The acceleration function is the first derivative of the velocity function, or 

\(\displaystyle a(t)=v'(t)=\frac{d}{dt}(v(t))\).

We will use the power rule, \(\displaystyle \frac{d}{dt}(t^{c})=ct^{c-1}\), where \(\displaystyle c\) is a constant,

and one of the trigonometric rules, \(\displaystyle \frac{d}{dt}(cos(t))=-sin(t)\), to find the derivative of this velocity function.

For this problem:

\(\displaystyle a(t)=\frac{d}{dt}(v(t))=7\frac{d}{dt}(t)-7\frac{d}{dt}(cos(t))\)

\(\displaystyle a(t)=7+7sin(t)\)

The acceleration function is the second derivative of the position function, or 

\(\displaystyle a(t)=s''(t)=\frac{d^{2}}{dt^{2}}(s(t))=\frac{d}{dt}\left [ \frac{d}{dt}(s(t)) \right ]\).

We will use the power rule, \(\displaystyle \frac{d}{dt}(t^{c})=ct^{c-1}\), where \(\displaystyle c\) is a constant,

and the constant rule, \(\displaystyle \frac{d}{dt}(c)=0\), where \(\displaystyle c\) is a constant, to find the first and second derivative of this position function.

For this problem:

\(\displaystyle s(t)=t^{3}-t^{2}-15t-8\)

\(\displaystyle \frac{d}{dt}(s(t))=\frac{d}{dt}(t^{3})-\frac{d}{dt}(t^{2})-15\frac{d}{dt}(t)-\frac{d}{dt}(8)\)

\(\displaystyle \frac{d}{dt}(s(t))=3t^{2}-2t-15\)

\(\displaystyle a(t)=\frac{d}{dt}\left [ \frac{d}{dt}(s(t)) \right ]=3\frac{d}{dt}(t^{2})-2\frac{d}{dt}(t)-\frac{d}{dt}(15)\)

\(\displaystyle a(t)=3*2t-2\)

\(\displaystyle a(t)=6t-2\)

The acceleration function is the second derivative of the position function, or 

\(\displaystyle a(t)=s''(t)=\frac{d^{2}}{dt^{2}}(s(t))=\frac{d}{dt}\left [ \frac{d}{dt}(s(t)) \right ]\).

We will use the power rule, \(\displaystyle \frac{d}{dt}(t^{c})=ct^{c-1}\), where \(\displaystyle c\) is a constant,

and the trigonometric rules, \(\displaystyle \frac{d}{dt}(cos(at))=-asin(at)\) and \(\displaystyle \frac{d}{dt}(sin(at))=acos(at)\), where \(\displaystyle a\) is a constant, to find the first and second derivative of this position function.

For this problem:

\(\displaystyle s(t)=t^{3}+2sin(3t)\)

\(\displaystyle \frac{d}{dt}(s(t))=\frac{d}{dt}(t^{3})+2\frac{d}{dt}[sin(3t)]\)

\(\displaystyle \frac{d}{dt}(s(t))=3t^{2}+2*[3cos(3t)]\)

\(\displaystyle \frac{d}{dt}(s(t))=3t^{2}+6cos(3t)\)

\(\displaystyle a(t)=\frac{d}{dt}\left [ \frac{d}{dt}(s(t)) \right ]=3\frac{d}{dt}(t^{2})+6\frac{d}{dt}[cos(3t)]\)

\(\displaystyle a(t)=3*2t+6[-3sin(3t)]\)

\(\displaystyle a(t)=6t-18sin(3t)\)

To find the function representing acceleration, differentiate the position function twice. Use the power rule to differentiate.

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\)

The first derivative is taken as follows:

\(\displaystyle x(t)=\frac{1}{t}\)

\(\displaystyle x'(t)=-\frac{1}{t^2}\)

Then, differentate \(\displaystyle x'(t)\) to get the second derivative. 

\(\displaystyle x''(t)=\frac{2}{t^3}\) 

Therefore, the acceleration function is:

\(\displaystyle a(t)=\frac{2}{t^3}\)

To find the funciton representing the pendulum's acceleration, differentiate the velocity function.

\(\displaystyle v(t)=4sin(t)\)

\(\displaystyle v'(t)=4sin(t)\)

Therefore, the acceleration function is:

\(\displaystyle a(t)=4cos(t)\)

 Evaluate \(\displaystyle a(t)\) when \(\displaystyle t=2\pi\). In other words, evaluate \(\displaystyle a(2\pi)\).

\(\displaystyle a(2\pi)=4cos(2\pi)\)

\(\displaystyle a(2\pi)=4*1\)

\(\displaystyle a(2\pi)=4\)

The acceleration is the second derivative of the position and the derivative of the velocity, which in turn is the derivative of the position.

\(\displaystyle \\ a(t)=v'(t)=s''(t)\\ \\=(2t^3+t^4)''=(6t^2+4t^3)'\\ \\=12t+12t^2\)

At \(\displaystyle t=1\), we have 

\(\displaystyle a(1)=12(1)+12(1)^2=24\).

Note: these derivatives can be evaluated using the Power Rule --- which says that for 

\(\displaystyle p(x)=x^n\), \(\displaystyle p'(x)=nx^{n-1}\) --- and the linear properties of derivatives.

The acceleration of the particle is given by the first derivative of the velocity function, which is

\(\displaystyle v'(t)=a(t)=-10\sin(t)+e^t+te^t+4t\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}\), 

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

Now, plug in t=0 into the above function and we get

\(\displaystyle a(0)=0+1+0+0=1\)

The acceleration function is the second derivative of the position function, or the first derivative of the velocity function, which itself is the first derivative of the position function. So, we must first find the velocity function, which is

\(\displaystyle v(t)=s'(t)=e^t+3t^2+4\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}\), 

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

Now, take the derivative of the velocity function to find the acceleration:

\(\displaystyle v'(t)=a(t)=e^t+6t\)

and was found using the same rules as above.

Finally, plug in t=1 into the above function:

\(\displaystyle a(1)=e^1+6\)

To find the acceleration of the particle, find the second derivative of the position equation. 

For this particular function we will need to use the power rule, natural log rule, quotient rule, and exponential rule.

Power Rule: \(\displaystyle x^n \rightarrow nx^{n-1}\)

Natural Log Rule: \(\displaystyle ln(u)\rightarrow \frac{1}{u}\cdot \frac{du}{dx}\)

Quotient Rule: \(\displaystyle \frac{f(x)}{g(x)}\rightarrow \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\)

Exponential Rule: \(\displaystyle e^u\rightarrow e^u \cdot \frac{du}{dx}\)

First lets use the rule of logs to rewrite the position function as follows.

\(\displaystyle x(t)=\ln (\frac{t^2+1}{e^{2t}})=\ln (t^2+1)-2t\).

Now applying the above rules of differentiation the first derivative will give you the velocity of the particle. 

\(\displaystyle x'(t)=v(t)=\frac{2t}{t^2+1}-2\).

The derivative of the velocity will then give you the acceleration of the particle. 

\(\displaystyle v'(t)=a(t)=\frac{2(1-t^2)}{(t^2+1)^{2}}\).

Then set the accelartion equal to zero and solve for the time. Only choose the time that is applicable \(\displaystyle (t\geq0)\). 

\(\displaystyle a(t)=\frac{2(1-t^2)}{(t^2+1)}=0\).

And when you solve for it, \(\displaystyle t=1\).

If we consider the relation between acceleration and position, we can see that it is the rate of change of the rate of change of the position.

So this problem amounts to calculating the second derivative of our function at the given point. 

\(\displaystyle (x^n)' = n\cdot x^{n-1}\) and the derivative of a constant is 0, so:

\(\displaystyle s(t) = 3t^2+1\)

\(\displaystyle s'(t)=6t\)

\(\displaystyle s''(t)=6\)

Fortunately for us this is a constant function, and so no more calculation is needed. 6 itself is the answer.