Acustomer sits in an amusement park ride in which the compartment is to be pulled downward in the negative direction of a y axis with an acceleration magnitude of 1.24g, with g 9.80 m/s2 . a 0.567 g coin rests on the customer’s knee. once the motion be- gins and in unit-vector notation, what is the coin’s acceleration rel- ative to (a) the ground and (b) the customer? (c) how long does the coin take to reach the compartment ceiling, 2.20 m above the knee? in unit-vector notation, what are (d) the actual force on the coin and (e) the apparent force according to the customer’s meas- ure of the coin’s acceleration?

(a) The coin is accelerated only by gravity (until it hits the ceiling). Its acceleration is -9.80 m/s²

(b) The acceleration relative to the customer is
  (-1.0 g) - (-1.24 g) = 0.24 g ≈ 2.352 m/s²

(c) The distance (d) covered is
  1/2at² = d . . . . for some acceleration a
  2.2 m = (1/2)(2.352 m/s²)t²
Solving for the time to cover the distance, we get
  t = √(2·2.2/2.352) s ≈ 1.368 s

(d) The actual force on the coin is that due to gravity, as found in part (a). (The first m is "mass"; the second m is "meters".)
  F = mg = (5.67·10^-4 kg)·(-9.8 m/s²) = -5.5566·10^-3 N

(e) From the customer's point of view, the apparent force on the coin is
  F = ma = (5.67·10^-4 kg)·(2.352 m/s²) = 1.333584·10^-3 N
(The positive sign means directed upward.)