Acceleration is equal to the second derivative of position with respect to time.

Recall the following rules of differentiation to help solve this problem.

Power Rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}\) 

Product Rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(u(t)v(t)) = u(t)v'(t)+u'(t)v(t)\)

Exponentials Rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(ae^{bt}) = abe^{bt}\)

Therefore, by the product rule, the power rule, and the rule for differentiating exponentials,

\(\displaystyle \\a(t) = \frac{\mathrm{d} }{\mathrm{d} t}\left(\frac{\mathrm{d} }{\mathrm{d} t}(e^{4t^2} + t)\right)\\ \\ a(t) = \frac{\mathrm{d} }{\mathrm{d} t}(8te^{4t^2}+ 1)\\ \\ a(t) = 64t^2e^{4t^2} + 8e^{4t^2}\\ \\ a(t) = 8e^{4t^2}(8t^2 + 1)\)

The position is given by the function \(\displaystyle s(t)=3sin(\frac{1}{2}t)\). First find the function for velocity by finding the derivative using the chain rule:

\(\displaystyle v(t)=3cos(\frac{1}{2}t)*\frac{1}{2} = 1.5cos(\frac{1}{2}t)\)

Then find the function for acceleration by finding the derivative using the chain rule:

\(\displaystyle a(t)=1.5*-sin(\frac{1}{2}t)*\frac{1}{2}=-\frac{3}{4}sin(\frac{1}{2}t)\)

We are seeking the acceleration at \(\displaystyle \frac{\pi}{3}\) seconds, so plug in \(\displaystyle \frac{\pi}{3}\) for t:

\(\displaystyle a(\frac{\pi}{3})=-\frac{3}{4}sin(\frac{1}{2}*\frac{\pi}{3})= -\frac{3}{4}sin(\frac{\pi}{6})= -\frac{3}{4}*\frac{1}{2}= -\frac{3}{8}\)

Take the first derivative to determine the function for velocity, using the chain rule:

\(\displaystyle g'(x)=8cos(8x)+\frac{1}{2}sin(\frac{1}{2}x)\).

Now use the chain rule again to find the second derivative, the function for acceleration:

\(\displaystyle g''(x)=-64sin(8x)+\frac{1}{4}cos(\frac{1}{4}x)\)

We're looking for the acceleration at \(\displaystyle x=0\), so plug in 0:

\(\displaystyle g''(0)=-64sin(8*0)+\frac{1}{4}cos(\frac{1}{4}*0)\)

At 0, sine is 0 and cosine is 1, so this results in

\(\displaystyle g''(0)=0+\frac{1}{4}*1 = 0.25\)

The derivative of velocity is accleration.

The power rule states that the derivative of \(\displaystyle x^n\) is \(\displaystyle nx^{n-1}\).

Thus the acceleration function is

\(\displaystyle a(t)=3t^2-6t-18\).

The last step is to plug in 4 to the acceleration function.

\(\displaystyle a(4)=3(4)^2-6(4)-18\)

\(\displaystyle =48-24-18\)

\(\displaystyle =6\)

Velocity is the derivative of position. Acceleration is the derivative of velocity. That means that the second derivative of position is the acceleration.

The derivative of \(\displaystyle x^n\) is \(\displaystyle nx^{n-1}\).

So using these rules

\(\displaystyle v(t)=5t^4+8t^3-15t^2+2t-10\)

\(\displaystyle a(t)=20t^3+24t^2-30t+2\).

Acceleration can be found as the second time derivative of the position function, or the first time derivative of the velocity function:

\(\displaystyle a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)\)

Begin by finding the velocity function by deriving the postion function with respect to time:

\(\displaystyle v(t)=2e^{2t}t^2+2e^{2t}t\)

After that, derive once more to find acceleration:

\(\displaystyle a(t)=4e^{2t}t^2+4e^{2t}t+4e^{2t}t+2e^{2t}\)

\(\displaystyle a(t)=4e^{2t}t^2+8e^{2t}t+2e^{2t}\)

Acceleration is given as the time derivative of velocity:

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

Therefore, if velocity is given as:

\(\displaystyle v(t)=sin^2(t^2)\)

Then acceleration is

\(\displaystyle a(t)=4tsin(t^2)cos(t^2)\) 

which can also be written as 

\(\displaystyle a(t)=2tsin(2t^2)\)

If the derivative seems tricky, it may help to view the velocity function as

\(\displaystyle v(t)=sin(t^2)sin(t^2)\) 

and then to use the product rule of derivatives.

To find the acceleration function, we must find the second derivative of the position equation (which is the acceleration equation):

\(\displaystyle f'(x)=-e^x\sin(x)+4x^3\)

\(\displaystyle f''(x)=-e^x\sin(x)-e^{2x}\cos(e^x)+12x^2\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f' (g(x))\cdot g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)

Now, plug in the point x=0 into the second derivative function to find the acceleration of the particle:

\(\displaystyle f''(0)=a(0)=-sin(1)-cos(1)\)

Acceleration is the second time derivative of the positional function and the first time derivative of the velocity function:

\(\displaystyle a(t)=\frac{d}{dt}v(t)=\frac{d^2}{dt^2}x(t)\)

If position is expressed as:

\(\displaystyle x(t)=t^3 +4t^2-sin(2t)\)

Then acceleration can be found to be:

\(\displaystyle a(t)=6t+8+4sin(2t)\)

\(\displaystyle a(0)=8+0=8\)

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

For this problem:

\(\displaystyle a(t)=\frac{d}{dt}(v(t))=3\frac{d}{dt}(t^{2})+15\frac{d}{dt}(t)-\frac{d}{dt}(7)\)

\(\displaystyle a(t)=3*2t+15\)

\(\displaystyle a(t)=6t+15\)