#### Answer

(a) $v = 3.8~m/s$
$a_c = 0.96~m/s^2$
(b) The ratio of the apparent weight to the true weight is 0.90
(c) The ratio of the apparent weight to the true weight is 1.1

#### Work Step by Step

(a) We find the speed first;
$v = \frac{distance}{time}$
$v = \frac{2\pi~r}{t}$
$v = \frac{(2\pi)(15~m)}{25~s}$
$v = 3.8~m/s$
We then find the centripetal acceleration;
$a_c = \frac{v^2}{r}$
$a_c = \frac{(3.8~m/s)^2}{15~m}$
$a_c = 0.96~m/s^2$
(b) Let $F_N$ be the normal force of the seat pushing up on the person. Note that $F_N$ is equal to the person's apparent weight.
$\sum F = \frac{Mv^2}{r}$
$Mg-F_N = \frac{Mv^2}{r}$
$F_N = Mg-\frac{Mv^2}{r}$
$F_N = (M)(9.80~m/s^2)-\frac{M(3.8~m/s^2)^2}{15~m}$
$F_N = (M)(9.80~m/s^2-0.96~m/s^2)$
$F_N = (M)(8.84~m/s^2)$
We can find the ratio of $F_N$ to $mg$:
$ratio = \frac{F_N}{mg}$
$ratio = \frac{(M)(8.84~m/s^2)}{(M)(9.80~m/s^2)}$
$ratio = 0.90$
The ratio of the apparent weight to the true weight is 0.90
(c) We can find an expression for the normal force at the bottom.
$\sum F = \frac{Mv^2}{r}$
$F_N-Mg = \frac{Mv^2}{r}$
$F_N = Mg+\frac{Mv^2}{r}$
$F_N = (M)(9.80~m/s^2)+\frac{M(3.8~m/s^2)^2}{15~m}$
$F_N = (M)(9.80~m/s^2+0.96~m/s^2)$
$F_N = (M)(10.76~m/s^2)$
We can find the ratio of $F_N$ to $mg$:
$ratio = \frac{F_N}{mg}$
$ratio = \frac{(M)(10.76~m/s^2)}{(M)(9.80~m/s^2)}$
$ratio = 1.1$
The ratio of the apparent weight to the true weight is 1.1