2. a: 2m/s W, 1.88m/s S
4. a: 15m b: 'a' is a reference point, she starts at 'b'.
c: 3m/s d: 3m/s in the x-direction.
e: x = x0 + vt
x = 15m + (3m/s)t
11. a: a = -g, v = v0 - gt, y = v0t - (1/2)gt2.
c: easiest: use v2 = v02 +2ay where v0 is given and v = 0 (at the top) to get 16.5m.
d and e: it is possible to get t from the equation for y by solving the quadratic equation. It is much easier to do get the answer by using the separate segments of the flight. First the way up, then the way down to the starting height, and finally the next 420m. The answers are 11.3s and -92.5m/s.
3. b: zero c: 40m/s Here is a part d: what is the acceleration at t = 5s? For an instantaneous change in velocity the acceleration is infinite, which is impossible and unrealistic. For a nearly instantaneous change the acceleration would be correspondingly large and probably still unrealistic.
6. It starts at x = 100m and ends at x = 0. The initial velocity is -25m/s (toward the origin). Its magnitude decreases but it does not change sign. Since the velocity becomes less negative (increases) the acceleration is positive.
Note that the graphs for problems 12 and 13 are at the top of page 53. The grtaph for problem 17 is at the top of the right-hand column.
16. a: 120m b: 24m/s c:10m/s in the direction opposite to that of the initial velocity.
18. See the comments on GR 11.
22. 34m/s (not 35!).
1, 2, 3. Only one quantity in each of these three questions - which is it?
1. a: no, b: yes,