91Ó°ÊÓ

Initial velocity of the particle, \(v_0 = 10 \mathrm{~m/s}\) Uniform retardation is unknown, let's call it \(a\) Time taken to reach the initial position, \(t = 10\mathrm{~s}\)

The first equation of motion is:\[v = v_0 + at\] Since the particle returns to its initial position, its final velocity will be in the opposite direction, which is -10 m/s. Thus, \(v = -10\mathrm{~m/s}\). We can plug in these values to find the retardation: \[-10\mathrm{~m/s} = 10\mathrm{~m/s} + a \cdot 10\mathrm{~s}\]

Solving the equation for retardation, we have: \[a = \frac{-20\mathrm{~m/s}}{10\mathrm{~s}} = -2\mathrm{~m/s^2}\] So, the uniform retardation is -2 m/s².

We are now asked to find the distance traversed by the particle in 6 seconds. Let's use the third equation of motion: \[s = v_0t + \frac{1}{2}at^2\] Plug in the values for \(v_0 = 10\mathrm{~m/s}\), \(a = -2\mathrm{~m/s^2}\), and \(t = 6\mathrm{~s}\): \[s = 10\mathrm{~m/s} \cdot 6\mathrm{~s} + \frac{1}{2}(-2\mathrm{~m/s^2})(6\mathrm{~s})^2\]

Now, we will calculate the distance: \[s = 60\mathrm{~m} - \frac{1}{2}(12\mathrm{~m}) = 60\mathrm{~m} - 6\mathrm{~m} = 54\mathrm{~m}\] However, as the particle is in the positive direction for 6 seconds, it will reach its maximum distance, and then begin to return towards its initial position. To find the actual distance traversed, we need to find the maximum distance covered before it starts coming back. The time taken to reach the maximum distance is half of the total time taken to return, that is, \(t/2 = 10\mathrm{~s}/2 = 5\mathrm{~s}\). Thus, we can calculate the distance covered in the first 5 seconds. \[s_{max} = 10\mathrm{~m/s} \cdot 5\mathrm{~s} + \frac{1}{2}(-2\mathrm{~m/s^2})(5\mathrm{~s})^2 = 50\mathrm{~m} - 25\mathrm{~m} = 25\mathrm{~m}\] Now, we know that in the first 5 seconds, it covers 25 m, and in the next second, it covers 24 m (as the retardation is 2 m/s²). Therefore, the total distance traversed by the particle in the first 6 seconds is: \[s_{total} = s_{max} + (s - s_{max}) = 25\mathrm{~m} + (54\mathrm{~m} - 25\mathrm{~m}) = 25\mathrm{~m} + 29\mathrm{~m} = 54\mathrm{~m} \] However, since the 29 m covered in the next second is in the opposite direction (negative x-direction), we need to subtract it from the maximum distance, which gives us the correct answer: \[s_{net} = s_{max} - (54\mathrm{~m} - 25\mathrm{~m}) = 25\mathrm{~m} - 29\mathrm{~m} = -4\mathrm{~m}\] Because the net distance in the sixth second is -4 m (in the negative x-direction), we consider only the maximum distance reached in the positive x-direction. Therefore, the distance traversed by the particle in 6 seconds is: \[25 \mathrm{~m}\] The correct option is (B).