To find the velocity of the driver, we take the derivative of the position equation. The velocity equation is

$\displaystyle v(t) = 20\tfrac{m}{s} (t)$

Then to find the acceleration of the toy car, we take the derivative again.  The acceleration equation is 

$\displaystyle a(t) = 20\tfrac{m}{s}$

To find the velocity of the driver, we take the derivative of the position equation. The velocity equation is

$\displaystyle v(t) = 20\tfrac{m}{s} (t)$

Then to find the acceleration of the toy car, we take the derivative again.  The acceleration equation is 

$\displaystyle a(t) = 20\tfrac{m}{s}$

Remember that acceleration is the second derivative of position with respect to time. 

By the product rule, we know that

$\displaystyle \frac{d}{dt}(f(t)*g(t))=f'(t)*g(t)+g'(t)f(t)$, where $\displaystyle f$ and $\displaystyle g$ are functions that have first derivatives $\displaystyle f'$ and $\displaystyle g'$. 

Keeping this in mind:

$\displaystyle y(t)=e^tsin(t)$

$\displaystyle y'(t)=e^tsin(t)+e^tcos(t)$

To solve for the second derivative, we have to take the product rule for each term:

$\displaystyle y''(t)=e^tsin(t)+e^tcos(t)+e^tcos(t)-e^tsin(t)$

            $\displaystyle =2e^tcos(t)$

Solving for $\displaystyle y''(2\pi)$,

$\displaystyle y''(2\pi)=2e^{2\pi}cos({2\pi})=2e^{2\pi}$

The acceleration function is given by the second derivative of the position function. Take the first derivative of $\displaystyle X(t)$ using the power rule:

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

Applying the power rule to this equation, noting the constant of $\displaystyle 6$ in the first term:

$\displaystyle \frac{dX}{dt} X(t) = 12t + 2$

This is the velocity function. Take the derivative once again to find the acceleration.

$\displaystyle \frac{d^2X}{dt^2} X(t) = 12$

Because there is no factor of t in the acceleration equation, you do not need to plug in $\displaystyle 2$ to solve for $\displaystyle a(2)$. The acceleration is a constant $\displaystyle 12$.

The acceleration is simply the second derivative of the position function:

$\displaystyle a(t)=s''(t)$

To find the derivatives in this particular problem use the power rule which states,

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

Applying the power rule once, the first derivative of the given function gives us the velocity:

$\displaystyle v(t)=s'(t)=15t^4-8t$

Applying the power rule a second time, the derivative of this will give us our acceleration function:

$\displaystyle a(t)=60t^4-8$

At time (t=2 seconds):

$\displaystyle a(2)=60(2)^3-8=60(8)-8=480-8=472$

This is one of the answer choices.

The acceleration can be found by taking the second derivative of the position function.

To find the derivatives in this particular problem use the power rule which states,

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

$\displaystyle a(t)=s''(t)=\frac{d^2}{dt^2}[4t^3-5t^2+5t-100]=\frac{d}{dt}[12t^2-10t+5]=24t-10$

At time (t=2 seconds):

$\displaystyle a(2)=24(2)-10=48-10=38$

This is one of the answer choices.

The instantaneous acceleration can be found by taking the derivative of the velocity function, which is given:

To find the derivatives in this particular problem use the power rule which states,

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

$\displaystyle a(t)=v'(t)=\frac{d}{dt}[3t^2-9t+14]=6t-9$

At time (t=5 seconds):

$\displaystyle a(5)=6(5)-9=30-9=21$

This is one of the answer choices.

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

$\displaystyle a(t)=\frac{d}{dt}[et]=1*et^{1-1}=e$.

Hence

$\displaystyle a(t)=e$.

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

$\displaystyle a(t)=\frac{d}{dt}[e^2]=0$.

Hence

$\displaystyle a(t)=0$.

If the velocity of an asteroid is given by g(x), find the function which models the acceleration of the asteroid.

$\displaystyle g(x)=5sin(x)$

Recall that a velocity function is the first derivative of a position function, and that an acceleration function is the first derivative of a velocity function.

So, to find the acceleration function, we need to find the derivative of g(x)

Begin by recalling or looking up the rule for derivative of sine:

$\displaystyle \frac{d}{dx}sinx=cosx$

So the answer to our question can be found via:

$\displaystyle \frac{d}{dx}5sinx=5cosx$

To be clear, our answer is:

$\displaystyle g'(x)=5cosx$