Take the second derivative of $\displaystyle s(t)$ by using the Power Rule $\displaystyle x^n=nx^{n-1}$ twice .

Applying the power rule once will give us our velocity function,

 $\displaystyle s'(t)=-8t+8$.

Applying the power rule a second time results in our acceleration function,

$\displaystyle s''(t)=-8$.

Therefore, the acceleration of the apple is $\displaystyle -8m/s^2$.

Take the second derivative by using the Power Rule $\displaystyle x^n=nx^{n-1}$ of $\displaystyle s(t)$,

$\displaystyle \\s'(t)=v(t)\\ v(t)=5(2)t-10\\v(t)=10t-10$.

Applying the power rule a second time we find the acceleration function,

$\displaystyle s''(t)=10$.

Take the second derivative by using the Power Rule $\displaystyle x^n=nx^{n-1}$ of $\displaystyle s(t)$,

$\displaystyle \\s'(t)=-3(2)t+5\\s'(t)=-6t+5$.

Applying the power rule again we find our acceleration function,

$\displaystyle \\s''(t)=a(t)\\s''(t)=-6$.

Take the second derivative by using the Power Rule $\displaystyle x^n=nx^{n-1}$ of $\displaystyle s(t)$,

$\displaystyle s'(t)=-4(2)t+12$.

Applying the power rule again we get the acceleration function,

$\displaystyle \\s''(t)=a(t)\\a(t)=-8$.

Take the second derivative by using the Power Rule $\displaystyle x^n=nx^{n-1}$ of $\displaystyle s(t)$,

$\displaystyle s'(t)=-8(2)t+20$.

Applying the power rule a second time we find the acceleration function,

$\displaystyle \\s''(t)=a(t)\\a(t)=-16$.

In order to find the acceleration function from the position function we need to take the second derivative of the position function.

$\displaystyle a(t)=\frac{d^2}{dt^2}[s(t)]$

When taking the derivative, we will use the power rule which states

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^{n}=nx^{n-1}$ 

and by applying this rule to each term we get

$\displaystyle a(t)=\frac{d^2}{dt^2}[s(t)]$

$\displaystyle =\frac{d^2}{dt^2}[t^3+5t^2+2t+4]$

$\displaystyle =\frac{d}{dt}[3t^{3-1}+2\cdot 5t^{2-1}+2t^{1-1}+0]$

$\displaystyle =\frac{d}{dt}[3t^2+10t+2]$.

Next we find the second derivative,
$\displaystyle a(t)=\frac{d}{dt}[3t^2+10t+2]$

$\displaystyle =2\cdot 3t^{2-1}+10t^{1-1}+0$

$\displaystyle =6t+10$.

Hence,

$\displaystyle a(t)=6t+10$.

In order to find the acceleration function from the velocity function we need to take the derivative of the velocity function.

$\displaystyle a(t)=\frac{d}{dt}[v(t)]$

 When taking the derivative, we will use the power rule which states

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^{n}=nx^{n-1}$

and by applying this rule to each term we get

$\displaystyle \frac{d}{dt}[v(t)]=\frac{d}{dt}[5t^2+3]=2\cdot 5t^{2-1}+0$.

As such, 

$\displaystyle a(t)=10t$.

Finally, to solve for $\displaystyle a(4)$ we set $\displaystyle t=4$ to get

$\displaystyle a(4)=40$.

The position function of the particle is given as follows:

$\displaystyle f(x)=2x^2+3x+1$

The first derivative of the position function describes the velocity of the particle:

$\displaystyle f'(x)=4x+3$
This derivative was found by using the power rule $\displaystyle \frac{d}{dx}x^n=nx^{n-1}$, the derivative of x=1, and the derivative of a constant equaling zero.

The second derivative of the function - the derivative of the first derivative of the position function - describes the acceleration of the particle:

$\displaystyle f''(x)=4$

The derivative was found by the derivative of x=1, and the derivative of the constant equaling zero.

Thus, the acceleration of the particle is 4.

The acceleration of the spaceship is given by the second derivative of the position equation.

The first derivative - which describes the velocity of the spaceship - is given by 

$\displaystyle v(t)=16t^3+56t$.

This derivative was found using the power rule This derivative was found by using the power rule 

$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$. 

The derivative of this function (the second derivative of the initial function)- the acceleration of the spaceship - is given by 

$\displaystyle a(t)=48t^2+56$.

This derivative was also found using the power rule, and the rule that the derivative of a constant (in this case, 56) equals zero.

At $\displaystyle t=10$, the acceleration 

$\displaystyle a(10)=48(10)^2+56$

$\displaystyle a(10)=4856$.

Take the first derivative by using the Power Rule ($\displaystyle x^n=nx^{n-1}$) of $\displaystyle s(t)$:

$\displaystyle s'(t)=-28t+6$

Using the Power Rule again, find the second derivative of $\displaystyle s(t)$:

$\displaystyle s''(t)=-28$