50 \\ & = \frac{100}{3}. \end{align} \]
So the distance between the two particles at time \(t=0\) is \(\frac{100}{3}.\)
2) This problem can be solved in the same way as question one, but you must use a variable, call it \(t',\) as the upper bound of integration, then set the result as equal to \(0\) and solve the equation for \(t'.\) Using the same steps as before, you will get that:
\[ \begin{align} \int_0^5 v_A(t) - v_B(t) \, \textrm{d}t & = \int_0^5 2t^2 - 4t \, \textrm{d}t \\ & = \left[ \frac{2}{3} t^3 - 2 t^2 \right]_0^{t'} \\ & = \frac{2}{3} t'^3 - 2 t'^2. \end{align} \]
Now you can set this equal to 0, and solve for \(t'.\)
\[ \begin{align} \frac{2}{3} t'^3 - 2 t'^2 & = 0 \\ & \implies t' (\frac{2}{3} t' - 2 ) = 0. \end{align} \]
For this equation to be true, either \(t'=0\) or \(\frac{2}{3} t' - 2 = 0.\) \(t'=0\) is simply the starting point, and this is not the answer that the question is looking for. Hence, it must be that
\[ \begin{align} \frac{2}{3} t' - 2 & = 0 \\ \implies t' & = 2 \cdot \frac{3}{2} \\ \implies t' & = 3\ \end{align} \]
Hence, the particles will be in the same position again at \(t=3.\)
Integrals of Motion Use
Integrals of motion are essential in areas of physics such as classical mechanics and general relativity. These relations between position, velocity and acceleration allow scientists and engineers to model anything that moves such as cars, rockets and objects in space. Of course, the equations used in these models tend to be much more complicated than just polynomials or trigonometric functions.
Integrals of Motion - Key takeaways
- The following formulas exist, relating displacement \(s(t),\) velocity \(v(t),\) and acceleration \(a(t)\) together \[ \begin{align} \int v(t) \, \mathrm{d}t & = s(t) + c \\ \int a(t) \, \mathrm{d}t & = v(t) + c. \end{align} \]
- The total displacement between times \(a\) and \(b\) of an object is the difference between it's start and end points. The formula for finding total displacement from velocity \(v(t)\) is \[ \text{total displacement} = \int_a^b v(t) \, \mathrm{d}t. \]
- The total distance travelled between times \(a\) and \(b\) of an object is how far the object travelled during this time, irrespective of direction. \[ \text{total distance} = \int_a^b |v(t)| \, \mathrm{d}t. \]
- If the velocity is always constant, total displacement and total distance travelled will always be equal.
Explanations 
Exams 
Magazine 
