Problem 2/126 Page 65, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: Driver traveling at an initial speed of v0 uniformly decelerates to a speed of v1 over a distance L. At a point midway between these two, the road has a radius of curvature of R.

Find: The magnitude of the acceleration of the car at this midway point.

Parameter values:
v0 = 250 km/hr
v1 = 200 km/hr
L = 300 m
R = 500 m

Recall the expression for the acceleration in path coordinates:
a = at t + an n

The tangent acceleration is prescribed to be a constant, A.
at = dv/dt = dv/ds ds/dt = v dv/ds = A

Separating variables:

v dv = A ds

Integrating both sides:

1/2 v² = A s + C

Knowing that at the starting point v=v0 permits the evaluation of C.

1/2 v² = A s + 1/2 v0²

Knowing that at s=L, v=v1 permits the evaluation of A:

1/2 v1² = A L + 1/2 v0²

A = 1/2 (v1² - v0²) / L = -2.89 m/s²

We can now evaluate the speed of the car at the midway point, s = L/2

1/2 vm² = A L/2 + 1/2 v0²

vm = { v0² + A L }^1/2 = 62.9 m/s

Knowing this speed and the radius of curvature, we can evaluate the normal acceleration:

a = A t + vm²/R n = -2.89 m/s² t + 7.91 m/s² n

The magnitude of the acceleration is the square root of the sum of the squares of the components:

| a | = { 2.89² + 7.91² }^1/2 m/s² = 8.42 m/s²