The object acceleration can be found by integrating the object's jerk. This can be done using the power rule where

\(\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}\).

Therefore the acceleration of the object is

\(\displaystyle a(t) = \int (11t)dt = \frac{11t^{(1+1)}}{1+1} +C = \frac{11t^2}{2} +C\)

Solving for the constant C using the initial condition given in the question is as follows.

\(\displaystyle a(0) = \frac{11(0)^2}{2} +C = 0 +C = 10\)

Therefore \(\displaystyle C =10\) and \(\displaystyle a(t) = \frac{11t^2}{2} + 10\) .

The acceleration of the object can be found by differentiating the position equation twice, using the product rule. The product rule is where

\(\displaystyle f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)\).

Using the product rule, the velocity of the object is

\(\displaystyle v(t) = (\cos(t))\cos(t) + \sin(t)(-\sin(t)) = \cos^2(t) - \sin^2(t)\).

To differentiate the velocity equation we must use the power rule and the chain rule. 

The power rule is

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

The chain rule is

\(\displaystyle f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)\).

Therefore the acceleration equation is then

\(\displaystyle \\a(t) = v'(t) = (2)\cos^{(2-1)}(t)(-\sin(t)) - (2)\sin^{(2-1)}(t)(\cos(t)) \\ \Rightarrow -2\cos(t)\sin(t) - 2\sin(t)\cos(t) = -4\cos(t)\sin(t)\).

The acceleration of the object is the derivative of the velocity. We can differentiate the velocity using the power rule where if

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

Therefore the acceleration equation of the object is

\(\displaystyle a(t) = 2t^1 - 5t^0 =2(1)t^{(1-1)} - 5(0)t^{(0-1)} = 2\).

Because the acceleration does not have depend on the variable \(\displaystyle t\) the function is constant. Therefore the acceleration of the object is \(\displaystyle 2\).

The acceleration of the object can be found by differentiating the position equation of the object twice. This can be done using the power rule where if

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

Therefore the velocity equation is 

\(\displaystyle v(t) = 5(5)t^{(5-1)} - 3(2)t^{(2-1)} = 25t^4 - 6t^\).

Repeating this process gives the acceleration equation

\(\displaystyle a(t) = 25(4)t^{(4-1)} - 6(1)t^{(1-1)} = 100t^3 - 6\).

Now we can find the acceleration at \(\displaystyle t= 1\).

\(\displaystyle a(1) = 100(1)^3 - 6 = 100 -6 = 94\)

To find acceleration we integrate the jerk of the object. This can be done using the power rule where if

\(\displaystyle f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}\).

Therefore the acceleration is

\(\displaystyle a(t) = \int (60t^3)dt = \frac{60t^{(3+1)}}{3+1} + C = 15t^4 +C\).

The initial acceleration of the equation can now be used to find the value of \(\displaystyle C\).

\(\displaystyle a(0) = 15(0)^4 + C = 0 + C = 9\)

Therefore \(\displaystyle C = 9\) and \(\displaystyle a(t) = 15t^4 + 9\).

The acceleration at time \(\displaystyle t =2\) can now be found.

\(\displaystyle a(2) = 15(2)^4 + 9 = 249\)

The acceleration can be found by differentiating the velocity.

The velocity can be differentiated using the chain rule where

\(\displaystyle f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)\).

The acceleration is then

\(\displaystyle a(t) = v'(t) = e^{7t}(7) = 7e^{7t}\).

At \(\displaystyle t = 0\), the acceleration is

\(\displaystyle a(0) = 7e^{7(0)} = 7e^0 = 7\).

The acceleration can be found by differentiating the position equation twice using the power rule.

The power rule is where

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

Therefore the velocity of the object is

\(\displaystyle v(t) = x'(t) = 10(1)t^{(1-1)} - 17(0)t^{0-1} = 10\).

The acceleration of the object can be found using the same method of differentiation as before.

\(\displaystyle a(t) = v'(t) = 10(0)t^{(0-1)} = 0\)

Therefore the acceleration of the object is \(\displaystyle 0\).

The acceleration of the object is the derivative of the velocity equation. To differentiate this equation we can use the chain rule and the power rule where if

\(\displaystyle f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x) \times g'(x)\)

and if 

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

From these rules we find that the acceleration of the object is

\(\displaystyle a(t) = \cos(e^{2t})(e^{2t})(2) = 2e^{2t}\cos(e^{2t})\).

We can now find the acceleration at time  \(\displaystyle t = -1\).

\(\displaystyle a(-1) = 2e^{2(-1)}\cos(e^{2(-1)}) = \frac{2}{e^2}\cos(e^{-2})\)

The acceleration of the object can be found by differentiating the position twice. This can be done using the power rule where if

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

Using this we can differentiate the velocity equation.

\(\displaystyle v(t) = 4(2)t^{(2-1)} - (3)t^{(3-1)} = 8t - 3t^2\)

Repeating this process we can find the acceleration equation.

\(\displaystyle a(t) = 8(1)t^{(1-1)} - 3(2)t^{(2-1)} = 8 - 6t\)

We can now find the acceleration at \(\displaystyle t = 2\).

\(\displaystyle a(2) = 8 - 6(2) = 8 - 12 = -4\)

To find the acceleration of the object we must differentiate the position of the object twice. We can differentiate the position using the power rule and the product rule where if

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

and if 

\(\displaystyle f(x)= j(x)k(x) \rightarrow f'(x) = j'(x)k(x) + j(x)k'(x)\).

Using these rules we obtian

\(\displaystyle v(t) = (6(3)t^{(3-1)})e^t + 6t^3(e^t) = 18t^2e^t+6t^3e^t\).

Using the same rules we can differentiate again to give the acceleration equation.

\(\displaystyle a(t) = 18(2)t^{(2-1)}e^t +18t^2(e^t) + 6(3)t^{(3-1)}e^t + 6t^3(e^t) = 36te^t+36t^2e^t+6t^3e^t\)