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 

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

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 :

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

If a car is travelling north at constant velocity 60 mph, it's possible to write a velocity function for this vehicle, where \(\displaystyle t\) is time in hours.

\(\displaystyle v(t)=60\)

To find the acceleration, take the derivative of the velocity function.

\(\displaystyle a(t)=v'(t)=0\)

The acceleration after an hour, or any time \(\displaystyle t\), is zero.

Recall that velocity is the first derivative of position and acceleration is the second derivative of position.

So given:

\(\displaystyle p(t)=14t^3-14t^2+14t-14\)

Apply the power rule to each term to find the velocity.

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\)

\(\displaystyle v(t)=p'(t)=42t^2-28t+14\)

Applying the power rule a second time we arrive at the acceleration function.

\(\displaystyle a(t)=p''(t)=84t-28\)

The acceleration equation is the second derivative of the displacement equation.

\(\displaystyle f(t) = \frac{100}{t} = 100t^{-1}\)

Therefore the first derivative is equal to

 \(\displaystyle f'(t) = (-1)(100)t^{-1-1} = -100t^{-2}\)

Differentiating a second time gives

\(\displaystyle f''(t) = (-2)(-100)t^{-2-1} = 200t^{-3} = \frac{200}{t^3}\)

We know that acceleration \(\displaystyle a(t)\) is the derivative of velocity with respect to time. 

\(\displaystyle a(t)=v'(t)\)

We also know that the velocity function \(\displaystyle v(t)\) is given by

\(\displaystyle v(t)=3tan(t)sin(t)\)

We need to apply the product rule to solve for the derivative. 

Recall that the product rule is given by:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(f(t)*g(t))=f(t)*g'(t)+g(t)*f'(t)\)

In our case, \(\displaystyle f(t)=tan(t)\) and \(\displaystyle g(t)=sin(t)\)

Therefore,

\(\displaystyle f'(t)=sec^2(t)\) 

\(\displaystyle g'(t)=cos(t)\)

\(\displaystyle v'(t)=a(t)=3(tan(t)cos(t)+sec^2(t)sin(t))\)

We can reduce some terms in the acceleration function. 

\(\displaystyle tan(t)*cos(t)=\frac{sin(t)}{cos(t)}*cos(t)=sin(t)\)

\(\displaystyle sec^2(t)*sin(t)=\frac{sin(t)}{cos(t)}*sec(t)=tan(t)*sec(t)\)

The final answer can be given as

\(\displaystyle a(t)=3(sin(t)+sec(t)tan(t))\)

To solve, simply differentiate to find the acceleration function and then plug in \(\displaystyle 2\).

\(\displaystyle a(t)=v'(t)=4x+6\)

\(\displaystyle a(2)=4(2)+6=14\)

To find the acceleration of a particle at a specific time we will need to take the derivative of the poistion function twice. Taking the derivative of position onces gives us the velocity function. Taking the derivative a second time will result in the acceleration function.

Since \(\displaystyle p(t)=t^3-2t^2-6t-9\),

we will apply the power rule for differentiation which state, \(\displaystyle (x^n)'=nx^{n-1}\).

Therefore applying it once we get,

\(\displaystyle v(t)=p'(t)=(t^3-2t^2-6t-9)'\)

\(\displaystyle v(t)=3t^{3-1}-2(2t^{2-1})-6t^{1-1}-0\)

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

From here we will apply the power rule one more time to find the acceleration function.

\(\displaystyle \\a(t)=v'(t)=(3t^2-4t-6)'\\ \\a(t)=(2*3)t^{2-1}-(1*4)t^{1-1}-0\\ \\a(t)=6t-4\)

Now substitute \(\displaystyle x=4\) to find the specific acceleration the question is asking for.

\(\displaystyle a(4)=6(4)-4=24-4=20\ \frac{in}{sec^2}\)

To solve, simply differentiate \(\displaystyle v(t)\) using the power rule to find \(\displaystyle a(t)\).  Then plug in \(\displaystyle 1\) for \(\displaystyle t\). 

Recall the power rule:

\(\displaystyle f(x)=x^n\)

\(\displaystyle f'(x)=nx^{n-1}\)

Apply this to get

\(\displaystyle a(t)=v'(t)=6t^2-16t+14\)

\(\displaystyle a(1)=6(1)^2-16(1)+14=4\)

Acceleration of a particle can be found by taking the derivative of the velocity function with respect to time. Recall that a derivative gives the rate of change of some parameter, relative to the change of some other variable. When we take the derivative of velocity with respect to time, we are evaluating how velocity changes over time; i.e acceleration! This is just like finding velocity by taking the derivative of the position function.

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

\(\displaystyle v(t)=\frac{d}{dt}p(t)\)

To take the derivative of the function

\(\displaystyle p(t)=\tan(t)+\cos(t)\)

We'll need to make use of the following derivative rule(s):

Trigonometric derivative: 

\(\displaystyle d[\sin(u)]=\cos(u)du\)

\(\displaystyle d[\cos(u)]=-\sin(u)du\)

\(\displaystyle d[\tan(u)]=\frac{du}{\cos^2(u)}\)

Quotient rule: \(\displaystyle d[\frac{u}{v}]=\frac{vdu-udv}{v^2}\)

Note that u and v may represent large functions, and not just individual variables!

Using the above properties, the velocity function is

 \(\displaystyle v(t)=\frac{1}{\cos^2(t)}-\sin(t)\)

And the acceleration function is

\(\displaystyle a(t)=\frac{2\tan(t)}{\cos^2(t)}-\cos(t)\)

At time \(\displaystyle t=4\)

\(\displaystyle a(4)=\frac{2\tan(4)}{\cos^2(4)}-\cos(4)\)

\(\displaystyle a(4)=6.07\)