#### Answer

The tangential acceleration of car A is $9.80~m/s^2$
The tangential acceleration of car B is $12.9~m/s^2$
The tangential acceleration of car C is $6.68~m/s^2$

#### Work Step by Step

Note that the force of kinetic friction causes each car to decelerate. Let $M$ be the mass of each car.
We can find the tangential acceleration of car A;
$F_f = Ma$
$F_N~\mu_k = Ma$
$Mg~\mu_k = Ma$
$a = g~\mu_k$
$a = (9.80~m/s^2)(1.0)$
$a = 9.80~m/s^2$
The tangential acceleration of car A is $9.80~m/s^2$.
We can use the centripetal force to find the normal force $F_N$ of the road on car B;
$\sum F = \frac{Mv^2}{r}$
$F_N-Mg = \frac{Mv^2}{r}$
$F_N = \frac{Mv^2}{r}+Mg$
$F_N = M~(\frac{v^2}{r}+g)$
$F_N = M~(\frac{(25~m/s)^2}{200~m}+9.80~m/s^2)$
$F_N = M~(12.9~m/s^2)$
We can find the tangential acceleration of car B.
$F_f = Ma$
$F_N~\mu_k = Ma$
$M~(12.9~m/s^2)~\mu_k = Ma$
$a = (12.9~m/s^2)~\mu_k$
$a = (12.9~m/s^2)(1.0)$
$a = 12.9~m/s^2$
The tangential acceleration of car B is $12.9~m/s^2$.
We can use the centripetal force to find the normal force $F_N$ of the road on car C;
$\sum F = \frac{Mv^2}{r}$
$Mg-F_N = \frac{Mv^2}{r}$
$F_N = Mg - \frac{Mv^2}{r}$
$F_N = M~(g-\frac{v^2}{r})$
$F_N = M~(9.80~m/s^2 - \frac{(25~m/s)^2}{200~m})$
$F_N = M~(6.68~m/s^2)$
We can find the tangential acceleration of car C.
$F_f = Ma$
$F_N~\mu_k = Ma$
$M~(6.68~m/s^2)~\mu_k = Ma$
$a = (6.68~m/s^2)~\mu_k$
$a = (6.68~m/s^2)(1.0)$
$a = 6.68~m/s^2$
The tangential acceleration of car C is $6.68~m/s^2$.