S.O.S. Mathematics CyberBoard

A train is travelling on a curved track. At a point where the train is travelling at a speed of 132ft/s and the radius of curvature of the track is 3000 ft, the engineer hits the brakes to make the train slow down at a constant rate of 7.5 ft/s^2.

a) Find the magnitude of the acceleration vector at the instant the engineer hits the brakes.

b) Approximate the angle between the acceleration vector and the unit tangent vector T at the instant the engineer hits the brakes.

Any help would be appreciated.

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mathematics is the future

radial acceleration...
a_R = (v_T)^2/r = (132 ft/s)^2/(3000 ft) = approx 5.81 ft/s^2

tangential acceleration = a_T = 7.5 ft/s^2

since radial and tangential accelerations are perpendicular, tha magnitude of the overall acceleration = sqrt[(a_R)^2 + (a_T)^2] =
approx 9.49 ft/s^2

direction of the acceleration vector is lagging behind the radial acceleration vector...
arctan(a_R/a_T) = approx 38 degrees
with respect to the unit tangential vector, the angle will be
180 - 38 = 142 degrees

Thank you skeeter

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mathematics is the future