In order to find the position function from the velocity function we need to take the integral of the velocity function.

$\displaystyle s(t)=\int v(t)dt$.

When taking the integral, we will use the inverse power rule which states,

$\displaystyle \int x^n=\frac{x^{n+1}}{n+1}$.

Applying this rule to each term we get, 

$\displaystyle \\s(t)= \int v(t)=\int (2t+3)dt \\ \\s(t)=\frac{2t^2}{2}+3t+c\\ \\s(t)= t^2+3t+c$

To find the value of the constant c we will use the initial condition given in the problem.

Setting the initial condition $\displaystyle t=0$,

$\displaystyle s(0)= 0^2+3(0)+c=4$ yields $\displaystyle c=4$.

Therefore the position function becomes,

$\displaystyle s(t)= t^2+3t+4$.

The vertical acceleration is given by $\displaystyle a(t)=-9.8$.

Initial vertical velocity is $\displaystyle 0$.

To integrate we do,

$\displaystyle \int x^n=\frac{x^{n+1}}{n+1}$.

Integrating once gives our velocity function, $\displaystyle v(t)=-9.8t$ given our initial vertical velocity. 

Integrating again gives a vertical position of $\displaystyle y(t)= -4.9t^2+C$.

Since we know the boulder starts at a height of $\displaystyle 700 m$, we determine that $\displaystyle C=700$.

Solving $\displaystyle 0=-4.9t^2+700$ gives $\displaystyle t=11.95$.

Then, determine a horizontal position equation for the boulder, with a starting position of $\displaystyle 0m$ and a constant velocity of $\displaystyle 140m/s$.

$\displaystyle x(t)=140t$

Evaluate $\displaystyle x(t)$ when $\displaystyle t=11.95$.

$\displaystyle x(t)=140(11.95)=1673.3$

The car has a constant of acceleration of $\displaystyle 1.5m/s^2$.

Represent acceleration with $\displaystyle a(t)=1.5$.

To integrate we do,

$\displaystyle \int x^n=\frac{x^{n+1}}{n+1}$.

Integrate the acceleration function to get the velocity function.

$\displaystyle v(t)=1.5t+C$

Since initial velocity is $\displaystyle 12m/s$, you can solve for $\displaystyle C$.

$\displaystyle v(0)=12=1.5(0)+C$

$\displaystyle C=12$

Thus, the function for velocity is  $\displaystyle v(t)=1.5t+12$.

Integrating the velocity equation gives the position function.

$\displaystyle x(t)= 0.75t^2+12t+D$

Since the initial position is $\displaystyle 80m$, you can solve for $\displaystyle D$ in the same way.

$\displaystyle x(0)=80= 0.75(0)^2+12(0)+D$

$\displaystyle D=80$

Finally, plugging $\displaystyle D$ back in gives the answer:

$\displaystyle x(t)=0.75t^2+12t+80$

By definition, a given vector $\displaystyle (a,b)$ has a perpendicular vector $\displaystyle (b,-a)$. Given a vector $\displaystyle (10,2)$, its perpendicular vector will be $\displaystyle (2,-10)$. We can further verify this result by noting that the product of two perpendicular vectors is $\displaystyle 0$; since $\displaystyle (10,2)\times (2,-10)=(10\times 2)+(2\times -10)=20+(-20)=0$, we know the two vectors are perpendicular to each other. 

By definition, a given vector $\displaystyle (a,b)$ has a perpendicular vector $\displaystyle (b,-a)$. Given a vector $\displaystyle (4,4)$, its perpendicular vector will be $\displaystyle (4,-4)$. We can further verify this result by noting that the product of two perpendicular vectors is $\displaystyle 0$; since $\displaystyle (4,4)\times (4,-4)=(4\times 4)+(4\times -4)=16+(-16)=0$, we know the two vectors are perpendicular to each other. 

To find the position of particle given its velocity function,we must integrate it.

$\displaystyle s(t) = \int v(t) dt = \int (\frac{3}{8}t^2 + 6t - 9) dt = \frac{1}{8}t^3 + 3t^2 - 9t + C$

To solve for the constant of integration, we use the given initial position.

$\displaystyle s(0) = \frac{1}{8}(0)^3 + 3(0)^2 - 9(0) + C$

$\displaystyle s(0)= 0 + 0 + C = 8$

Therefore, the final equation is

$\displaystyle s(t) = \frac{1}{8}t^3 + 3t^2 -9t + 8$

This problem can be done using the fundamental theorem of calculus. We hvae

$\displaystyle \small \small \int_2^4 v(t)dt=\int_2^4 s'(t)dt=s(4)-s(2)$

where $\displaystyle \small s(t)$ is the position at time $\displaystyle \small t$. So we have

$\displaystyle \small \small \small s(4)=s(2)+\int_2^4 v(t)dt$

so we need to solve for $\displaystyle \small \small \small \small \int_2^4 v(t)dt$:

$\displaystyle \small \small \small \small \small \small \int_2^4 v(t)dt=\int_2^4 (t^3-3t^2)dt=\frac{t^4}{4}-t^3|_2^4$

$\displaystyle \small =\left(\frac{4^4}{4}-4^3\right)-\left(\frac{2^4}{4}-2^3\right)=(4^3-4^3)-\left(\frac{16}{4}-8\right)=-(4-8)=4$

So then the position at $\displaystyle \small t=4$ is

 $\displaystyle \small \small s(4)=s(2)+\int_2^4 v(t)dt=4+4=8$

We can use the fundamental theorem of calculus for this problem. We have

$\displaystyle \small \small \small \int_0^{\pi/2} v(t)dt =s(\pi/2)-s(0)$

so we have

$\displaystyle \small \small \small s(\pi/2)=s(0)+ \int_0^{\pi/2} v(t)dt$

This means we just need to solve $\displaystyle \small \int_0^{\pi/2} v(t)dt$:

$\displaystyle \small \small \int_0^{\pi/2} v(t)dt=\int_0^{\pi/2} \sin 2tdt=-\frac{1}{2}\cos 2t|_0^{\pi/2}$

$\displaystyle \small \small =-\frac{1}{2}\left[ \cos \pi-cos 0 \right ]=-\frac{1}{2}[-1-1]=1$

So then the position of the particle at $\displaystyle \small t=\pi/2$ is 

$\displaystyle \small \small \small \small s(\pi/2)=s(0)+ \int_0^{\pi/2} v(t)dt=3+1=4$

Begin by finding the velocity function, the integral of acceleration with respect to time:

$\displaystyle v(t)=\int a(t)dt =\int 5dt$

$\displaystyle v(t)=5t + C$

To find the constant of integration, the fact that the car has a velocity of $\displaystyle 10 \frac{m}{s}$ when it begins to accelerate:

$\displaystyle v(0)=10=5(0)+C$

$\displaystyle C=10$

$\displaystyle v(t)=5t+10$

To find the distance travel, integrate the velocity function with respect to the lower and upper bounds, the start and end times respectively:

$\displaystyle d=\int_{t_i}^{t_f}v(t)dt=(2.5(t_f^2)+10(t_f))-(2.5(t_i^2)+10(t_i))$

$\displaystyle d=(2.5(4^2)+10(4))-(2.5(1^2)+10(1))=(2.5(16)+40)-(2.5+10)$

$\displaystyle d=(80)-(12.5)=67.5$

V(t) models the velocity of a spaceship. Find a function to model the spaceship's position as a function of time  if P(t) contains the point (2,15).

$\displaystyle V(t)=8t^3-3t^2+5$

Remember how position/velocity/acceleration are all related: Velocity is the first derivative of position, and acceleration is the derivative of velocity. 

So to get to position, we need to integrate velocity.

$\displaystyle P(t)=\int V(t)dt=\int 8t^3-3t^2+5dt$

To integrate polynomials, we increase the exponent by 1, and then divide  by the new exponent.

$\displaystyle \int 8t^3-3t^2+5dt=2t^4-t^3+5t+c$

Thus, our function so far looks like:

$\displaystyle 2t^4-t^3+5t+c$

However, we know that it must pass through the point (2,15), so we have to plug the point in and solve for C

$\displaystyle 15=2(2)^4-(2)^3+5(2)+c$

$\displaystyle 15=32-8+10+c$

$\displaystyle c=-19$

So our answer is:

$\displaystyle 2t^4-t^3+5t-19$