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.