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(D) Let the total time of travel be $T$ seconds and acceleration be $a$. The initial velocity $u = 0$.Distance covered in the first second ($t=1$ s) i

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$400$

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(D) Let the total time of travel be $T$ seconds and acceleration be $a$. The initial velocity $u = 0$.Distance covered in the first second ($t=1$ s) is given by $s_1 = ut + \frac{1}{2}at^2$.$1 = 0(1) + \frac{1}{2}a(1)^2 \Rightarrow a = 2 \,m/s^2$.Distance covered in the last second is given by the formula for distance in the $n^{th}$ second: $s_n = u + \frac{a}{2}(2n - 1)$.Here, $s_n = 39 \,m$, $u = 0$, $a = 2 \,m/s^2$.$39 = 0 + \frac{2}{2}(2T - 1) \Rightarrow 39 = 2T - 1 \Rightarrow 2T = 40 \Rightarrow T = 20 \,s$.The total distance $S$ covered in $T = 20 \,s$ is given by $S = ut + \frac{1}{2}aT^2$.$S = 0(20) + \frac{1}{2}(2)(20)^2 = 400 \,m$.

$A$ body travels in a straight line from point $A$ to point $B$ with an initial velocity zero and uniform acceleration, covering $1 \,m$ during the first second and $39 \,m$ during the last second. The distance between $A$ and $B$ in metre is

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