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

Since we are given a vector \(\displaystyle (5,3)\), we therefore know that its perpendicular vector is \(\displaystyle (3,-5)\). 

Perpendicular vectors have a dot product of zero therefore,

\(\displaystyle (x_1,y_1), (x_2,y_2)\rightarrow (x_1\cdot x_2) +(y_1 \cdot y_2)=0\)

\(\displaystyle (5\cdot 3)+(3\cdot -5)=15+-15=0\)

By definition, a given vector \(\displaystyle (a,b)\) has a perpendicular vector \(\displaystyle (b,-a)\).  Since we are given a vector \(\displaystyle (-7,9)\), we therefore know that its perpendicular vector is \(\displaystyle (9,7)\). 

Perpendicular vectors have a dot product of zero therefore,

\(\displaystyle (x_1,y_1), (x_2,y_2)\rightarrow (x_1\cdot x_2) +(y_1 \cdot y_2)=0\)

Plugging in our values, we can verify the two vectors are perpendicular.

\(\displaystyle (-7\cdot 9)+(9\cdot 7)=-63+63=0\)

By definition, a given vector \(\displaystyle (a,b)\) has a perpendicular vector \(\displaystyle (b,-a)\).  Since we are given a vector \(\displaystyle (6,2)\), we therefore know that its perpendicular vector is \(\displaystyle (2,-6)\). 

Perpendicular vectors have a dot product of zero therefore,

\(\displaystyle (x_1,y_1), (x_2,y_2)\rightarrow (x_1\cdot x_2) +(y_1 \cdot y_2)=0\)

Plugging in our values, we can verify the two vectors are perpendicular.

\(\displaystyle (6\cdot 2)+(2\cdot -6)=12+-12=0\)

To begin, let's recognize that since acceleration is the second derivative of position with respect to time, we can in turn, integrate an acceleration function twice with respect to time to find position.

Since accleration is a constant, 14 feet per second, our acceleration function is:

\(\displaystyle a(t) = 14 feet/s^2\)

Integrating this once gives us a velocity function

\(\displaystyle v(t) = (14 feet/s^2) \cdot t + v_0\)

where

\(\displaystyle v_0\)

is our initial velocity.  Since the car started at rest, v0 will be equal to zero, leaving us with:

\(\displaystyle v(t) = (14 feet/s^2) \cdot t\)

To get our position function, we can in turn integrate our velocity function:

\(\displaystyle x(t) = (1/2)(14 feet/s^2) \cdot t^2 + x_0\)

where x₀ is our initial position. We are told that our initial position is 10 feet away from the wall, so we can rewrite this equation as:

\(\displaystyle x(t) = (1/2)(14 feet/s^2) \cdot t^2+ 10 feet\)

To find where the car is at 6 seconds, we need only plug in the value of 6 seconds where ever our time variable t occurs, giving our answer of 262 feet.

By definition, a given vector \(\displaystyle (a,b)\) has a perpendicular vector \(\displaystyle (b,-a)\). Therefore, the vector perpendicular to \(\displaystyle (5,4)\) is \(\displaystyle (4,-5)\).

To verify that two vectors are perpendicular their dot product must equal zero.

The dot product is as follows.

\(\displaystyle (a_1,b_1),(a_2,b_2)\rightarrow (a_1\times a_2)+(b_1\times b_2)=0\)

Substituting in our vector values we get:

\(\displaystyle (5,4),(4,-5)\rightarrow (5\times 4)+(4\times -5)=20+-20=0\)

By definition, a given vector \(\displaystyle (a,b)\) has a perpendicular vector \(\displaystyle (b,-a)\). Therefore, the vector perpendicular to \(\displaystyle (7,2)\) is \(\displaystyle (2,-7)\).

To verify that two vectors are perpendicular their dot product must equal zero.

The dot product is as follows.

\(\displaystyle (a_1,b_1),(a_2,b_2)\rightarrow (a_1\times a_2)+(b_1\times b_2)=0\)

Substituting in our vector values we get:

\(\displaystyle (7,2), (2,-7)\rightarrow (7\times 2)+(2\times -7)=14+-14=0\)

By definition, a given vector \(\displaystyle (a,b)\) has a perpendicular vector \(\displaystyle (b,-a)\). Therefore, the vector perpendicular to \(\displaystyle (-3,-8)\) is \(\displaystyle (-8,3)\).

To verify that two vectors are perpendicular their dot product must equal zero.

The dot product is as follows.

\(\displaystyle (a_1,b_1),(a_2,b_2)\rightarrow (a_1\times a_2)+(b_1\times b_2)=0\)

Substituting in our vector values we get:

\(\displaystyle (-3,-8),(-8,3)\rightarrow (-3\times -8)+(-8\times 3)=24+-24=0\)

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

Given a vector \(\displaystyle (7,-5)\), its perpendicular vector will be \(\displaystyle (-5,-7)\).

We can further verify this result by noting that the product of two perpendicular vectors is \(\displaystyle 0\). 

Since \(\displaystyle (7,-5)\times(-5,-7) = (7\times -5) + (-5\times -7)=-35+35=0\), then \(\displaystyle (-5,-7)\)  is perpendicular to \(\displaystyle (7,-5)\).

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

Given a vector \(\displaystyle (8,4)\), its perpendicular vector will be \(\displaystyle (4,-8)\).

We can further verify this result by noting that the product of two perpendicular vectors is \(\displaystyle 0\).

Since \(\displaystyle (8,4)\times (4,-8) = (8\times 4) + (4\times -8)=32-32=0\), then \(\displaystyle (4,-8)\) is perpendicular to \(\displaystyle (8,4)\).

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

Given a vector \(\displaystyle (2,2)\), its perpendicular vector will be \(\displaystyle (2,-2)\).

We can further verify this result by noting that the product of two perpendicular vectors is \(\displaystyle 0\). 

Since \(\displaystyle (2,2)\times (2,-2) = (2\times 2) + (2\times -2)=4-4=0\), then \(\displaystyle (2,-2)\) is perpendicular to \(\displaystyle (2,2)\).