To find the position of the object we must first find the position equation of the object. The position equation can be found by integrating the velocity equation. This can be done using the power rule where if

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

Using this rule we find that 

\(\displaystyle x(t) = \int (6t-3)dt = \frac{6t^{1+1}}{1+1} - \frac{3t^{0+1}}{0+1} +C = 3t^2 - 3t +C\)

Using the position of the object at time \(\displaystyle t = 1\) we can solve for \(\displaystyle C\)

\(\displaystyle x(1) = 3(1)^2 - 3(1) + C = 3 - 3 + C =0+C= 6\)

Therefore \(\displaystyle C = 6\) and \(\displaystyle x(t) = 3t^2 - 3t +6\)

We can now find the position at time \(\displaystyle t = 4\).

\(\displaystyle x(4) = 3(4)^2 - 3(4) +6 = 48-12 +6 = 42\)

The position of the object can be found by integrating the velocity of the object. This can be done using the power rule where if

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

Using the power rule the position of the object is

\(\displaystyle x(t) = \int (5t^4 + 3t^2 + 5)dt = \frac{5t^{(4+1)}}{4+1} +\frac{3t^{(2+1)}}{2+1} + \frac{5t^{(0+1)}}{0+1} +C= t^5 +t^3 +5t +C\).

The value of \(\displaystyle C\) can be found using the initial position of the object.

\(\displaystyle x(0) = (0)^5 +(0)^3 + 5(0) +C = 0\)

Therefore \(\displaystyle C = 0\) and \(\displaystyle x(t) = t^5 + t^3 +5t\).

The position of the object at \(\displaystyle t = 3\) can now be found,

\(\displaystyle x(3) = (3)^5 + (3)^3 + 5(3) =243+ 27+ 15 = 285\).

The position of the object can be found by integrating the velocity 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 position equation of the object is

\(\displaystyle x(t) = \int (12t-39)dt = \frac{12t^{(1+1)}}{1+1} -\frac{39t^{(0+1)}}{0+1}+C = 6t^2 - 39t +C\)

We must now solve for the constant \(\displaystyle C\). We can do this using the initial position of the object.

\(\displaystyle x(0) = 6(0)^2 - 39(0) +C = 0 - 0 + C = 0\)

Therefore \(\displaystyle C = 0\) and \(\displaystyle x(t) = 6t^2 - 39t\)

We can now find the position of the object at \(\displaystyle t = 2\).

\(\displaystyle x(2) = 6(2)^2 - 39(2) = 24 - 78 = -54\)

The position of the object can be found by integrating the object's velocity. 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 position of the object is

\(\displaystyle x(t) = \int (1t^0-9t^2)dt = \frac{1t^{(0+1)}}{0+1} + \frac{9t^{(2+1)}}{2+1}+C = t + 3t^3 + C\).

We can find the value of \(\displaystyle C\) using the position at \(\displaystyle t =2\).

\(\displaystyle x(2) = (2) - 3(2)^3 + C = 2-24+C=-22+C= 10\)

Therefore \(\displaystyle C= 32\) and \(\displaystyle x(t) = t - 3t^3 + 32\).

\(\displaystyle x(1) = (1) - 3(1)^3 +32 = -2 + 32 = 30\)

The position is the integral of the velocity. By integrating with the power rule we can find the object's position.

The power rule is where 

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

Therefore the position of the object is

\(\displaystyle x(t) = \int (3t^2)dt = \frac{3t^{(2+1)}}{2+1} +C = t^3 +C\).

We can solve for the constant \(\displaystyle C\) using the object's initial position.

\(\displaystyle x(0) = (0)^3 + C = 0+C =1000\)

Therefore \(\displaystyle C= 1000\) and \(\displaystyle x(t) = t^3 +1000\).

The position of the object can be found by integrating the velocity. This can be done using the power rule where if

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

Using this rule we find that 

\(\displaystyle x(t) = \int (18t^2-11)dt = \frac{18t^{(2+1)}}{2+1} - \frac{11t^{(0+1)}}{0+1} +C = 6t^3 - 11t +C\).

We can find the value of \(\displaystyle C\) using the initial position of the object.

\(\displaystyle x(0) = 6(0)^3 - 11(0) +C = 0 - 0 +C =10\)

Therefore \(\displaystyle C = 10\) and \(\displaystyle x(t) = 6t^3 - 11t +10\).

\(\displaystyle x(3) = 6(3)^3 - 11(3) + 10 = 162 - 11(3) + 10 = 139\)

The position of the object can be found by integrating the velocity. This can be done using the power rule where if

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

Using this rule to integrate the velocity gives us

\(\displaystyle x(t) = \int (44t^3 + t^2 + 2)dt = \frac{44t^{(3+1)}}{3+1} + \frac{t^{(2+1)}}{2+1} +\frac{2t^{(0+1)}}{0+1} +C\)

\(\displaystyle x(t) = 11t^4 + \frac{t^3}{3} + 2t +C\).

The value of \(\displaystyle C\) can be found by using the initial position of the object.

\(\displaystyle x(0)= 11(0)^4 +\frac{(0)^3}{3} + 2(0) +C = 0 + 0+0+C = 2\)

Therefore \(\displaystyle C = 2\) and \(\displaystyle x(t) = 11t^4 + \frac{t^3}{3} + 2t + 2\).

Using the position equation we find the position at \(\displaystyle t = 1\).

\(\displaystyle x(1) = 11(1)^4 + \frac{(1)^3}{3} + 2(1) +2 = 11 + \frac{1}{3} + 2 +2 = 15 \frac{1}{3}\)

To determine the position of an object given the velocity function, integrate once to obtain the position function.

\(\displaystyle x(t)=\int v(t)\: dt=\int cos(t)\:dt= sin(t)\)

Substitute the value \(\displaystyle t=2\pi\) into the position function.

\(\displaystyle sin(2\pi)=0\)

Obtain the position function by integrating the acceleration twice.

\(\displaystyle a(t)=sin(2t)+cos(t)\)

\(\displaystyle v(t)= \int a(t)\:dt = -\frac{1}{2}cos(2t)+sin(t)\)

Integrate again to obtain the position function.

\(\displaystyle x(t)= \int v(t)\:dt= \int -\frac{1}{2}cos(2t)+sin(t)\:dt\)

\(\displaystyle x(t)=-\frac{1}{4}sin(2t)-cos(t)\)

Substitute \(\displaystyle t=\frac{\pi}{2}\).

\(\displaystyle x(t=\frac{\pi}{2})=-\frac{1}{4}sin\left(2\left(\frac{\pi}{2}\right)\right)-cos\left(\frac{\pi}{2}\right)=0-0=0\)

By definition, any vector \(\displaystyle (a,b)\) has a perpendicular vector \(\displaystyle (b,-a)\). Given a vector \(\displaystyle (6,5)\), the perpendicular vector is \(\displaystyle (5,-6)\). 

We can verify this further by noting that the product of any vector \(\displaystyle x\) and its perpendicular vector \(\displaystyle y\) is equal to \(\displaystyle 0\), or \(\displaystyle x\times y=0\). Taking the product of \(\displaystyle (6,5)\) and \(\displaystyle (5,-6)\), we get:

\(\displaystyle (6,5)\times (5,-6)=(6\times 5)+(5\times -6)=30+(-30)=0\)