Answers for Problem Set 2

11.

(a) (v₁ + v₂) / 2 = 45 mi/hr
(b) 2 / [ 1/v₁ + 1/v₂) = 43 mi/hr
(c) Apply the formula in (b) to the results in (a) and (b), getting 44 mi/hr
(d) 0
(e) Graph

12.

(a) 0, -2, 0, 12 m
(b) 12 m
(c) 7 m/s
(d) Graph

15. The number of trips is infinite. The unphysical number of trips is an artifact caused by neglecting the time required for the bird to turn around every time it reaches a train.

The distance the bird travels is

where d₀ is the initial distance between the trains. Note that the fact that the distance the bird flies is the same as the initial distance between the trains is specific to the speeds chosen for the bird and trains.

17. All times in seconds.

(a) 2 < = t < = 4
(b) 0 < = t < 3
(c) 3 < t < 7
(d) t = 3

19. The runner travels 100 m. There are 25 squares under the curve and each square represents 4 meters.

21. GRAPH. The acceleration starts at or slightly above 0, peaks at about t = 1 s, goes through zero at or about t = 2 s, has its most negative value at or about t = 3 s, vanishes again at t = 4 s, and remains positive through the rest of the graph.

23. (a)

Interval	velocity is	acceleration is
AB	positive	negative
BC	zero	zero
CD	positive	positive
DE	positive	near zero

(b) If the acceleration must be continuous, then it must change somewhere just short of B, just after C, and perhaps just before D. If the acceleration is permitted to be discontinuous, then there is no finite interval where it is obvious that the acceleration is changing.

(c) No.

48.

a = 2 (d - v₀ t ) / t² = - 4.61 x 10⁴ km / hr².
v_f = - v₀ + d/(2t) = 30.4 km/hr = 18.9 mi/hr

87. The time required to fall is

t = (1/g) [ v_0y + sqrt(v_0y² + 2gy₀) ] = 5.45 s

and the speed when the package hits the ground is

v_y = - sqrt( v_0y² + 2 g y₀ ) = -41.4 m/s

III. x = v₁ d / (v₁ - v₂ )