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Step-by-Step Solution
Step 1: Identify the Type of Motion
Since a constant force acts on the particle from rest, its acceleration is constant. Hence, the motion is uniformly accelerated, starting from rest.
Step 2: Write Down the Relevant Equation
For uniform acceleration starting from rest (initial velocity $u=0$), the distance covered $S$ in time $t$ is given by:
$S = \frac{1}{2} a t^2$
Step 3: Express $S_1$ and $S_2$
Let $S_1$ be the distance traveled in the first 10 seconds ($t_1 = 10\,\text{s}$) and $S_2$ be the distance traveled in the first 20 seconds ($t_2 = 20\,\text{s}$). Using the formula:
$S_1 = \frac{1}{2} a \times (10)^2 = \frac{1}{2} a \times 100$
$S_2 = \frac{1}{2} a \times (20)^2 = \frac{1}{2} a \times 400$
Step 4: Find the Ratio of $S_1$ and $S_2$
Taking the ratio of $S_1$ to $S_2$:
$\frac{S_1}{S_2} = \frac{\frac{1}{2}a \times 100}{\frac{1}{2}a \times 400}
= \frac{100}{400}
= \frac{1}{4}$
Hence, $S_1 : S_2 = 1 : 4$, so
$S_2 = 4 \, S_1$
Step 5: Conclusion
Therefore, the distance covered in the first 20 seconds is four times the distance covered in the first 10 seconds.
