Step 1: Analyze Phase 1 (0 to 3 seconds):
Initial velocity, \( u = 0 \).
Acceleration, \( a_1 = 2 \, \text{ms}^{-2} \).
Time, \( t_1 = 3 \, \text{s} \).
Let's calculate the displacement (\( S_1 \)) and final velocity (\( v_1 \)) at the end of this phase.
\[ S_1 = ut_1 + \frac{1}{2}a_1t_1^2 = 0 + \frac{1}{2}(2)(3)^2 = 9 \, \text{m} \]
\[ v_1 = u + a_1t_1 = 0 + 2(3) = 6 \, \text{ms}^{-1} \]
So, at \( t=3 \, \text{s} \), the particle is at position \( x=9 \, \text{m} \) moving with velocity \( 6 \, \text{ms}^{-1} \).
Step 2: Analyze Phase 2 (t>3 seconds):
The direction of acceleration is reversed.
New acceleration, \( a_2 = -2 \, \text{ms}^{-2} \).
The particle starts this phase with initial velocity \( u' = v_1 = 6 \, \text{ms}^{-1} \) from position \( x = 9 \, \text{m} \).
We want the particle to return to its initial position (\( x = 0 \)).
Required displacement for Phase 2, \( S' = \text{Final Pos} - \text{Initial Pos} = 0 - 9 = -9 \, \text{m} \).
Step 3: Calculate Time for Phase 2 (\( t' \)):
Using the equation of motion \( S = ut + \frac{1}{2}at^2 \):
\[ -9 = 6(t') + \frac{1}{2}(-2)(t')^2 \]
\[ -9 = 6t' - (t')^2 \]
\[ (t')^2 - 6t' - 9 = 0 \]
Solving this quadratic equation for \( t' \):
\[ t' = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ t' = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-9)}}{2(1)} \]
\[ t' = \frac{6 \pm \sqrt{36 + 36}}{2} = \frac{6 \pm \sqrt{72}}{2} \]
\[ t' = \frac{6 \pm 6\sqrt{2}}{2} = 3 \pm 3\sqrt{2} \]
Since time cannot be negative, we take the positive root:
\[ t' = 3 + 3\sqrt{2} = 3(1+\sqrt{2}) \, \text{s} \]
Step 4: Calculate Total Time:
Total time from the beginning = Time of Phase 1 + Time of Phase 2
\[ T = t_1 + t' = 3 + 3(1+\sqrt{2}) \]
\[ T = 3 + 3 + 3\sqrt{2} = 6 + 3\sqrt{2} \]
\[ T = 3(2 + \sqrt{2}) \, \text{s} \]