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# A particle starts from the position of rest under a constant acceleration. If travels a distance x in the first 10 seconds and distance y in the next 20 seconds . Then

$\begin{array}{1 1}(A)\;y=x \\(B)\;y=2x\\(C)\;y=9x \\(D)\;y=4x \end{array}$

$x=\large\frac{1}{2}$$a (10)^2 =\large\frac{1}{2}$$a(100)$
$x+y =\large\frac{1}{2} $$a(30)^ 2 =\large\frac{1}{2}$$a (900)$
$\qquad= 9x$
Hence c is the correct answer.