Differentiation in Kinematics of Linear Motion

8.2

Differentiation in Kinematics of Linear Motion

In differentiation, for a function \(y=f(x)\), its derivative \(\dfrac{dy}{dx}\) can be considered as the rate of change of \(y \text{ with respect to }x\)
This concept can be used for the movement of a particle along a straight line

Let the displacement function, \(s=f(t)\), the velocity function, \(v=g(t)\) and the acceleration function, \(a=h(t)\)
The relationship of those three functions can be simplified as follows:

Example

	A particle moves along a straight line. Its displacement, \(s \text{ m}\), from the fixed point \(O\) is given by \(s=3+2t-t^2\), where \(t\) is time, in seconds, after it starts moving. Determine the velocity function, \(v\) and acceleration function, \(a\) of the particle.


	Solution: Given the displacement function, \(s=3+2t-t^2\) \(\begin{aligned} \text{Then, the velocity function at time }t, v&=\dfrac{ds}{dt}\\\\ v&=2-2t \end{aligned}\) \(\begin{aligned} \text{and the acceleration function at time }t, \ a&=\dfrac{dv}{dt}\\\\ a&=-2 \end{aligned}\)

Initial velocity of particle is the velocity at time \(t=0\)
The maximum or minimum displacement occurs when the gradient of tangent or instantaneous velocity of the particle is zero, that is

An instantaneous velocity of a particle that moves along a straight

line from a fixed point from a displacement function, \(s=f(t)\) can be

determined by substituting the value of \(t\) in the velocity function, \(v=\dfrac{ds}{dt}\)

Constant acceleration is when the gradient or the rate of change of velocity with respect to time at any moment is the same
In general,

Instantaneous acceleration, \(a\) of a particle moving along a straight

line and passes through a fixed point can be determined from a

velocity function \(v = f(t)\) or a displacement function, \(v = f(t)\)

by substituting the value of \(t\) into the acceleration function \(a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}\)