Let the acceleration of particle N be \( a \). Then, the acceleration of particle M will be \( 2a \).
Since both particles start from rest and cover the same distance \( s \), we can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken.
For particle N:
\[ s = 0 + \frac{1}{2} a t_N^2 \] \[ s = \frac{1}{2} a t_N^2 \quad \text{(1)} \]
For particle M:
\[ s = 0 + \frac{1}{2} (2a) t_M^2 \] \[ s = a t_M^2 \quad \text{(2)} \]
Both equations represent the same distance \( s \). Setting them equal gives:
\[ \frac{1}{2} a t_N^2 = a t_M^2 \]
We can cancel \( a \) from both sides (assuming \( a \neq 0 \)):
\[ \frac{1}{2} t_N^2 = t_M^2 \]
Rearranging this equation:
\[ t_N^2 = 2 t_M^2 \]
Taking the square root of both sides:
\[ t_N = \sqrt{2} t_M \]
To find the ratio of the times taken by M and N, we calculate:
\[ \frac{t_M}{t_N} = \frac{t_M}{\sqrt{2} t_M} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]
Thus, the ratio of the time taken by M to that taken by N is:
\[ \frac{t_M}{t_N} = \frac{1}{\sqrt{2}} \quad \text{or} \quad 1 : \sqrt{2} \]
In conclusion, the time taken by M to that taken by N is in the ratio:
\[ \boxed{1 : \sqrt{2}} \quad \text{or simply} \quad \boxed{1 : 1.414} \]
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