Question

A particle moves a distance 'x' in time t according to the equation x = (t + 5)-¹. Find the instantaneous acceleration​

Answer 1

Answer:

a=2(t+5)^-3

Explanation:

dx/dt=v=-1(t+5)^-2

dv/dt=a=-1*-2(t+5)^-3

            =2/(t+5)^3

Answer 2

Answer :

• The instantaneous accelaration of the particle is 2(t + 5)⁻³ m/s².

Explanation :

Given :

• Position of the particle, x = (t + 5)⁻¹ m or 1/(t + 5)

To find :

• Instantaneous accelaration of the particle, a = ?

Knowledge required :

• Quotient rule of differentiation :

⠀⠀⠀⠀⠀⠀⠀$\boxed{\sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dx} - (u)\dfrac{d(v)}{dx}}{(v^{2})}}}$

• Derivative of a constant term is 0.

⠀⠀⠀⠀⠀⠀⠀$\boxed{\sf{\dfrac{d(c)}{dx} = 0}}$

• If we differentiate the position of a particle, we will get the instantaneous velocity of that particle.

• If we differentiate the velocity of a particle, we will get the instantaneous acceleration of that particle.

Solution :

First let us find the instantaneous velocity of that particle.By using the quotient rule of differentiation and substituting the values in it, we get : [w.r.t t]

$:\implies \sf{\dfrac{d}{dt}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dt} - (u)\dfrac{d(v)}{dt}}{(v^{2})}} \\ \\ \\ :\implies \sf{v = \dfrac{d}{dt}\bigg[\dfrac{1}{(t + 5)}\bigg] = \dfrac{(t + 5)\dfrac{d(1)}{dt} - (1)\dfrac{d(t + 5))}{dt}}{(t + 5)^{2}}} \\ \\ \\ :\implies \sf{v = \dfrac{d}{dt}\bigg[\dfrac{1}{(t + 5)}\bigg] = \dfrac{(t + 5)(0) - (1)\bigg[\dfrac{d(t)}{dt} + \dfrac{d(5)}{dt}\bigg]}{(t + 5)^{2}}} \\ \\ \\ :\implies \sf{v = \dfrac{d}{dt}\bigg[\dfrac{1}{(t + 5)}\bigg] = \dfrac{(t + 5)(0) - (1)[1 + 0]}{(t + 5)^{2}}} \\ \\ \\:\implies \sf{v = \dfrac{d}{dt}\bigg[\dfrac{1}{(t + 5)}\bigg] = \dfrac{0 - 1}{(t + 5)^{2}}} \\ \\\\ :\implies \sf{v = \dfrac{d}{dt}\bigg[\dfrac{1}{(t + 5)}\bigg] = \dfrac{(-1)}{(t + 5)^{2}}} \\ \\ \\ \boxed{\therefore \sf{v = \dfrac{d}{dt}\bigg[\dfrac{1}{(t + 5)}\bigg] = -(t + 5)^{(-2)}}} \\ \\ \\ \sf{Thus, the \: derivative \: of \: (x + 5)^{(-1)} \: is \: -(x + 5)^{(-2)}}$

Hence the velocity of that particle is -(x + 5)⁻² m/s.

Now let's find out the accelaration of the particle.

By using the quotient rule of differentiation and substituting the values in it, we get : [w.r.t t]

$:\implies \sf{\dfrac{d}{dt}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dt} - (u)\dfrac{d(v)}{dt}}{(v^{2})}} \\ \\ \\ :\implies \sf{a = \dfrac{d}{dt}\bigg[(-)\dfrac{1}{(t + 5)^{2}}\bigg] = (-)\bigg[\dfrac{[(t + 5)^{2}]\dfrac{d(1)}{dt} - (1)\dfrac{d[(t + 5)^{2}]}{dt}}{[(t + 5)^{2}]^{2}}\bigg]} \\ \\ \\ :\implies \sf{a = \dfrac{d}{dt}\bigg[(-)\dfrac{1}{(t + 5)^{2}}\bigg] = (-)\bigg[\dfrac{[(t + 5)^{2}]\dfrac{d(1)}{dt} - (1)\dfrac{d(t^{2} + 10t + 5)}{dt}}{[(t + 5)^{2}]^{2}}\bigg]} \\ \\ \\ :\implies \sf{a = \dfrac{d}{dt}\bigg[(-)\dfrac{1}{(t + 5)^{2}}\bigg] = (-)\bigg[\dfrac{[(t + 5)^{2}]\dfrac{d(1)}{dt} - (1)\dfrac{d(t^{2})}{dt} - \dfrac{d(10t)}{dt} - \dfrac{d(5)}{dt}}{[(t + 5)^{2}]^{2}}\bigg]} \\ \\ \\ :\implies \sf{a = \dfrac{d}{dt}\bigg[(-)\dfrac{1}{(t + 5)^{2}}\bigg] = (-)\bigg[\dfrac{0 - 2t - 10 - 0}{[(t + 5)^{2}]^{2}}\bigg]} \\ \\ \\ :\implies \sf{a = \dfrac{d}{dt}\bigg[(-)\dfrac{1}{(t + 5)^{2}}\bigg] = (-)\bigg[\dfrac{- 2(t + 5)}{(t + 5)^{4}}\bigg]} \\ \\ \\ :\implies \sf{a = \dfrac{d}{dt}\bigg[(-)\dfrac{1}{(t + 5)^{2}}\bigg] = \dfrac{2(t + 5)}{(t + 5)^{4}}} \\ \\ \\ :\implies \sf{a = \dfrac{d}{dt}\bigg[(-)\dfrac{1}{(t + 5)^{2}}\bigg] = \dfrac{2}{(t + 5)^{3}}} \\ \\ \\ \boxed{\therefore \sf{a = \dfrac{d}{dt}\bigg[(-)\dfrac{1}{(t + 5)^{2}}\bigg] = 2(t + 5)^{(-3)}}} \\ \\ \\ \sf{Thus, the \: derivative \: of \: (-)(t + 5)^{(-2)} \: is \: 2(t + 5)^{(-3)}}$

Hence the instantaneous accelaration of the particle is 2(t + 5)⁻³ m/s².