p11_098.pdf

98. (a) The linear speed at t =15 . 0sis

v = a t t = 0 . 500 m / s 2 (15 . 0s)=7 . 50 m / s .

The radial (centripetal) acceleration at that moment is

a r = v 2

= (7 . 50 m / s) 2

30 . 0m

=1 . 875 m / s 2 .

Thus, the net acceleration has magnitude:

a = a t + a r = 0 . 500 m / s 2 2

+ 1 . 875 m / s 2 2

=1 . 94m / s 2 .

(b)Wenotethat 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.