Step 1: Understanding the Concept:
This problem requires the application of Newton's Second Law of Motion combined with integral calculus.
Newton's Second Law states that the net force acting on an object is equal to the product of its mass and its acceleration (\(F = ma\)).
In standard introductory problems, the force is often constant, which leads to constant acceleration.
However, in this problem, the force is given as \(F = 6t\). This means the force is a function of time.
As time increases, the force pushing the object increases.
Consequently, the acceleration is also a function of time, meaning we cannot use the standard kinematic equations (SUVAT equations like \(v = u + at\)).
To find the velocity when acceleration is changing, we must sum up the infinitesimal changes in velocity over a period, which is the definition of integration.
Key Formula or Approach:
First, we find the acceleration as a function of time:
\[ a(t) = \frac{F(t)}{m} \]
Since acceleration is the derivative of velocity (\(a = \frac{dv}{dt}\)), we can rearrange this to find the change in velocity:
\[ dv = a(t) dt \]
To find the total velocity at a specific time, we integrate both sides of the equation from the initial state to the final state:
\[ \int_{u}^{v} dv = \int_{0}^{t} a(t) dt \]
Given the body starts from rest, the initial velocity \(u\) is 0.
Step 2: Detailed Explanation:
Let's list the given parameters:
Mass (\(m\)) = 2 kg
Force (\(F\)) = \(6t\)
Initial state: \(t = 0\), \(v = 0\) (at rest)
Target: Find \(v\) at \(t = 2\text{ s}\).
Step 1: Calculate the acceleration function.
\[ a(t) = \frac{F}{m} = \frac{6t}{2} = 3t \text{ m/s}^{2} \]
Set up the definite integral for velocity.
\[ v = \int_{0}^{2} a(t) dt = \int_{0}^{2} 3t dt \]
Step 2: Perform the integration using the power rule \(\int t^n dt = \frac{t^{n+1}}{n+1}\).
\[ v = 3 \times \int_{0}^{2} t^1 dt = 3 \times \left[ \frac{t^2}{2} \right]_{0}^{2} \]
\[ v = \frac{3}{2} \times [t^2]_{0}^{2} \]
Step 3: Evaluate the integral at the boundaries (limits).
\[ v = \frac{3}{2} \times (2^2 - 0^2) \]
\[ v = \frac{3}{2} \times (4 - 0) \]
\[ v = \frac{3 \times 4}{2} = \frac{12}{2} = 6 \text{ m/s} \]
The physics behind this result shows that even though the force is increasing, the velocity also increases at a non-linear rate.
If a student incorrectly used the constant acceleration formula with the final force, they would get 12 m/s, which is a common mistake.
Step 3: Final Answer:
By integrating the acceleration function \(a = 3t\) from 0 to 2 seconds, we find the final velocity to be 6 m/s.
This corresponds to option (B).