Howdy all, I had a quick question: An example problem in my physics book (Physics for Scientists and Engineers, 6th edition, Serway and Jewett) is going over tangential and radial acceleration (it is example 4.9, if you've got a copy lying around). The question is as follows: A car exhibits a constant acceleration of .3 m/s^2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius 500m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.0 m/s. What is the direction of the total acceleration vector for the car at this instant? If someone requests the picture, I can use my scanner to help the discussion along, just lemme know. My question comes from their use of the radial acceleration - they plugged the tangential speed (6 m/s) into v^2/r, it is in the negative radial direction, so acceleration(radial)=-.0720m/s^2. That is the tangential acceleration coming from moving in a circle at a constant speed. Yeah. But if this car was on Earth, wouldn't we add that to the acceleration due to gravity, which is also in the negative radial direction? That is how I did the problem, then looked at their answer and was confused as to why the hell they wouldn't include acceleration due to gravity here. Thanks for the help all!