As far as I know, calculus is about rates of change, or slopes of lines at a particular point.

That is only one half of caclulus, differentiation. The other half is 'integration' and the basic statement of calculus is that integration and differentiation are opposite processes.

What you need to look at are the 'equations of motion'. There are 4 or 5 basic equations involving initial velocity (u), final velocity (v), acceleration (a), displacement (s), and time (t). Of the top of my head I can only remember two:

v^2 = u^2 + 2as

s = ut + (1/2)at^2

There are a couple more too. They are all derived using calculus, something which you should probably explore on your own...it will get you much more comfortable with understanding basic calculus and its applications.

For your application, look at what variables you are provided with, and which variable you need to obtain:

The acceleration of a moving object has a constant magnitude of a = 3.00 m/s^2. The object starts out from a dead stop at t = 0.00 s and moves in a straight-line path. How far will it have traveled from its starting point at t = 5.00 s?

u = 0 m/s

a = 3 m/s^2

t = 5s (time interval)

s = ?

s = ut + (1/2)at^2

= 0*5 + (1/2)*3*25

= (1/2)*3*25

= (1/2)*75

= 37.5m

That is using the equation of motion stated above. For the calculus method, look at velocity-time graphs ( in this case the graph will be a straight line, because the gradient, v/t = a, is constant). Considering that integration will give you the area under the graph, you will arrive at a statement for v*t giving you the expression above for displacement. You will also find that it's not as simple as s=v*t, because it is specific to your graph - which will be a triangle, and thats where the (1/2) comes into it. As I said though, its better to derive the formulas yourself to get a better understanding.