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Question

(C) Given that the particle starts from rest,so initial velocity $u = 0$.Let the constant acceleration be $a$.The distance covered in time $t$ is give

Vedclass · Accepted Answer

$y = 3x$

Vedclass · Answer

(C) Given that the particle starts from rest,so initial velocity $u = 0$.Let the constant acceleration be $a$.The distance covered in time $t$ is given by $s = ut + \frac{1}{2}at^2$.For the first $2$ seconds ($t = 2$ s),the distance $x$ is:$x = 0(2) + \frac{1}{2}a(2)^2 = 2a$.For the first $4$ seconds ($t = 4$ s),the total distance covered is $x + y$:$x + y = 0(4) + \frac{1}{2}a(4)^2 = 8a$.Substituting $x = 2a$ into the equation:$2a + y = 8a$$y = 6a$.Now,comparing $x$ and $y$:$\frac{y}{x} = \frac{6a}{2a} = 3$.Therefore,$y = 3x$.

$A$ particle starting from rest and moving with constant acceleration travels a distance $x$ in the first $2$ seconds and a distance $y$ in the next $2$ seconds. Then:

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