In order to find the acceleration of a certain point, you must first find the derivative of the position function which gives us the velocity function and then the derivative of the velocity function which gives us the acceleration function: \(\displaystyle p''(t)= v'(t)= a(t)\)

In this case, the position function is: \(\displaystyle p(t)= -\cos t + \sin t - 45e^t + 5t^2\)

The velocity function is found by taking the derivative of the position function: \(\displaystyle v(t)= \sin t + \cos t - 45e^t + 10t\)

The acceleration function is found by taking the derivative of the velocity function: \(\displaystyle a(t)= \cos t - \sin t - 45e^t + 10\)

Finally, to find the accelaration, substitute \(\displaystyle t=0\) into the acceleration function: \(\displaystyle a(0)= \cos (0) - \sin (0) - 45e^{(0)} + 10\)

Therefore, the answer is: \(\displaystyle -34\)

In order to find the acceleration of a certain point, you must first find the derivative of the position function which gives us the velocity function and then the derivative of the velocity function which gives us the acceleration function: \(\displaystyle p''(t)= v'(t)= a(t)\)

In this case, the position function is: \(\displaystyle p(t)= 2t^{\frac{1}{2}} + 3e^t +4\)

The velocity function is found by taking the derivative of the position function: \(\displaystyle v(t)= t^{-\frac{1}{2}} + 3e^t\)

The acceleration function is found by taking the derivative of the velocity function: \(\displaystyle a(t)= {-\frac{1}{2}}t^{(-\frac{3}{2})} + 3e^t\)

Finally, to find the accelaration, substitute \(\displaystyle t=4\) into the acceleration function: \(\displaystyle a(4)= {-\frac{1}{2}}(4)^{(-\frac{3}{2})} + 3e^{(4)}\)

Therefore, the answer is: \(\displaystyle 163.7\)

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: \(\displaystyle p''(t)=v'(t) = a(t)\)

In this case, the position function is: \(\displaystyle p(t) = (t-4)^3\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t) = 3(t-4)^2\)

Then take the dervivative of the velocity function to get the acceleration function: \(\displaystyle a(t) = 6(t-4)\)

Then, plug \(\displaystyle t= 2\) into the acceleration function: \(\displaystyle a(2) = 6({\color{Blue} 2}-4)\)

Therefore, the answer is: \(\displaystyle -12\)

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: \(\displaystyle p''(t)=v'(t) = a(t)\)

In this case, the position function is: \(\displaystyle p(t) = (t-4)^3\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= \frac{1}{t}\)

Then take the dervivative of the velocity function to get the acceleration function: \(\displaystyle a(t)= -\frac{1}{t^2}\)

Then, plug \(\displaystyle t= 1\) into the acceleration function: \(\displaystyle a(1)= -\frac{1}{{\color{Blue} 1}^2}\)

Therefore, the answer is: \(\displaystyle -1\)

We'll need to first figure out how to express the rocket's position in terms of its acceleration, \(\displaystyle a\). This will give us an equation we can solve for \(\displaystyle a\).

The rocket begins at rest, so 

\(\displaystyle \begin{align*} p_0&=0\,m \\ v_0 &= 0\,m/s\end{align*}\)

Initial and final time are given as 

\(\displaystyle \begin{align*} t_i &=0\,s \\ t_f &= 5\,s \end{align*}\)

We're also given that

\(\displaystyle p(5)=150\,m\)

Acceleration is the rate of change in velocity, which is the rate of change in position, so to find the equation for position, we'll ned to integrate and use initial conditions twice. 

\(\displaystyle \int a(t)dt=\int a\,dt\)

To integrate, we use the following rule:

\(\displaystyle \int a\,dt = ax+C\)

To solve for C, we'll use our initial condition, \(\displaystyle v(0)=0\).

\(\displaystyle \\ v(0)=a(0)+C=0 \\ C=0 \\ v(t)=at\)

We'll need to integrate again to find position:

\(\displaystyle \int v(t)=\int at \, dt\)

using the following rule:

\(\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C\)

Remember, when integrating we have to add the constant, C, to indicate that there's multiple functions that could be our answer.

Integrating then gives us:

\(\displaystyle \begin{align*} \int v(t)\,dt &= \int at\,dt \\ &=\frac{at^2}{2}+C \\ &=\frac{a}{2}t^2+C \end{align*}\)

To solve for C, we'll use our initial condition, \(\displaystyle p(0)=0\).

\(\displaystyle \\ p(0)=\frac{a}{2}(0^2)+C=0 \\ C=0 \\ p(t)=\frac{a}{2}t^2\)

Now, we know that \(\displaystyle p(5)=150\), so we can solve for \(\displaystyle a\).

\(\displaystyle \begin{align*} \frac{a}{2}(5^2) &=150\\ 25a &= 300 \\ a &= 12\,m/s^2 \end{align*}\)

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: \(\displaystyle p''(t)=v'(t) = a(t)\)

In this case, the position function is: \(\displaystyle p(t)= e^{3t}\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= 3e^{3t}\)

Then take the dervivative of the velocity function to get the acceleration function: \(\displaystyle a(t)= 9e^{3t}\)

Then, plug \(\displaystyle t= 1\) into the acceleration function: \(\displaystyle a(1)= 9e^{3({\color{Blue} 1})}\)

Therefore, the answer is: \(\displaystyle 180.8\)

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: \(\displaystyle p''(t)=v'(t) = a(t)\)

In this case, the position function is: \(\displaystyle p(t)= \sin(3t^2)\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)=6t \cos(3t^2)\)

Then take the dervivative of the velocity function to get the acceleration function: \(\displaystyle a(t)=6 \cos(3t^2) - 36t^2\sin(3t^2)\)

Then, plug \(\displaystyle t= 0\) into the acceleration function: \(\displaystyle a(0)=6 \cos(3({\color{Blue} 0})^2) - 36({\color{Blue} 0})^2\sin(3({\color{Blue} 0})^2)\)

Therefore, the answer is: \(\displaystyle 6\)

Acceleration is the rate of change in velocity, which is the rate of change in position, so to answer this question, we'll need to differentiate this equation twice. First, we find the velocity, by applying the product rule

\(\displaystyle (fg)'=(f'g)+(fg')\)

This gives us:

 \(\displaystyle \begin{align*}v(t)&=(\sin t \cos t)'\\&= (\cos t \cdot \cos t)+(\sin t\cdot -\sin t)\\&= \cos^2 t-\sin^2 t\end{align*}\)

Differentiating that using the chain rule

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

 on each term gives us

\(\displaystyle \begin{align*} a(t)& =(\cos^2 t)'-(\sin^2 t)'\\ &=(-\sin t)(2\cos t)-(\cos t)(2\sin t)\\&=-4\sin t\cos t\end{align*}\)

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: \(\displaystyle p''(t)=v'(t) = a(t)\)

In this case, the position function is: \(\displaystyle p(t)= 2t^4 + cos(t^2) - ln(2t)\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= 8t^3 - 2t\sin(t^2) - \frac{1}{t}\)

Then take the dervivative of the velocity function to get the acceleration function: \(\displaystyle a(t)= 24t^2 - 2\sin(t^2) + 4t^2\cos(t^2) + \frac{1}{t^2}\)

Then, plug \(\displaystyle t= 1\) into the acceleration function: \(\displaystyle a(1)= 24({\color{Blue} 1})^2 - 2\sin({\color{Blue} 1}^2) + 4({\color{Blue} 1})^2\cos({\color{Blue} 1}^2) + \frac{1}{{\color{Blue} 1}^2}\)

Therefore, the answer is: \(\displaystyle 25.5\)

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: \(\displaystyle p''(t)=v'(t) = a(t)\)

In this case, the position function is: \(\displaystyle p(t)= e^{2t} + \frac{1}{ln(2t)}\)

Then take the derivative of the position function to get the velocity function: \(\displaystyle v(t)= 2e^{2t} - \frac{1}{t ln^2(2t)}\)

Then take the dervivative of the velocity function to get the acceleration function: \(\displaystyle a(t)= 4e^{2t} + \frac{1}{t^2 ln^2(2t)} + \frac{2}{t^2ln^3(2t)}\)

Then, plug \(\displaystyle t= 1\) into the acceleration function: \(\displaystyle a(1)= 4e^{2({\color{Blue} 1})} + \frac{1}{{\color{Blue} 1}^2 ln^2(2({\color{Blue} 1}))} + \frac{2}{{\color{Blue} 1}^2ln^3(2({\color{Blue} 1}))}\)

Therefore, the answer is: \(\displaystyle 37.6\)