If the velocity of an asteroid is given by g(x), find the acceleration of the asteroid when $\displaystyle x=\frac{9\pi}{4}$.

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

Begin by recalling that acceleration is the first derivative of velocity.

Next, we can put that knowledge to the use by finding the derivative of g(x)

The derivative of sine is cosine, so we get

$\displaystyle \frac{d}{dx}5sin(x)=5cos(x)$

So now we have the acceleration function, but we need the acceleration when $\displaystyle x=\frac{9\pi}{4}$. 

We can see that 

$\displaystyle \frac{9\pi}{4}=\frac{\pi}{4}+2 \pi$

So...

$\displaystyle 5cos(\frac{9\pi}{4})=\frac{5 \sqrt{2}}{2}$

So our answer must be  

$\displaystyle \frac{5 \sqrt{2}}{2}$.

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

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

The acceleration of the mail is equal to the second derivative of its postion. The first derivative of the position gives the velocity, since it is how the position of the mail is changing over time. The derivative of the velocity gives how the velocity is changing over time, which is the acceleration. 

$\displaystyle f(t) = 5t^2+3t+5\ m$

In order to take the derivative of the position equation, use the power rule. Bring the exponent to the front of the term, multiple it by the constant in front of $\displaystyle t$, and reduce the exponent by one. If the term does not include the $\displaystyle t$ variable, the derivative of the constant will be equal to zero. 

$\displaystyle f'(t) = v(t) = (5*2)t^{2-1} + (3*1)t^{1-1}\ ^m/s$

$\displaystyle f'(t) = v(t) = 10t + 3\ ^m/s$

Now that the velocity has been derived from the position function, the acceleration can be found by taking the derivative of the velocity function. Once again using the power rule to differentiate:

$\displaystyle f''(t) = a(t) = (10*1)t^{1-1}} \ ^m/s^2$

$\displaystyle f''(t) = a(t) = 10 \ ^m/s^2$

This gives the acceleration function.

No integration is required for this question.  Simply convert three minutes to seconds.  There are 60 seconds in a minute, which means that there are 180 seconds in three minutes.

Substitute 180 seconds into the acceleration equation.

$\displaystyle a(t)=3(180)+1 = 540+1=541$

To solve, simply differentiate the position function twice and evaluate at $\displaystyle t=1$. Thus,

$\displaystyle a(t)=p''(t)=12$

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 acceleration at a point, simply differentiate the position function twie and then plug in $\displaystyle 4$. Thus,

$\displaystyle a(t)=d''(t)=12t$

$\displaystyle a(4)=12(4)=48$

Integrate the acceleration function twice to obtain the position function.

$\displaystyle \int\int a(t)\: dt \:dt = \int\int 1\: dt \:dt$

$\displaystyle v(t)=\int 1\: dt = t+C_1$

Integrate again for position.

$\displaystyle s(t) = \int t+C_1\: dt =\frac{t^2}{2} + C_1t + C_2$

There are two unknown constants $\displaystyle C_1$ and $\displaystyle C_2$.  Without the initial conditions for velocity and position, we cannot find the position of the ball at $\displaystyle t=1$.

The answer is:  $\displaystyle \textup{Cannot be solved.}$

The acceleration of the ball is given by the first derivative of the ball's velocity function:

$\displaystyle a(t)=v'(t)=t-2$

The derivative was found using the following rule:

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

Now, find the acceleration at $\displaystyle t=2$ by plugging in $\displaystyle 2$ into the function:

$\displaystyle a(2)=0$

We know that acceleration is the derivative of velocity which is the derivative of position therefore, for this particular problem we will need to take the derivative of the position twice.

For this particular function we will need to use the power rule to find the derivative.

The power rule states, $\displaystyle (x^n)'=nx^{n-1}$.

Also recall, the derivative of a constant is always zero.

Applying the above rules to the position function we can first find our velocity.

$\displaystyle \\x(t)=4t^2+5t+3\\ \\ x'(t)=(4*2)t^{2-1}+(1*5)t^{1-1}+0\\ \\v(t)=x'(t)=8t^1+5t^0\\ \\v(t)=8t+5$

Now to find the acceleration we will need to apply the power rule to the velocity function.

$\displaystyle \\ a(t)=v'(t)=x''(t) \\ \\ a(t)=(8t+5)'\\ \\a(t)=1*8t^{1-1}+0\\ \\a(t)=8t^0=8$

Hence, the acceleration is a constant $\displaystyle 8\ \frac{in}{sec^2}$ for all times, $\displaystyle t$.