To find the position function, you must integrate the velocity function. To integrate, add 1 to the exponent and then put that result on the denominator. $\displaystyle \int 3t-1 dt=\frac{3t^2}{2}-t+C$. Then, plug in your initial values to find your C: $\displaystyle \frac{3}{2}(1^2)-1+C=2$. Then, $\displaystyle C=\frac{3}{2}$. Therefore, your answer is: $\displaystyle x(t)=\frac{3}{2}t^2-t+\frac{3}{2}$.

We'll need to find the acceleration, and then work backwards to find the position. We can rewrite Newton's Second Law as $\displaystyle a=\frac{f}{m}$, which in this case gives us an acceleration of 

$\displaystyle a(t)=\frac{30000 N}{1500 kg} = 20\,m/s^2$.

The car's initial position and velocity are

$\displaystyle \begin{align*} p_0&=0\,m \\ v_0 &= 25\,m/s\end{align*}$

Initial and final time are given as 

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

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. 

$\displaystyle \int a(t)dt=\int20 dt$

To integrate, we use the following rule:

$\displaystyle \int a\,dt = ax+C$

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:

$\displaystyle \int a(t)dt=\int 20dt =20t+C$

To solve for C, we'll use our initial condition, $\displaystyle v(0)=25$.

$\displaystyle \\ v(0)=20(0)+C=25 \\ C=25 \\ v(t)=20t+25$

We'll need to integrate again to find position:

$\displaystyle \int v(t)=\int 20t+25 \, dt$

using the following rule in addition to the earlier one:

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

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:

$\displaystyle \begin{align*} \int v(t)\,dt &= \int 20t+25\,dt \\ &=\frac{20t^2}{2}+25t+C \\ &=10t^2+25t+C \end{align*}$

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

$\displaystyle \\ p(0)=10(0^2)+25(0)+C=0 \\ C=0 \\ p(t)=10t^2+25t$

which, evaluated at $\displaystyle t=5$, gives us

$\displaystyle \begin{align*} p(4)&=10(5^2)+25(5) \\ &=10(25)+125 \\ &=375\,m\end{align*}$

The equation for position is $\displaystyle \frac{at^2}{2}$, since we're starting from rest. Evaluating at $\displaystyle t=20$ gives us our answer, $\displaystyle 3,000\,m$

We'll need to work backwards from the acceleration to find the position. We're given the car's acceleration,

$\displaystyle \begin{align*} a(t) &=15\,m/s^2\end{align*}$

The car 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 

$\displaystyle \begin{align*} t_i &=0\,s \\ t_f &= 20\,s \end{align*}$

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. 

$\displaystyle \int a(t)dt=\int15 dt$

To integrate, we use the following rule:

$\displaystyle \int a\,dt = ax+C$

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:

$\displaystyle \int a(t)dt=\int 15dt =15t+C$

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

$\displaystyle \\ v(0)=15(0)+C=0 \\ C=0 \\ v(t)=15t$

We'll need to integrate again to find position:

$\displaystyle \int v(t)=\int 15t \, dt$

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:

$\displaystyle \begin{align*} \int v(t)\,dt &= \int 15t\,dt \\ &=\frac{15t^2}{2}+C \\ &=7.5t^2+C \end{align*}$

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

$\displaystyle \\ p(0)=7.5(0^2)+C=0 \\ C=0 \\ p(t)=7.5t^2$

which, evaluated at $\displaystyle t=20$, gives us

$\displaystyle \begin{align*} p(4)&=7.5(20^2) \\ &=7.5(400) \\&=3000\,m\end{align*}$

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

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

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t)= 2(3t^2 + e^{2t}) (6t+2e^{2t})$

Then, plug $\displaystyle t= 1$ into the velocity function: $\displaystyle v(1)= 2(3({\color{Blue} 1})^2 + e^{2({\color{Blue} 1})}) (6({\color{Blue}1 })+2e^{2({\color{Blue} 1})})$

Therefore, the answer is: $\displaystyle 431.7$

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

In this case, the position function is: $\displaystyle p(t)= \sin(t^2+3t^4)$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t)= \cos(t^2+3t^4) (2t+12t^3)$

Then, plug $\displaystyle t= 0$ into the velocity function: $\displaystyle v(t)= \cos({\color{Blue} 0}^2+3({\color{Blue} 0})^4) (2({\color{Blue} 0})+12({\color{Blue} 0})^3)$

Therefore, the answer is: $\displaystyle 0$

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

In this case, the position function is: $\displaystyle p(t)= \sin(t^2+3t^4)$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t)= \cos(t^2+3t^4) (2t+12t^3)$

Then, plug $\displaystyle t= 0$ into the velocity function: $\displaystyle v(t)= \cos({\color{Blue} 0}^2+3({\color{Blue} 0})^4) (2({\color{Blue} 0})+12({\color{Blue} 0})^3)$

Therefore, the answer is: $\displaystyle 0$

We'll need to work backwards from the acceleration to find the position. We're given the rocket's acceleration, which is opposed by gravity, so the total acceleration will be

$\displaystyle \begin{align*} a(t) &=15\,m/s^2-9.8\,m/s^2 \\ &=5.2\,m/s^2\end{align*}$

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 

$\displaystyle \begin{align*} t_i &=0\,s \\ t_f &= 4\,s \end{align*}$

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. 

$\displaystyle \int a(t)dt=\int5.2 dt$

To integrate, we use the following rule:

$\displaystyle \int a\,dt = ax+C$

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:

$\displaystyle \int a(t)dt=\int 5.2dt =5.2t+C$

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

$\displaystyle \\ v(0)=5.2(0)+C=0 \\ C=0 \\ v(t)=5.2t$

We'll need to integrate again to find position:

$\displaystyle \int v(t)=\int 5.2t \, dt$

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:

$\displaystyle \begin{align*} \int v(t)\,dt &= \int 5.2t\,dt \\ &=\frac{5.2t^2}{2}+C \\ &=2.6t^2+C \end{align*}$

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

$\displaystyle \\ p(0)=2.6(0^2)+C=0 \\ C=0 \\ p(t)=2.6t^2$

which, evaluated at $\displaystyle t=4$, gives us

$\displaystyle \begin{align*} p(4)&=2.6(4^2) \\ &=2.6(16) \\&=41.6\,m\end{align*}$

The first step in solving this problem is integrating the velocity function to get your position function. When integrating, remember to raise the exponent by 1 and then also put that result on the denominator: $\displaystyle \int 2t-2dt=\frac{2t^2}{2}-2t=t^2-2t+C$. Now, to figure out what C is, plug in your initial conditions given: $\displaystyle (1^2)-2(1)+C=3; C=4$. Plug in 4 for C to get your answer: $\displaystyle x(t)=t^2-2t+4$.

First, you need to find the derivative of the function. When taling the derivative, multiply the exponent by the coefficient in front of the x term and then decrease the exponent by 1. Therefore: $\displaystyle f{}'(x)=9x^2-2x+4$. Now, plug in 1 for x to get your answer: 11.

To find the position function from the velocity function, you must integrate the velocity. When integrating, raise the exponent by 1 and then put that result on the denominator as well: $\displaystyle \frac{4x^2}{2}-x+C=2x^2-x+C$. Then, plug in your initial conditions to find C: $\displaystyle 2(1^2)-1+C=2; C=1$. Therefore, the position function is: $\displaystyle x(t)=2x^2-x+1$