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A particle moves a distance x in time t according to equation $x = (t + 5)^{-1}$. The acceleration of particle is proportional to
A
$(\text{Velocity})^{2/3}$
B
$(\text{Velocity})^{3/2}$
C
$(\text{Distance})^2$
D
$(\text{Distance})^{-2}$
Detailed Solution
$x = (t + 5)^{-1}$
Velocity $v = \frac{dx}{dt} = -(t + 5)^{-2}$
Acceleration $a = \frac{dv}{dt} = 2(t + 5)^{-3}$
From the velocity, $|v| = (t + 5)^{-2}$, so $(t + 5)^{-1} = |v|^{1/2}$
$a = 2\left[(t + 5)^{-1}\right]^3 = 2\left(|v|^{1/2}\right)^3 = 2|v|^{3/2}$
So $a \propto (\text{velocity})^{3/2}$
(In terms of distance, $a = 2x^3$, which is not among the choices.)
Velocity $v = \frac{dx}{dt} = -(t + 5)^{-2}$
Acceleration $a = \frac{dv}{dt} = 2(t + 5)^{-3}$
From the velocity, $|v| = (t + 5)^{-2}$, so $(t + 5)^{-1} = |v|^{1/2}$
$a = 2\left[(t + 5)^{-1}\right]^3 = 2\left(|v|^{1/2}\right)^3 = 2|v|^{3/2}$
So $a \propto (\text{velocity})^{3/2}$
(In terms of distance, $a = 2x^3$, which is not among the choices.)