Kinematics is a topic in physics that describes the motion of points, bodies and systems in space. Calculus kinematics can be used to derive equations for velocity and acceleration using derivatives and their integrals.
- What are kinematic equations?
- Displacement
- Velocity
- Acceleration
What are kinematic equations?
Kinematics is, broadly, the study of motion. Kinematic equations relate displacement, velocity and acceleration of a body through derivatives and integrals.
Displacement kinematic equations
The displacement of a particle simply shows how far a point has moved in space relative to a fixed origin point. This quantity will be referred to as x and is a vector, rather than a scalar since it takes into consideration the direction in which the particle moves, as well as the magnitude (or size) of that change in position.
A scalar number is a single value that represents a magnitude or quantity. Scalar numbers are used to describe quantities that have only magnitude and no direction, such as temperature, length, and time. Scalar numbers are usually represented by real numbers, and they can be added, subtracted, multiplied, and divided just like any other real number.
A vector is a mathematical representation of a quantity that has both magnitude and direction. Vectors are used to describe physical quantities such as velocity, force, and displacement. Vectors are represented by directed line segments, and they can be added, subtracted, and multiplied (scalar multiplication) to produce new vectors. The magnitude of a vector can be represented by its length, and its direction can be represented by its angle relative to a reference axis.
Fig. 1. The representation of a vector. Note that the line representing the vector (blue) has direction.
Considering a particle is in motion along a straight line,
If x > 0, the particle is to the right of the origin.
If x < 0, the particle is to the left of the origin.
When a particle changes direction during its movement, a motion diagram along a number line can be sketched, in order to understand the start point, turning point (s) and end point of the particle in space.
The difference between displacement and distance is that distance is irrespective of direction (it is a scalar quantity), while displacement considers the position of a particle respective to the origin point of the movement, hence also considering direction (it is a vector quantity).
An object travels with displacement function \(x = 10t^2 -7t+1\) meters, where t > 0 seconds.
(a) What is the initial displacement of the object?
(b) When does the object change direction?
(a) The initial displacement signifies that time is 0. Hence, we substitute 0 in for t in the equation for the displacement.
\[x = 10(0)^2 -7(0)+1 = 1\]
The initial displacement is 1 metre (to the right of the origin).
(b) The object changes direction when the value for x reaches its maximum, as any other point after this is closer to the origin signifying a change of direction.
Maximum displacement is reached at the vertex of the quadratic equation.
Vertex occurs when \(t = \frac{-b}{2a}\), so when \(t = \frac{7}{2(10)} = 0.35\)
Hence, the object changes direction at t = 0.35 seconds.
Velocity kinematic equations
Velocity describes how fast a point is moving in a certain direction. In other words,
Velocity is how the displacement of the particle changes with time or the rate of change of displacement of the particle.
Velocity will be referred to as v and \(v = \frac{dx}{dt}\), where t is the time and x is the displacement achieved by the particle in that specific amount of time. This means that if we have an expression for x in terms of t, we can take the derivative of this expression to find the velocity.
The unit for this quantity is displacement/time. If the displacement is in meters and the time in seconds, velocity would be in meters per second, or m/s.
Considering a particle:
If v > 0, the particle is moving to the right.
If v < 0, the particle is moving to the left.
If v = 0, the particle is stationary. If there is a change in the sign of v at this point, the particle has changed direction.
If you are asked a question that asks for a value of speed, instead of velocity, it is important to note that speed, unlike velocity, does not consider the direction of movement. In other words, speed is the change of distance over time and velocity is the change of displacement over time.
For example, if you were walking along the walls of a square room with a perimeter of 12 m in 36 s, and ended back at the point you started, the total distance travelled would be 12 m but the total displacement would be 0 m, as there there is no change between the start and end positions. Therefore, your speed would be \(\frac {12}{36} = 0.333 \space m/s\) and your velocity would be \(\frac{0}{36} = 0 \space m/s\).
The displacement in meters of a car moving between two points A and B is given by \(x = 40t^2 - 15\). Find an expression for the velocity of the car at a given point in time.
We know that \(v = \frac {dx}{dt}\) so we can differentiate the expression above with respect to t to find v. This gives: \(v = 80t\)
So, to go from displacement to velocity we need to differentiate, but how do we go from velocity to displacement? You might recall that integration is the reverse process of differentiation, so we will have to integrate our expression for velocity with respect to time to find the displacement.
\[x = \int v\space dt\]
A marathon runner is moving with a constant velocity of 6 m/s. What is the displacement of the runner in terms of t?
Let the displacement of the runner be x.
As \(x = \int v\space dt\), we will have to integrate the given expression in terms of t to find x. Therefore, \(x = 6t + c\) where c is the integration constant.
Sometimes, the integration constant c needs to be found. In this case, the question will need to provide a value for the displacement at a particular time. This value is often the initial displacement when t = 0. This value can then be substituted into the equation to solve for the unknown c, the integration constant.
Acceleration kinematic equations
Acceleration describes how much faster or slower a particle becomes over time. In other words, acceleration is the change in the velocity of a particle with time, or the rate of change of this velocity. Acceleration will be denoted with the letter a.
\[a = \frac {dv}{dt}\]
But we know already that \(v = \frac {dx}{dt}\), so we can therefore place an expression for a in terms of x: \(a = \frac {d^2v}{dt^2}\). This means that to find the acceleration, we will need to differentiate the displacement twice with respect to t.
The units for this quantity, if the displacement is in meters and the time in seconds, is metres per square second or m/s²
The displacement at time t of a bird is given by \(x = 3t^2 + 12t+-5\) m. What are the velocity and acceleration of the bird?
We know that \(v = \frac {dx}{dt}\) so \(v = 6t + 12 \space m/s\). Now we can find the acceleration by finding the derivative of this expression. Therefore: \(a = 6 \space m/s^2\).
Like before, we can find the velocity of a particle by integrating its acceleration with respect to time,
\[v = \int a \space dt\]
To find x, we will need to integrate a twice with respect to t. This can be represented with a double integral.
\[x = \iint a \space dt \space dt\]
The acceleration of a particle is given by \(a = 2t\). Find the velocity and displacement of this particle in terms of t.
For the velocity, we will need to integrate this expression once and for the displacement, twice.
\(v = \int 2t \space dt = t^2 + c\)
\(x = \int t^2 + c \space dt = \frac {t^3}{3} + ct +d \), where c and d are both constants.
To determine whether the speed of a particle is increasing or decreasing, the signs of both velocity and acceleration must be taken into account.
If velocity and acceleration have the same sign (both positive or both negative), the speed of the particle is increasing.
If velocity and acceleration have opposite signs (one is positive and the other negative), the speed of the particle is decreasing.
Therefore, it is possible to convert between displacement, velocity, and acceleration using calculus.
Derivation of kinematic equations using calculus
As we have seen throughout the article, kinematic equations can be obtained through differentiation or integration. The following table summarises the differentiation or integration that give each kinematic variable: displacement, velocity and acceleration.
| Kinematic Variable | Kinematic Derivative | Kinematic Integration |
| Displacement | --- | \[x = \int v\space dt\]\[x = \iint a \space dt \space dt\] |
| Velocity | \[v = \frac {dx}{dt}\] | \[v = \int a \space dt\] |
| Acceleration | \[a = \frac {dv}{dt}\]\[a = \frac {d^2v}{dt^2}\] | --- |
Table 1. Differentiation and integration of kinematic equations.
Calculus Kinematics - Key takeaways
- The displacement of a particle is the position of that particle relative to an origin point. It is denoted as x, and is a vector quantity, indicating that direction is significant.
- The velocity of a particle is the rate of change of its displacement with respect to time. It is denoted as v and can be obtained by differentiating x with respect to time t. v can be integrated to calculate the displacement x.
- The acceleration of a particle is the rate of change of its velocity with respect to time. It is denoted as a and can be obtained by differentiating x with respect to time t twice, or by differentiating v with respect to time t once. Similarly, a can be integrated once or twice to calculate velocity v, and displacement t, respectively.
- The speed of a particle is the magnitude of its velocity and is a scalar quantity. It is calculated either by taking the modulus of the velocity, or the derivative of the distance (rather than displacement for velocity).
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