A train slows down at a constant rate as it rounds a sharp circular horizontal turn. Its initial speed is not known. It takes 17.7 s to slow down from 68 km/h to 26 km/h. The radius of the curve is .184 km. As the train goes around the turn, (a) what is the magnitude to the tangential component of the acceleration and (b) at the moment the train's speed is 59 km/h, what is the magnitude of the total acceleration? Answer in units of m/s^2.

(a) Tangential acceleration is given by dlvl/dt. However, I am finding the average acceleration by subtracting 68 from 26, and then dividing it by 17.7 seconds. The negative sign would disappear. However, tangential acceleration is the derivative of velocity at that moment, not the average acceleration over a period of time. D:

(b) Total acceleration is the quadrature of radial acceleration and tangential acceleration. I suppose I could square my answer in (a), and then compute a[r] by squaring 59 and dividing it by .184 (and then tacking on the negative sign). Once I have a[r] and a[t], I will square them both, add them, and then take the square root. But, could I use the different part derived from different velocities (tangential acc. deals with a range of velocities, while radial acc. has the one given in the problem)?

I know, also, that I must convert everything to meters, but I will do that last.

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Recognitions: Homework Help I think they want you to assume tangential acceleration is constant over the turn... everything looks good.
 Wow. Thanks. (Yay for getting better at this. Even if it's just plugging in numbers.)