Question

10. A particle moves a distance x in time t
to equation x = (t + 5)^-1. The acceleration of paticle
is proportional to
[AIPMT (Prelims
(1) (Velocity)^3/2
(2) (Distance)^2
(3) (Distance)^-2
(4) (Velocity)^2/3​

Answer 1

Answer:

hey mate I think the answer should be

Explanation:

(4) this one

Answer 2

$\huge\underline\mathfrak\red{Answer}$

$x = {(t +5 )}^{ - 1} \\ \\ \frac{dx}{dt} = \frac{ - 1}{ {(t + 5)}^{2} } \\ \\ v = \frac{ - 1}{ {(t + 5)}^{2} } \\ \\ - \sqrt{v} = \frac{1}{t + 5} \\ \\ - {v}^{ \frac{1}{2} } = \frac{1}{t + 5} \\ \\ - (t + 5) = \frac{1}{ {v}^{ \frac{1}{2} } } \\ \\ \frac{d(t + 5)}{dt} = \frac{d( {v}^{ \frac{ - 1}{2} } )}{dt} \\ \\ 1 = - \frac{1}{2} ( {v}^{ \frac{ - 3}{2} }). \frac{dv}{dt} \\ \\ - 2. \frac{1}{ {v}^{ \frac{ - 3}{2} } } = \frac{dv}{dt} \\ \\ - 2 {v}^{ \frac{3}{2} } = a$

ANSWER IS OPTION A