The position of the object can be found by integrating the velocity equation and solving for $\displaystyle x(4)$. To integrate the velocity equation we first rewrite the equation. 

$\displaystyle v(t) = (4t^2)(3t) = 12t^3$

To integrate this equation we must use the power rule where,

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

Applying this to the velocity equation gives us,

$\displaystyle x(t) = \int (12t^3)dt = \frac{12t^{3+1}}{3+1} = \frac{12t^4}{4} = 3t^4 +C$.

We must solve for the value of $\displaystyle C$ by using the initial position of the object.

$\displaystyle x(0) = 3(0)^4 + C =0 + C= 0$

Therefore, $\displaystyle C = 0$ and $\displaystyle x(t) = 3t^4$.

$\displaystyle x(4) = 3(4)^4 = 3(256) = 768$

The position of the object can be found by integrating the equation for the object's velocity. Knowing that the derivative of $\displaystyle -\cos(t)$ is $\displaystyle \sin(t)$, the integral of $\displaystyle \sin(t)$ must be $\displaystyle -\cos(t)$. 

Integrating the velocity equation gives us,

$\displaystyle x(t) = \int (2\sin(t))dt = 2(-\cos(t)) +C = -2\cos(t) +C$

To find the complete solution of the position equation we must use the initial position.

$\displaystyle x(0) = -2\cos(0) +C = -2 +C = 0$

Therefore $\displaystyle C = 2$ and $\displaystyle x(t) = -2\cos(t) +2$

By definition, a vector $\displaystyle (a,b)$ has a perpendicular vector $\displaystyle (-b,a)$.

Therefore, the vector $\displaystyle (5,7)$ has a perpendicular vector $\displaystyle (-7,5)$. 

By definition, a vector $\displaystyle (a,b)$ has a perpendicular vector $\displaystyle (-b,a)$.

Therefore, the vector $\displaystyle (2,6)$ has a perpendicular vector $\displaystyle (-6,2)$. 

By definition, a vector $\displaystyle (a,b)$ has a perpendicular vector $\displaystyle (-b,a)$.

Therefore, the vector $\displaystyle (3,12)$ has a perpendicular vector $\displaystyle (-12,3)$. 

To find the position of the object we must use the power rule to integrate the acceleration equation twice. The power rule is such that

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

Therefore integrating the acceleration equation gives us

$\displaystyle v(t) = \int (4t+1)dt = \frac{4t^{1+1}}{1+1} +\frac{1t^{0+1}}{0+1} +C = 2t^2 + t +C$

We can solve for the value of $\displaystyle C$ by using the initial velocity of the object.

$\displaystyle v(0) = 2(0)^2 +(0)+C = 10$

Therefore $\displaystyle C = 10$ and $\displaystyle v(t) = 2t^2 + t + 10$

To find the position of the object we integrate the velocity equation.

$\displaystyle x(t) = \int (2t^2+t+10)dt = \frac{2t^{2+1}}{2+1} + \frac{t^{1+1}}{1+1} + \frac{10t^{0+1}}{0+1} +C = \frac{2t^3}{3}+\frac{t^2}{2}+10t +C$

 We can solve for this new value of $\displaystyle C$ by using the object's initial position

$\displaystyle x(0) = \frac{2(0)^3}{3} + \frac{(0)^2}{2} + 10(0) +C = 0 + C = 0$

Therefore $\displaystyle C = 0$ and $\displaystyle x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

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 $\displaystyle x(t) = t - 3t^3 + 32$.

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