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# A particle moves a distance x in time t according to equation $x=(t+5)^{-1}$ acceleration of particle is proportional to

$(a)\;(velocity)^{3/2} \quad (b)\;(velocity)^2 \quad (c)\;(velocity)^{-2} \quad(d)\;(velocity)^{2/3}$

+1 vote
distance $x=(t+5)^{-1}$
velocity $v=\large\frac{dx}{dt}$$=-(t+5)^{-2} acceleration a=\large\frac{dv}{dt}$$=2(t+5)^{-3}$
Therefore $v^{3/2}=-(t+5)^{-3}$
Therefore $a=-2 v^{3/2}$
$a\; \alpha\; v^{3/2}$
Hence a is the correct answer.
edited May 26, 2014