Yes; the quantity you're interested in is called "tangential acceleration."
And now for a more thorough answer than you wanted:
Velocity is a vector, so it has both a magnitude and a direction. (The magnitude of the velocity is called "speed," as you are already aware.) Acceleration is a change in velocity. Thus if either the speed or the direction of a velocity changes, an acceleration is taking place.
There's a theorem in differential geometry which says that a particle's acceleration vector can always be expressed as the sum of two vectors, called the tangential acceleration and the centripetal acceleration. Furthermore, the tangential acceleration points in the same direction as the velocity (where "same direction" includes the possibility of pointing exactly backwards, for a decelerating particle) and has a magnitude equal to the derivative of speed with respect to time, and the normal acceleration always points perpendicular to the particle's path (unless it's zero) and has a magnitude equal to the particle's instantaneous speed squared over the radius of curvature of the particle's path.
This theorem is a big deal because the magnitude of the centripetal acceleration, when written in symbols, is v^2/r. You should remember that as the formula for centripetal acceleration when a particle moves in a circle with a constant speed. What this theorem says is that if a particle *isn't* moving in a circle with constant speed, then at any instant in time you can still approximate it as a particle moving in some circle but with a changing speed, and then its acceleration is the same old centripetal acceleration PLUS a tangential acceleration corresponding to the change in speed.
This fact is a lot more exciting when it's written as an equation, but you need three or four different equations to define all of the symbols that show up.