P17_052.PDF

52. (a) This distance is determined by the longitudinal speed:

d = v t = (2000 m / s) 40

10 − 6 s =8 . 0

10 − 2 m .

(b) Assuming the acceleration is constant (justiﬁed by the near-straightness of the curve a = 300 / 40

10 − 6 ) we ﬁnd the stopping distance d :

v 2 = v o +2 ad =

⇒

d =

(300) 2 40

10 − 6

2(300)

10 − 3 m. This and the radius r form the legs of a right triangle (where r is

opposite from θ =60 ◦ ). Therefore,

tan 60 ◦ = r

⇒

r = d tan 60 ◦ =1 . 0

10 − 2 m .

which gives d =6 . 0