Sol. We observe that the radius of curvature appears in the expression for the normal component of acceleration, so we use n– and t-components of the total acceleration are obtained by resolving the given horizontal and vertical accelerations into there n- and t-components and then combining.

From the figure we get

a n = 30 cos 15º – 20 sin 15º = 23.8 ft/sec 2 a t = 30 sin 15º + 20 cos 15º = 27.1 ft/sec

2 We may now compute the radius of curvature from

[a n = v

2 / ρ ] ρ = =

= 13.01 (10 6 ) ft Ans.

The rate at which v is increasing is simply the t-component of acceleration.

[ = a t ] = 27.1 ft/sec 2 Ans.

The angular rate of line GC depends on v and ρ and is given by

[v = ρ ] = v/ ρ = = 13.53(10 –4 ) rad/sec . Ans.

With unit vectors e n and e t for the n- and t-directions, respectively, the total acceleration becomes

a = 23.8e n + 27.1e t ft/sec 2 Ans.

Helpful Hints:

1. Alternatively we could find the resultant acceleration and then resolve it into n- and t-components.

2. To convert from mi/hr to ft/sec., multiply by = which is easily remembered, as 30 mi/hr is the same as 44 ft/sec.