First, here is a site talking a bit about a VERY exciting event happening Wednesday in Geneva, Switzerland. The largest particle super-collider (considered the most expensive science experiment ever built) will be turned on and tested tomorrow. This is a very big deal - read about it, and also check out this rap below that does a good job of showing what it will be able to do:
Solutions from the handout:
5.
a) 4 m/s2
b) Speeder displacement = 300 meters, police displacement = 200 meters
c) 100 meters
d) 20 seconds (They meet 10 seconds after the police officer stops accelerating, which is 10 seconds after the problem starts.)
6.
b) 4.5 seconds
c) The acceleration is the same at all three times, and has magnitude .267 m/s2
d) Area under the velocity vs. time graph during the acceleration = 1.2 meters
e) The total displacement during the trip is the total area under the curve. Since the area above the t-axis is the same as the area below, the net area is zero.
6(the second one)
a)5 m/s2
b) 1.25 m/s2
c) -2.5 m/s2
d) -5 m/s2
e) 0 m/s2
7.
a) Cart is at rest when velocity is zero, at t = 4 seconds, 18 seconds
b) The cart speeds up when the velocity and acceleration have the same sign. This occurs between t = 4 to 9 seconds, and t = 18 - 20 seconds
c) Displacement from t = 0 to t = 9 seconds is 1.6 meters - 2.5 meters = -0.9 meters, as found from the area above and below the t-axis. Since this is the displacement, and the displacement is x - x0, x = displacement + x0 = 2 meters + -0.9 meters = 1.1 meters.
8. velocity before entering the water is 8.85 m/s, which becomes v0 for the acceleration in the water. Applying v^2 = v0^2 + 2a*delta_x, a = 19.58 m/s2.
Tuesday, September 9, 2008
Monday, September 8, 2008
Free Fall and other Animals
HANDOUT PROBLEMS:
4. 706.6 meters
5. 4.9 meters
6. 400 meters
7. 1.078 seconds
8. 7.877 m/s
HW SOLUTIONS: Page 43,Problems 34, 37, 38, 41, 44, 46
34. 60.03 meters
37. From the max height, you can get the initial velocity of the kangaroo. (7.275 m/s) You can then use this with a kinematics equation to get the time required to reach this height. Double it to get the answer of 1.485 s.
38. Since the total time is 3.3 seconds, it takes 1.65 seconds to reach the top. You can use this to find the initial velocity of the ball (16.17 m/s) from v = v0 + a*delta_t. The height can then be found using v^2 = vo^2 + 2a*delta_x. The answer is 13.06 meters.
41. Set up a quadratic with delta_y = -105 meters, vo = 5.5 m/s, and a = -9.8 m/s2. Solving numerically, delta_t = 5.224 seconds.
44. You can use v^2 = vo^2 + 2a*delta_x to find the initial velocity of 12.837 m/s. Again, set up a quadratic to find the times. There are two times because the object has the 12 meter height both when it goes up and goes down. These times are at 0.7309 s and 3.351 seconds.
46. Follow the hint given on the sheet - first try to find the initial velocity of the stone at the top of the window. Since acceleration is constant, we can use the equation stating that average velocity = (v + v0)/2 = delta_x/delta_t. This means the velocity of the stone halfway through its fall past the window is 7.333 m/s. This means that after 0.15 s (halfway of the fall time), the speed of the stone is 7.333 m/s.
For the interval of time from when the stone passes the top of the window to when its speed is 7.333 m/s, v = 7.333, delta_t = 0.15 s, a = 9.8 m/s2 downwards, and v0 is unknown. Calculating, v0 = 5.863 m/s.
Now consider the interval of time between when the stone is dropped and it FIRST reaches the top of the window. During this interval, v0 = 0 m/s, a = 9.8 m/s2, and v = 5.863 m/s. From these quantities, you can use ONE of the kinematics equations to find the displacement of the stone over this interval. The final answer is 1.754 meters.
4. 706.6 meters
5. 4.9 meters
6. 400 meters
7. 1.078 seconds
8. 7.877 m/s
HW SOLUTIONS: Page 43,Problems 34, 37, 38, 41, 44, 46
34. 60.03 meters
37. From the max height, you can get the initial velocity of the kangaroo. (7.275 m/s) You can then use this with a kinematics equation to get the time required to reach this height. Double it to get the answer of 1.485 s.
38. Since the total time is 3.3 seconds, it takes 1.65 seconds to reach the top. You can use this to find the initial velocity of the ball (16.17 m/s) from v = v0 + a*delta_t. The height can then be found using v^2 = vo^2 + 2a*delta_x. The answer is 13.06 meters.
41. Set up a quadratic with delta_y = -105 meters, vo = 5.5 m/s, and a = -9.8 m/s2. Solving numerically, delta_t = 5.224 seconds.
44. You can use v^2 = vo^2 + 2a*delta_x to find the initial velocity of 12.837 m/s. Again, set up a quadratic to find the times. There are two times because the object has the 12 meter height both when it goes up and goes down. These times are at 0.7309 s and 3.351 seconds.
46. Follow the hint given on the sheet - first try to find the initial velocity of the stone at the top of the window. Since acceleration is constant, we can use the equation stating that average velocity = (v + v0)/2 = delta_x/delta_t. This means the velocity of the stone halfway through its fall past the window is 7.333 m/s. This means that after 0.15 s (halfway of the fall time), the speed of the stone is 7.333 m/s.
For the interval of time from when the stone passes the top of the window to when its speed is 7.333 m/s, v = 7.333, delta_t = 0.15 s, a = 9.8 m/s2 downwards, and v0 is unknown. Calculating, v0 = 5.863 m/s.
Now consider the interval of time between when the stone is dropped and it FIRST reaches the top of the window. During this interval, v0 = 0 m/s, a = 9.8 m/s2, and v = 5.863 m/s. From these quantities, you can use ONE of the kinematics equations to find the displacement of the stone over this interval. The final answer is 1.754 meters.
Sunday, September 7, 2008
Constant Acceleration Equations
Answers from the problems on the back of the handout:
1. delta_x = 300 m
2. final velocity = 8 m/s
3. initial speed = 6.61 m/s
4. t = 0.1714 s
5. time to meet = .636 hrs, position is .184 miles west of flagpole.
HW solutions:
16. acceleration = - 6.25 m/s2 (in opposite direction to initial velocity.) This is the same as .638 g's.
21. delta_x = 150 m
23. delta_x = 62.5 m
25. displacement = -156.25 m, time = 25 s
28.
29. time = 23.96 seconds, speed = 66.61 m/s
1. delta_x = 300 m
2. final velocity = 8 m/s
3. initial speed = 6.61 m/s
4. t = 0.1714 s
5. time to meet = .636 hrs, position is .184 miles west of flagpole.
HW solutions:
16. acceleration = - 6.25 m/s2 (in opposite direction to initial velocity.) This is the same as .638 g's.
21. delta_x = 150 m
23. delta_x = 62.5 m
25. displacement = -156.25 m, time = 25 s
28.
29. time = 23.96 seconds, speed = 66.61 m/s
Thursday, September 4, 2008
Velocity and Acceleration as Functions of Time
HW: Read Sections 2-1 to 2-4, Do QUESTIONS 2 - 4 and 10 and PROBLEMS 5, 8, 9, 13
Questions:
2. Velocity is a vector. An object could travel at constant speed but change direction (as in an object traveling in a circle) which means the velocity changes.
3. No - constant velocity means constant magnitude (speed) and direction.
4. No - if velocity is constant, then the average and instantaneous velocities are equal during that time interval.
10. If down is positive, then an object thrown upwards will have a negative initial velocity with a positive acceleration downwards. By switching up to be the positive direction, the reverse will be true.
Problems:
5. Part I of trip takes (130 mi)/(65 mi/h) = 2 hrs. The trip takes 3 1/3 hours total, so you drive at 55 mi/h for 1 1/3 hours. (55 mi/h)(1 1/3 hrs) = 73.3 miles.
a) total distance = 130 mi + 73.3 miles = 203.3 miles
b) average speed = total distance / total time = (203.3 mi) / (3 1/3 hrs) = 60./9 mi/h
8. average speed = 10.37 m/s, average velocity = 3.46 m/s
9. We have to write position functions for each car. Set x = 0 at train A at t = 0. This makes Xa,o = 0 and Xb,o = 8.5 km.
Train A: Xa = Xa,o + va*delta-t
Train B: Xb = Xb,o - vb*delta-t (negative since train B travels in the opposite direction to Train A.
When the trains meet, their position functions are equal.
Solving, delta-t = .0447 hours or about 2 minutes and 41 seconds.
13. 4.256 m/s2
Questions:
2. Velocity is a vector. An object could travel at constant speed but change direction (as in an object traveling in a circle) which means the velocity changes.
3. No - constant velocity means constant magnitude (speed) and direction.
4. No - if velocity is constant, then the average and instantaneous velocities are equal during that time interval.
10. If down is positive, then an object thrown upwards will have a negative initial velocity with a positive acceleration downwards. By switching up to be the positive direction, the reverse will be true.
Problems:
5. Part I of trip takes (130 mi)/(65 mi/h) = 2 hrs. The trip takes 3 1/3 hours total, so you drive at 55 mi/h for 1 1/3 hours. (55 mi/h)(1 1/3 hrs) = 73.3 miles.
a) total distance = 130 mi + 73.3 miles = 203.3 miles
b) average speed = total distance / total time = (203.3 mi) / (3 1/3 hrs) = 60./9 mi/h
8. average speed = 10.37 m/s, average velocity = 3.46 m/s
9. We have to write position functions for each car. Set x = 0 at train A at t = 0. This makes Xa,o = 0 and Xb,o = 8.5 km.
Train A: Xa = Xa,o + va*delta-t
Train B: Xb = Xb,o - vb*delta-t (negative since train B travels in the opposite direction to Train A.
When the trains meet, their position functions are equal.
Solving, delta-t = .0447 hours or about 2 minutes and 41 seconds.
13. 4.256 m/s2
Wednesday, September 3, 2008
Unit Systems - HW and Solutions
HW for tonight: Read p. 8-14, do PROBLEMS (not questions) 15, 17, 20, 26, 33.
Solutions:
15. You should do this for your own height, but Mr. Weinberg is 5'7", or 67 inches. This works out to 1.702 meters.
17.
a) 3.9 x 10-9 inches
b) Conversion factor is that 1 atom = 1.0 x 10-10 meters. You then use the 1 cm length given to find how many atoms there are. Answer: 108 atoms.
20. 7.3% longer
26. There are many correct answers! Here's a reasonable way to go about it.
Say your heart beats 64 beats per minute, and that the average time a person lives is around 75 years. 64 beats/min * (60 min/hr)*(24 hr/ 1 day) * (365 days/year)*(75 years/lifetime) = 2.523 x 109 beats per lifetime.
33.
a) 31,536,000 seconds/year
b) 3.154 x 1016 nanoseconds
c) 3.171 x 10-8 years
WP1. units of A: m/s4
Solutions:
15. You should do this for your own height, but Mr. Weinberg is 5'7", or 67 inches. This works out to 1.702 meters.
17.
a) 3.9 x 10-9 inches
b) Conversion factor is that 1 atom = 1.0 x 10-10 meters. You then use the 1 cm length given to find how many atoms there are. Answer: 108 atoms.
20. 7.3% longer
26. There are many correct answers! Here's a reasonable way to go about it.
Say your heart beats 64 beats per minute, and that the average time a person lives is around 75 years. 64 beats/min * (60 min/hr)*(24 hr/ 1 day) * (365 days/year)*(75 years/lifetime) = 2.523 x 109 beats per lifetime.
33.
a) 31,536,000 seconds/year
b) 3.154 x 1016 nanoseconds
c) 3.171 x 10-8 years
WP1. units of A: m/s4
Tuesday, September 2, 2008
Welcome back!
Hi everyone,
Glad to know you found the site. I'm very excited to get going this year with you all.
As promised, I'd like you to respond to the following questions for HW tonight:
1. What are you MOST interested in learning during the course?
2. What are you most concerned about in taking this course?
3. What are your strengths and weaknesses as a math and science student?
I ask that you send your responses by email ASAP.
See you all tomorrow!
EMW
Glad to know you found the site. I'm very excited to get going this year with you all.
As promised, I'd like you to respond to the following questions for HW tonight:
1. What are you MOST interested in learning during the course?
2. What are you most concerned about in taking this course?
3. What are your strengths and weaknesses as a math and science student?
I ask that you send your responses by email ASAP.
See you all tomorrow!
EMW
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