Step 1: Understanding the Concept:
Kinematics is the study of motion without considering the forces that cause it.
The three primary variables in kinematics are displacement (\(s\)), velocity (\(v\)), and acceleration (\(a\)).
These variables are intrinsically linked through the process of differentiation and integration with respect to time (\(t\)).
Velocity is defined as the rate of change of displacement, and acceleration is defined as the rate of change of velocity.
In graphical terms, if you have a velocity-time (\(v-t\)) graph, the "slope" of the tangent to the curve at any point gives the instantaneous acceleration.
When the relationship between velocity and time is linear (expressed as a first-degree equation), the slope is constant.
A constant slope implies that the velocity is changing at a uniform rate, which is the definition of uniform or constant acceleration.
Key Formula or Approach:
The standard mathematical definition of instantaneous acceleration (\(a\)) is the first derivative of the velocity function with respect to time:
\[ a = \frac{dv}{dt} \]
If we look at the general equation for a straight line passing through the origin, \(y = mx\), where \(m\) is the slope.
In the context of motion, if \(v = at\), then \(a\) is the slope.
We can also compare the given equation to the first equation of motion: \(v = u + at\), where \(u\) is initial velocity and \(a\) is acceleration.
Step 2: Detailed Explanation:
The question provides a specific velocity function: \(v(t) = 4t\).
This indicates that as time increases, the velocity increases proportionally. For example:
At \(t = 0\text{s}\), \(v = 0\text{ m/s}\).
At \(t = 1\text{s}\), \(v = 4\text{ m/s}\).
At \(t = 2\text{s}\), \(v = 8\text{ m/s}\).
To find the exact acceleration, we apply calculus by differentiating the function \(v = 4t\) with respect to time \(t\).
Using the power rule of differentiation, where \(\frac{d}{dt}(t^n) = nt^{n-1}\):
\[ a = \frac{d}{dt}(4t) \]
\[ a = 4 \times \frac{d}{dt}(t^1) \]
\[ a = 4 \times (1 \times t^{1-1}) \]
\[ a = 4 \times 1 = 4 \]
The result is a constant numerical value, 4.
Since the derivative does not contain the variable \(t\), the acceleration does not change as time passes.
This means the particle is undergoing uniform acceleration of \(4 \text{ m/s}^{2}\).
Graphically, this corresponds to a straight line on a \(v-t\) plot with a constant gradient of 4.
If the acceleration were "variable," the velocity equation would have contained higher powers of \(t\) (like \(t^2\) or \(t^3\)).
Step 3: Final Answer:
By differentiating the linear velocity function \(v = 4t\), we obtain a constant value of 4.
Thus, the acceleration is \(4 \text{ m/s}^{2}\), which matches option (B).