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(B) Let $u$ be the initial velocity and $a$ be the constant acceleration.Using the first equation of motion,$v = u + at$,at $t = 10 \, s$,$v = 20 \, m

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$20$

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(B) Let $u$ be the initial velocity and $a$ be the constant acceleration.Using the first equation of motion,$v = u + at$,at $t = 10 \, s$,$v = 20 \, m/s$:$20 = u + 10a \quad \dots(i)$Using the formula for distance traveled in the $n^{th}$ second,$s_n = u + \frac{a}{2}(2n - 1)$,for $n = 10$ and $s_{10} = 10 \, m$:$10 = u + \frac{a}{2}(2 	imes 10 - 1)$$10 = u + \frac{19a}{2} \quad \dots(ii)$Subtracting equation $(ii)$ from equation $(i)$:$(20 - 10) = (u - u) + (10a - 9.5a)$$10 = 0.5a$$a = \frac{10}{0.5} = 20 \, m/s^2$.

$A$ particle is moving along a straight line with constant acceleration. At the end of the $10^{th}$ second,its velocity becomes $20 \, m/s$,and in the $10^{th}$ second,it travels a distance of $10 \, m$. Then the acceleration of the particle will be ........ $m/s^2$.

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