Question

a particle moves distance x in t time to equation x = ( t +5 ) raise to -1 . the accelration of particle is proportional to ​

Answer 1

Given that,

$\sf x(t) = (t + 5) {}^{ - 1} \\ \\ \dashrightarrow \sf x(t) = \dfrac{1}{(t + 5)}$

Differentiating w.r.t t, we get velocity of the particle,

$\dashrightarrow \sf v(t) = \dfrac{ - 1}{(t + 5) {}^{2} }$

Differentiating w.r.t t, we get acceleration of the particle,

$\dashrightarrow \sf a(t) = \dfrac{2}{(t + 5) {}^{3}} \\ \\ \dashrightarrow \sf a(t) = 2(t + 5) {}^{ - 3} \\ \\ \dashrightarrow \boxed{ \boxed{ \sf a(t) \propto x{}^{ - 3} }}$