Additional: Kinematics using Calculus

Velocity as Derivative of Displacement ($ v = \frac{dx}{dt} $)

In kinematics, the relationship between position and velocity is fundamental. When motion is described by a continuous function of position with respect to time, $ x(t) $, the instantaneous velocity is given by the derivative of the position function with respect to time.

Definition: Instantaneous velocity ($ v $) is the rate of change of position ($ x $) with respect to time ($ t $).

Mathematical Representation:

$ v(t) = \frac{dx}{dt} $

Here:

• $ v(t) $ is the instantaneous velocity at time $ t $.
• $ x(t) $ is the position of the object at time $ t $.
• $ \frac{dx}{dt} $ represents the derivative of the position function $ x(t) $ with respect to $ t $. Geometrically, this derivative is the slope of the tangent line to the position-time graph at time $ t $.

Interpretation: This equation tells us that if we know how the position of an object changes over time, we can find its velocity at any given instant by differentiating the position function. For example, if the position is given by $ x(t) = 5t^2 + 3t + 2 $, then the velocity is $ v(t) = \frac{d}{dt}(5t^2 + 3t + 2) = 10t + 3 $. This means the velocity is not constant but changes linearly with time.

Acceleration as Derivative of Velocity ($ a = \frac{dv}{dt} $)

Similarly, acceleration relates to the rate of change of velocity. Just as velocity is the derivative of position, instantaneous acceleration is the derivative of velocity with respect to time.

Definition: Instantaneous acceleration ($ a $) is the rate of change of velocity ($ v $) with respect to time ($ t $).

Mathematical Representation:

$ a(t) = \frac{dv}{dt} $

Since velocity itself is the derivative of position ($ v = \frac{dx}{dt} $), acceleration can also be expressed as the second derivative of position with respect to time:

$ a(t) = \frac{d}{dt} \left( \frac{dx}{dt} \right) = \frac{d^2x}{dt^2} $

Here:

• $ a(t) $ is the instantaneous acceleration at time $ t $.
• $ v(t) $ is the instantaneous velocity at time $ t $.
• $ \frac{dv}{dt} $ represents the derivative of the velocity function $ v(t) $ with respect to $ t $. Geometrically, this is the slope of the tangent line to the velocity-time graph at time $ t $.

Interpretation: This equation allows us to find the acceleration if we know the velocity as a function of time. For instance, if velocity is given by $ v(t) = 10t + 3 $, then the acceleration is $ a(t) = \frac{d}{dt}(10t + 3) = 10 $. This indicates a constant acceleration of 10 units (e.g., $ \text{m/s}^2 $).

Displacement as Integral of Velocity ($ \Delta x = \int v \, dt $)

Calculus provides the inverse relationship between differentiation and integration. Just as velocity is the derivative of position, displacement can be found by integrating the velocity function with respect to time. Integration essentially sums up infinitesimal changes in position over time.

Definition: Displacement ($ \Delta x $) is the accumulation of velocity over a time interval.

Mathematical Representation:

If the velocity $ v(t) $ is known as a function of time, the displacement $ \Delta x $ from time $ t_1 $ to time $ t_2 $ is given by the definite integral of the velocity function:

$ \Delta x = \int_{t_1}^{t_2} v(t) \, dt $

Here:

• $ \Delta x $ is the displacement over the time interval $ [t_1, t_2] $.
• $ v(t) $ is the instantaneous velocity as a function of time.
• $ dt $ represents an infinitesimal interval of time.
• $ \int_{t_1}^{t_2} \dots \, dt $ denotes the definite integral from $ t_1 $ to $ t_2 $.

Interpretation: This equation allows us to find the total change in position by summing up all the small displacements ($ v(t) \cdot dt $) that occur over the given time interval. For example, if $ v(t) = 10t + 3 $, the displacement from $ t=1 $ to $ t=3 $ is:

$ \Delta x = \int_{1}^{3} (10t + 3) \, dt = \left[ 5t^2 + 3t \right]_{1}^{3} = (5(3)^2 + 3(3)) - (5(1)^2 + 3(1)) $

$ \Delta x = (45 + 9) - (5 + 3) = 54 - 8 = 46 $ units.

Velocity as Integral of Acceleration ($ \Delta v = \int a \, dt $)

In a similar manner, the change in velocity can be found by integrating the acceleration function with respect to time. This process sums up all the infinitesimal changes in velocity that occur over a given time interval.

Definition: Change in velocity ($ \Delta v $) is the accumulation of acceleration over a time interval.

Mathematical Representation:

If the acceleration $ a(t) $ is known as a function of time, the change in velocity $ \Delta v $ from time $ t_1 $ to time $ t_2 $ is given by the definite integral of the acceleration function:

$ \Delta v = \int_{t_1}^{t_2} a(t) \, dt $

Here:

• $ \Delta v $ is the change in velocity ($ v_f - v_i $) over the time interval $ [t_1, t_2] $.
• $ a(t) $ is the instantaneous acceleration as a function of time.
• $ dt $ represents an infinitesimal interval of time.
• $ \int_{t_1}^{t_2} \dots \, dt $ denotes the definite integral from $ t_1 $ to $ t_2 $.

Interpretation: This equation allows us to find the total change in velocity by summing up all the small changes in velocity ($ a(t) \cdot dt $) that occur over the given time interval. For example, if $ a(t) = 10 $, the change in velocity from $ t=1 $ to $ t=3 $ is:

$ \Delta v = \int_{1}^{3} 10 \, dt = [10t]_{1}^{3} = 10(3) - 10(1) = 30 - 10 = 20 $ units.

If the initial velocity $ v_i $ is known, the final velocity $ v_f $ can be found as $ v_f = v_i + \Delta v $. This is essentially deriving the first kinematic equation ($ v = u + at $) using calculus, where $ a $ is constant.