98.
(a) The linear speed at t = 15.0 s is
v = a
t
t =
0.500 m/s
2
(15.0 s) = 7.50 m/s .
The radial (centripetal) acceleration at that moment is
a
r
=
v
2
r
=
(7.50 m/s)
2
30.0 m
= 1.875 m/s
2
.
Thus, the net acceleration has magnitude:
a =
a
2
t
+ a
2
r
=
0.500 m/s
2
2
+
1.875 m/s
2
2
= 1.94m/s
2
.
(b) We note that a
t
v. Therefore, the angle between v and a is
tan
−1
a
r
a
t
= tan
−1
1.875
0.5
= 75.1
◦
so that the vector is pointing more toward the center of the track than in the direction of motion.