Sol. The dimensions of the car are small compared with those of the path, so we will treat the car as a particle. The velocities are

v A = = 27.8 m/s

v C = 50 = 13.89 m/s

We find the constant deceleration along the path from

a t = (v C 2 – v A 2 ) = = –2.41 m/s 2 Condition at A.

With the total acceleration given and a t determined, we can easily compute a n and hence ρ from

[a

2 = a n 2 + a t 2 ] a n 2 = 3 2 – (2.41) 2 = 3.19 a n = 1.785 m/s 2 [a n = v

2 / ρ ] ρ = v 2 /a n = (27.8) 2 /1.785 = 432 m . Ans.

Condition at B.

Since the radius of curvature is infinite at the inflection

point, a n = 0 and a = a t = –2.41 m/s 2 Condition at C.

The normal acceleration becomes

[a n = v

2 / ρ ] a n = (13.89) 2 /150 = 1.286 m/s 2 With unit vectors e n and e t in the n-and t-directions, the acceleration may be written

a = 1.286e n – 2.41 e t m/s

2 where the magnitude of a is

[a = ] a= = 2.73 m/s

2 Ans.

The acceleration vectors representing the conditions at each of the three points are shown for clarification.