Question

A particle initially at rest is moving along a straight line with an acceleration of 2 ms⁻². At a time of 3 s after the beginning of motion, the direction of acceleration is reversed. The time from the beginning of the motion in which the particle returns to its initial position is

• A. \((3+\sqrt{3})\)s
• B. \((2+\sqrt{2})\)s
• C. \(3(2+\sqrt{2})\)s
• D. \(2(3+\sqrt{3})\)s

Hint:

For problems with changing conditions (like a reversal in acceleration), always break the problem into segments. The final conditions of one segment become the initial conditions for the next. Draw a simple diagram to visualize the particle's path.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves motion with non-uniform acceleration. The motion must be broken down into two distinct phases: the first phase with constant positive acceleration and the second phase with constant negative acceleration. We will use the standard kinematic equations.
Step 2: Phase 1 (from t = 0 to t = 3 s):
The particle starts from rest (\(u = 0\)) with acceleration \(a = 2\) ms⁻².
First, we find the velocity (\(v_1\)) and displacement (\(s_1\)) at the end of this phase (at \(t_1 = 3\) s).
\[ v_1 = u + at_1 = 0 + (2)(3) = 6 \text{ m/s} \] \[ s_1 = ut_1 + \frac{1}{2}at_1^2 = 0 + \frac{1}{2}(2)(3)^2 = 9 \text{ m} \] So, at t = 3 s, the particle is at a position of 9 m from the start and is moving with a velocity of 6 m/s.
Step 3: Phase 2 (for t>3 s):
Now, the acceleration is reversed, so \(a' = -2\) ms⁻². The initial velocity for this phase is the final velocity from Phase 1, i.e., \(u' = v_1 = 6\) m/s. The particle is currently at \(s_1 = 9\) m. It needs to return to its initial position, which means its displacement in this second phase must be \(\Delta s = -9\) m.
Let \(t_2\) be the time taken for this second phase. We use the displacement equation:
\[ \Delta s = u't_2 + \frac{1}{2}a't_2^2 \] \[ -9 = (6)t_2 + \frac{1}{2}(-2)t_2^2 \] \[ -9 = 6t_2 - t_2^2 \] Rearranging this into a quadratic equation:
\[ t_2^2 - 6t_2 - 9 = 0 \] We solve for \(t_2\) using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t_2 = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-9)}}{2(1)} = \frac{6 \pm \sqrt{36 + 36}}{2} = \frac{6 \pm \sqrt{72}}{2} \] \[ t_2 = \frac{6 \pm \sqrt{36 \times 2}}{2} = \frac{6 \pm 6\sqrt{2}}{2} = 3 \pm 3\sqrt{2} \] Since time cannot be negative, we take the positive root:
\[ t_2 = 3 + 3\sqrt{2} = 3(1+\sqrt{2}) \text{ s} \] Step 4: Final Answer:
The total time from the beginning of the motion is the sum of the times for both phases:
\[ T = t_1 + t_2 = 3 + (3 + 3\sqrt{2}) = 6 + 3\sqrt{2} = 3(2 + \sqrt{2}) \text{ s} \] Thus, the correct option is (C).