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(B) Let the body start from point $A$ at $t=0$ with initial velocity $u=0$ and constant acceleration $a$. At $t=2 \ s$,it reaches point $B$. The dista

Vedclass · Accepted Answer

$(4+2 \sqrt{2}) \ s$

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(B) Let the body start from point $A$ at $t=0$ with initial velocity $u=0$ and constant acceleration $a$. At $t=2 \ s$,it reaches point $B$. The distance covered is $s_1 = \frac{1}{2} a (2)^2 = 2a$. The velocity at $B$ is $v_B = a(2) = 2a$.At $t=2 \ s$,the body reverses its direction but the acceleration $a$ remains in the same direction (acting as deceleration). It stops at point $C$ where its velocity becomes $0$. Let the time taken from $B$ to $C$ be $t'$. Using $v = u + at$,we have $0 = 2a - a(t')$,which gives $t' = 2 \ s$. The distance $BC$ is $s_2 = (2a)(2) - \frac{1}{2} a (2)^2 = 4a - 2a = 2a$.The total distance from $A$ to $C$ is $AC = s_1 + s_2 = 2a + 2a = 4a$.Now,the body starts from rest at $C$ and moves towards $A$ with constant acceleration $a$. Let the time taken to cover distance $AC$ be $T$. Using $s = ut + \frac{1}{2}at^2$,we have $4a = 0 + \frac{1}{2} a T^2$,which gives $T^2 = 8$,so $T = 2\sqrt{2} \ s$.The total time $t_0$ is the sum of time to reach $C$ and time to return to $A$. Time to reach $C$ is $2 \ s + 2 \ s = 4 \ s$. Thus,$t_0 = 4 + 2\sqrt{2} \ s$.

$A$ body starting from rest at $t=0$ moves along a straight line with a constant acceleration. At $t=2 \ s$,the body reverses its direction keeping the acceleration same. The body returns to the initial position at $t=t_0$,then $t_0$ is

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