Since the velocity function is \(\displaystyle 0\). The acceleration function is also \(\displaystyle 0\), since the derivative of \(\displaystyle 0\) is \(\displaystyle 0\).

Therefore,

\(\displaystyle a(t)=0\).

We are given a function dealing with distance and asked to find an acceleration. recall that velocity is the first derivative of position and acceleration is the derivative of velocity. Find the second derivative of h(t) and evaluate at t=6.

\(\displaystyle h(t)=\frac{4t^3}{3}+\frac{4.5t^2}{3}+3t-7\)

\(\displaystyle h'(t)=4t^2+3t+3\)

\(\displaystyle h''(t)=8t+3\)

\(\displaystyle h''(6)=8*6+3=48+3=51\)

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

To derive a polynomial, simply decrease each exponent by one and bring the original number down in front to multiply.

So this

\(\displaystyle \small \small v(t)=t^5+3t^4-\frac{7t^2}{3}\)

Becomes:

\(\displaystyle \small \small \small \small \small v'(t)=5t^4+12t^3-\frac{14t}{3}\)

So our acceleration is given by

\(\displaystyle \small \small \small \small \small \small a(t)=5t^4+12t^3-\frac{14t}{3}\)

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

To derive a polynomial, simply decrease each exponent by one and bring the original number down in front to multiply.

So this

\(\displaystyle \small \small v(t)=t^5+3t^4-\frac{7t^2}{3}\)

Becomes:

\(\displaystyle \small \small \small \small \small v'(t)=5t^4+12t^3-\frac{14t}{3}\)

So our acceleration is given by

\(\displaystyle \small \small \small \small \small \small a(t)=5t^4+12t^3-\frac{14t}{3}\)

Now, to find the acceleration at 5 seconds, we need to plug in 5 for t

\(\displaystyle \small \small \small \small \small \small \small a(t)=5*5^4+12*5^3-\frac{14*5}{3}=3125+1500-\frac{70}{3}=4601\frac{2}{3}\)

Recall that acceleration is the first derivative of velocity. So, to find acceleration we need to find and evaluate the following:

\(\displaystyle v'(5)\)

So, if 

\(\displaystyle v(t)=4t^3-5t^2+6\)

Then,

\(\displaystyle v'(t)=3\cdot4t^{3-1}-2\cdot5t^{2-1}=12t^2-10t\)

\(\displaystyle v'(5)=12\cdot 5^2-10\cdot 5=250\)

So our acceleration is \(\displaystyle 250\frac{m}{s^2}\).

Recall that acceleration is the second derivative of position, so we need p''(7).

\(\displaystyle p(t)=6t^4-5t^3+6t-18\)

Taking the first derivative we get:

\(\displaystyle p''(t)=72t^2-30t\)

Taking the second derivative and plugging in 7 we get:

\(\displaystyle p''(7)=72\cdot7^2-30\cdot7=3318\)

So our acceleration after 7 seconds is 

\(\displaystyle 3318 \frac{m}{s^2}\).

To find the expression of the acceleration we need to differentiate the expression of the position vector twice with respect to time.

Note that we do differentiation componentwise.

We have the following:

\(\displaystyle (tsin(t))'=sin(t)+tcos(t)\)

\(\displaystyle (tcos(t))'=cos(t)-tsin(t)\).

We differentiate one more time:

 \(\displaystyle (tsin(t))''=2cos(t)-tsin(t)\)

\(\displaystyle (tcos(t))''=-tcos(t)-2sin(t)\)

and the third component is simply zero since it is a constant.

This gives us the expression for the acceleration.

 \(\displaystyle \vec{a}=(2cos(t)-tsin(t),-tcos(t)-2sin(t),0)\)

To find the correct expression of the acceleration, we will need to differentiate the expression of the position twice with respect to time. 

We recall the following:

\(\displaystyle arctan(t)'=\frac{1}{1+t^2}\rightarrow (artan(t))''=\left(-\frac{2t}{(1+t^2)^2}\right)\)

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

and

\(\displaystyle e(t)'=e^t \rightarrow (e^{t})''=e^{t}\). 

This gives us

\(\displaystyle \vec{a}=\left(-\frac{2t}{(1+t^2)^2},-\frac{1}{t^2},e^{t}\right)\)

To find the expression of the acceleration we need to find the second derivative of each of the components with respect to time.

Note that we have the following:

The first component is zero since its second derivative is zero.

\(\displaystyle x(t)=(t+1)'=1 \rightarrow 1'=0\)

\(\displaystyle y(t)=(ln(t)+1)'=\frac{1}{t}\rightarrow (ln(t)+1)''=-\frac{1}{t^2}\)

\(\displaystyle z(t)=(t ln(t))'=\frac{t}{t}+ln(t)\rightarrow (t ln(t))''=\frac{1}{t}\) 

This gives the required expression.

\(\displaystyle \left(0,\frac{-1}{t^2},\frac{1}{t}\right)\)

All we need to do to find the components of the velocity is to differentiate the components of the position vector with respect to time.

We have :

\(\displaystyle (tcos(t))'=cos(t)-tsin(t)\)

\(\displaystyle t'=1\)

\(\displaystyle (tsin(t))'=sin(t)+tcos(t)\)

Collecting the components we have :

 \(\displaystyle \vec{v}=(cos(t)-tsin(t),1,sin(t)+tcos(t))\)