Alright. First, read the question and collect all information :

- it is aequation, so it has the form (I believe it is useless to check if the line is parallel to the y-axis, that would be boring )linear

- we are talking about velocity. Therefore, we are talking of.feet per second

-the bullet leaves the gun atDoesn't this remind us of something ? Yes, I guess that would be the slope of the line : if the bullet goes up 100 feet (y-axis) for 1 second (x-axis), then the slope will be . Therefore, in the equation, .100 feetper[1] second.

-After 2 seconds the velocity of the bullet is 36 feet per second.Hmm ... maybe this could actually be apointof the linear equation. Therefore, the point belongs to the equation.

Good. We know that , and that is a point of the equation. But we are still missing to complete the equation. Let's try to substitute the values of the point that we haveintothe current equation to find (since the point belongs to the equation) :

We found ! Now our equation is complete :

Let's conclude : the velocity of the bullet (in feet) at time (in seconds) is given by the linear equation

Note that I used and to denote the velocity and time. You can change following the question, therefore the equation is :

The second question should be easy now