| Last revised 1999/09/14 |
24. Given that v = dx/dt, how do you get the velocity from a graph? Do the entire v(t) curve before starting a(t).
28. What can you say about the position of a particle if its velocity dx/dt = 0 ? What can you say about the velocity of a particle if its acceleration vanishes?
57. What is the condition for just avoiding a collision. How is this condition different than the condition for avoiding a collision without specifying by how much the collision is avoided? What is the condition for not avoiding a collision?
58. Try writing an equation for the distance between the trains, the rate of change of the distance, and the "acceleration" of the distance. Make sure you get the signs right. What are the initial conditions for the distance between the trains and the rate of change of that distance? What is the condition for avoiding a collision using these variables?
It is more straightforward to just calculate the distance each train moves, but it is also more work.
72. Definitely not an easy problem. Notice that you don't know the distance to the bottom of the tube. If you calculate the time required to return to an arbitrary level, you can use your formula twice, once for the lower and once for the upper height.
The convention in physics problems is that you know a value for any quantity mentioned and not asked for, unless, of course, you are specifically told otherwise. Hence in this problem you assume you have a value for the time spent above each level and the distance between levels, but you do not know exactly what g is in the laboratory in which you make the measurement.
| Problem # | Answer |
|---|---|
| 27 | vave = [x(3) - x(0)] / (3 s) = 80 m/s, vinstantaneous =50 + 20 t = 110 m/s, a = 20 m/s2, all at t = 3 s |
| 52 | (a) You can stop in 35.4 m or reach the intersection in 2.56 sec,
so you can stop or continue successfully (b) You can still stop in 35.4 m [too far] or you can reach the intersection in 2.05 sec; neither attempt is successful. The time that the yellow light is on is too short. |
| 56 | Distance 82 meters and speed 19 meters/second after 8.64 sec. The t=0 solution of the quadratic equation is simply the starting time. |
Q9. What quantity, besides acceleration and x0 , is constant as you move from trajectory to trajectory? [The magnitude of the initial velocity is not constant.]
38. A suicide is not "accidental".
53. If you work symbolically, do part (b) first; you will then only have to solve the problem once. If you have trouble, however, part (a) is simpler and might make good practice before doing the harder part.
56. The problem has its own hint, and both angles will come from the quadratic equation the text mentions. Have you drawn a diagram showing the trajectory for the largest possible angle and the smallest possible angle so that you understand the physics of getting two solutions?
IV. This problem also comes with a hint. Look at HRW sections 4-8 and 4-9.
| Problem # | Answer [All from HRW 4th edition] |
|---|---|
| 18 | (a) 18 cm (b) 1.9 m |
| 20 | (a) 5.4 x 10 -13 m (b) Decreases |
| 22 | (a) 0.50 s (b) 10 ft/s |
| 26 | (a) 16.9 m horizontally, 8.21 m vertically (b) 27.6 m horizontally, 7.26 m vertically (c) 40.2 m horizontally, 0.00 m vertically |
| 28 | (a) 1.15 s (b) 12.0 m (c) 19.2 m/s horizontally, 4.80 m/s vertically (d) No |