Granger, Problem 1.58

A fluid parcel moves along a river of shape y = x^2 + 5 and speed v = 10 m/s

Determine where acceleration is a maximum:

Solution:

  • Acceleration is a maximum when y' at some critical point = 0 and when y'' at some critical point < 0
    (the second derivative test)
    y' = 2x = 0 when x = 0 and y'' = 2 when x = 0

    Calculate the maximum acceleration:

    Solution:

    r = (x^2 + 5)j
    x = rcosQ, where Q = theta
    r = (rcosQ)^2 + 5
    V = velocity = dr/dt = -2(rsinQ)
    {QUESTION: Did I differentiate w.r.t. x instead of t?}
    {QUESTION: Since speed v = 10 m/s, would V = v*e = (10)e be a better way to solve?
    a = acceleration = dV/dt = -2(rcosQ)

    Problem posted 6/28/05 -- LAL