Question

X=(t+2)-1acceleration and velocity me relation

Answer 1

AnswEr :

Given that,

$\sf \: x = {(t + 2)}^{ - 1}$

We have to find out the relation between Velocity and Acceleration

Differentiating x w.r.t t,we get velocity :

$\sf \: v = \dfrac{dx}{dt} \\ \\ \longrightarrow \: \sf \: v = \dfrac{d \bigg( (t + 2) {}^{ - 1} \bigg)}{dt} \\ \\ \longrightarrow \: \sf \: v = - (t + 2) {}^{ - 2} - - - - - - (1)$

Differentiating a w.r.t t,we get :

$\sf \: a = \dfrac{dv}{dt} \\ \\ \longrightarrow \: \sf \: a = - \dfrac{d \bigg( (t + 2) {}^{ - 2} \bigg)}{dt} \\ \\ \longrightarrow \: \sf \: a= (t + 2) {}^{ - 3} - - - - - - - (2)$

Comparing equations (1) and (2),

$\sf \: a = - \bigg( - (t + 2) {}^{ - 3} \bigg) \\ \\ \dashrightarrow \: \sf \: a = - \bigg( - (t + 2) {}^{ -2 } \bigg) \bigg((t + 2) {}^{ - 1} \bigg) \\ \\ \dashrightarrow \: \sf \: a = - v(t + 2) {}^{ - 1} \\ \\ \dashrightarrow \boxed{ \boxed{ \sf \: a = - vx}} \:$

Answer 2

Given

$x = (t-2)^{-1}$

b) Velocity

Velocity is the rate of change of displacement.

$\sf Velocity = \dfrac{Displacement }{Time}$

$\implies \sf v = \sf \dfrac{dx}{dt}$

$\implies \sf v = \dfrac{d( (t-2)^{-1}))}{dt}$

$\implies \boxed{\boxed{\bold{\sf v = - (t-2)^{-1}}}}}}$

a) Acceleration

Acceleration is the rate of change of velocity

$\sf Acceleration = \dfrac{Velocity }{Time}$

$\implies \sf a = \dfrac{dv}{dt}$

$\implies \sf a = \dfrac{d( (t-2)^{-2}))}{dt}$

$\implies \boxed{\boxed{\bold{\sf a = - (t+2)^{-2}}}}}}$

Comparing, we get,

$\sf a = -(- (t-2)^{-3}})$

$\sf a = -(- (t+2)^{-2})((t+2)^{-1}$

$\sf a = -v(t+2)^{-1}$

$\boxed{\boxed{\bold{a=vx}}}}}$