Question

If $x={{t}^{4}}-8{{t}^{3}}+16t+8$ then find the acceleration when it is minimum.

Answer 1

Acceleration (a) is equal to the second derivative position (x) of a particle with respect to time. When a function, say $y=f(x)$ is minimum, its first derivative with respect to x is zero and the second derivative of the function at that point is positive.

Formula used:
$a=\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$

Complete step by step answer:
In the above question, let us assume that x is the position of a particle. We can see that x is a function of time and we are supposed to find the minimum acceleration of the body. Let us first find the expression for the acceleration of the particle. Acceleration (a) is equal to the second derivative position (x) of a particle with respect to time.
i.e. $a=\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$.
Therefore, differentiate x with respect to time t.
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{4}}-8{{t}^{3}}+16t+8 \right)$
$\Rightarrow \dfrac{dx}{dt}=4{{t}^{3}}-24{{t}^{2}}+16$ …. (i).
Now, differentiate (i) with respect to t.
$\Rightarrow \dfrac{d}{dt}\left( \dfrac{dx}{dt} \right)=\dfrac{d}{dt}\left( 4{{t}^{3}}-24{{t}^{2}}+16 \right)$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=12{{t}^{2}}-48t$.
Therefore, the acceleration of the particle is $a=12{{t}^{2}}-48t$ … (i)
When a function, say $y=f(x)$ is minimum, its first derivative with respect to x is zero and the second derivative of the function at that point is positive.
In this case, acceleration (a) is a function of time t.
Now, differentiate (i) with respect to t.
$\Rightarrow \dfrac{da}{dt}=\dfrac{d}{dt}\left( 12{{t}^{2}}-48t \right)$
$\Rightarrow \dfrac{da}{dt}=24t-48$ …. (ii)
Now, equate it to zero.
$\Rightarrow \dfrac{da}{dt}=24t-48=0$
$\Rightarrow t=2s$.
Now, differentiate (ii) with respect to t, to find the second derivative of a.
$\Rightarrow \dfrac{d}{dt}\left( \dfrac{da}{dt} \right)=\dfrac{d}{dt}\left( 24t-48 \right)=24$
$\Rightarrow \dfrac{{{d}^{2}}a}{d{{t}^{2}}}=24$
This means that the acceleration at time $t=2s$ is minimum.
Substitute $t=2s$ in (i).
$a =12{{(2)}^{2}}-48(2)\\\ \Rightarrow a =48-96\\\ \therefore a = -48m{{s}^{-2}}$

Therefore, the minimum acceleration of the particle is $-48m{{s}^{-2}}$.

Note: When a function $y=f(x)$ is at its maximum value, its first derivative with respect to x at that point is zero and the second derivative of the function at that point is negative. The first derivative of position of a particle with respect to x is defined as velocity of the particle. This means that acceleration is equal to the first derivative of the position.