Question

Acceleration varies as $a = 6t$. Starting from rest, the velocity of the particle after $t = 2\text{ s}$ is:

• A. $6\text{ ms}^{-1}$
• B. $12\text{ ms}^{-1}$
• C. $18\text{ ms}^{-1}$
• D. $24\text{ ms}^{-1}$

Hint:

Always check if the acceleration is constant before applying $v = u + at$. Here, acceleration depends on time ($a = 6t$), so using $v = 0 + (6 \times 2) \times 2$ or similar configurations incorrectly treats acceleration as uniform, leading to wrong answers. Integration is mandatory for variable acceleration!

Answer 1

The Correct Option is B

Concept: When acceleration is given as a explicit function of time, $a(t)$, the equations of motion with constant acceleration cannot be used directly. Instead, calculus must be applied. By definition, acceleration is the time rate of change of velocity: $$a = \frac{dv}{dt}$$ Rearranging this differential equation enables integration to determine the velocity: $$dv = a \cdot dt$$ $$\int_{v_1}^{v_2} dv = \int_{t_1}^{t_2} a(t) \, dt$$

Step 1: The problem states that the particle is "starting from rest." This provides our baseline initial conditions at time $t = 0$: $$\text{Initial time } t_1 = 0\text{ s}$$ $$\text{Initial velocity } v_1 = 0\text{ ms}^{-1}$$ We need to compute the final velocity $v_2 = v$ at the target time limit: $$\text{Final time } t_2 = 2\text{ s}$$

Step 2: Substitute the functional form $a = 6t$ into our integral formulation: $$\int_{0}^{v} dv = \int_{0}^{2} 6t \, dt$$ Integrating both sides using the standard power rule formula $\int t^n dt = \frac{t^{n+1}}{n+1}$: $$\Big[ v \Big]_{0}^{v} = 6 \cdot \left[ \frac{t^2}{2} \right]_{0}^{2}$$ $$v - 0 = 3 \cdot \Big[ t^2 \Big]_{0}^{2}$$

Step 3: Evaluate the definite integral by inserting the specific boundary numbers: $$v = 3 \cdot (2^2 - 0^2)$$ $$v = 3 \cdot (4 - 0)$$ $$v = 12\text{ ms}^{-1}$$ Thus, the exact numerical velocity achieved after $2\text{ s}$ is $12\text{ ms}^{-1}$, matching Option (B).