4, 7, 11, 12, 14, 17, 19, 24, 25, 27, 28, 29, 32*, 63, 68, 73, 74 Note the correction to the hint to #68, Quiz 1. Scroll down to #68. |
| TURN IN: 7( see note) , 12, 14, 24, 28, 32, 68,
74. Note: Redo problem 7 using the following changed input
information: You are driving home steadily at 89 km/h for 135 km. It then begins to rain and you slow down to 62 km/h. You arrive home after driving 3 hours and 9 minutes. |
| * done in class |
| Complete hints given later here or in class/office hours; check back soon ! Note 11 was done in section 01 (Day session). I have repeated that hint below along with selected hints to at least get you started. |
| Lab1: BEFORE YOU LOOK AT THE HINTS BELOW, please
remember to read the following handouts before lab Friday August
28th. GENERAL MEASUREMENT THEORY: METER STICK, VERNIER CALIPERS and "TRIPLE BEAM" MASS SCALE: http://phoenix.phys.clemson.edu/tutorials/measure/index.html DETAIL ON VERNIER CALIPERS (CLICK LINK "vernier calipers" on the left.): http://phoenix.phys.clemson.edu/labs/cupol/index.html or go directly here: http://phoenix.phys.clemson.edu/labs/cupol/vernier/index.html Here is another link to check out for the first laboratory: THIS LINK IS DIRECTLY RELATED TO THE EXPERIMENT, INCLUDING THE DATA SHEET FOR MEASURING THE DENSITY OF VARIOUSLY SHAPED OBJECTS. http://wise.fau.edu/~dchen/1.pdf -------------------------------------------------------- OTHER LINKS: http://phoenix.phys.clemson.edu/tutorials/measure/index.html http://en.wikipedia.org/wiki/Caliper http://www.physics.ccsu.edu/LEMAIRE/genphys/virtual_physics_labs.htm http://www.upscale.utoronto.ca/PVB/Harrison/Micrometer/Micrometer.html and also THIS LINK ON SIGNIFICANT FIGURES http://library.thinkquest.org/10796/ch1/ch1.htm |
| Lab 2: Galileo's inclined planes : http://www.teachersdomain.org/resource/phy03.sci.phys.mfw.galileoplane/ and http://www.mcm.edu/academic/galileo/ars/arshtml/mathofmotion1.html |
| 4. Read Chapter 1, Section 1.6. |
| 7. (a) If you write your answer in meters,
please convert the velocities from km/h to m/s by
multiplying them by the conversion factor 5/18 which
we discussed in class. Note: 130 km = 130,000 m = 1.3x10 5
m. Also note that 3 hours and 20 minutes = 180 + 20 = 200 minutes
= 12, 000 seconds = total time.
You travel 130 km at 95*(5/18) m/s. You now need
to find the distance you travel at the smaller velocity
65*(5/18) m/s. Now, the total time you drive is 12,000
seconds. The time you drive at 95*(5/18) m/s is (b) Average velocity = total displacement /total time. Note that since the motion is always in the same direction in this case, the total displacement = total distance; thus average speed = total distance/ total time = average velocity. |
| 11. You can assume that where they meet, the
value of x is zero. They will meet at the same time t and
they will have the same position x = 0. Write down the equation
for each object:
Locomotive A, moving in the positive x direction: Set xA = xB and solve for the time t.
|
| 12. See hint to 73 below. The car and truck front
will have the same displacement (xf - xi)
when the car passes.
Write the equation for the position of each moving object: Then plug t into either equation to find each position x. |
| 14. First of all the average velocity is zero
since the total displacement = 0 for a round trip. So let us find the average speed = total distance/total time. Clearly, the total distance is 250 km + 250 km = 500 km, as reflected in your gasoline usage. Now we need the total time: Total time = 1 hour + (Outgoing time) + ( Return time) . Outgoing time = 250 km/(outgoing speed) Return time = 250 km/(return speed) . Unit conversion issues: Multiply the outgoing and return speeds by (5/18) to covert to m/s. Also convert 500 km to meters using 1 km = 1000 m. Finally, 1 hour = 3600 seconds. Thus you can get the total time in seconds and the total distance in m. The average speed is in m/s. |
| 17. average acceleration = (change in velocity)/(change in time) . For part (b), see section 1.6. |
| 19. average acceleration = (V2 - V1)/(t2 - t1) , where (t2 - t1)= 5.0 seconds. The second velocity is zero. The first velocity can be found from the 110-m distance and 4.0-s time interval given when the car is moving at constant speed. |
| 24. Use equation 2.11 (c) to find the acceleration and 2.11(a) to get the time. note the initial velocity is zero. |
| 25. Use 2.11 (a) to find the acceleration, which will be negative. Note the final velocity is zero, and the initial velocity is given. Once you find the acceleration, you can use 2.11 (c) to find the displacement. Remember, the final velocity is zero, and the initial velocity is given. |
| 27. Use 2.11 (c). Note: the final velocity is zero, and the initial velocity and displacement are both given. Once you get the acceleration, divide by 9.8 to get the number of g's. |
| 29. Here is a hint to 29 first, useful for #28
.There are two intervals:
Motion at constant velocity: Motion with negative acceleration Note that the final velocity is zero, and VR = Vo . Solve for (Xf - XR) . Write (XR - X1) + (Xf - XR) in terms of symbols. |
| 28. Use the formula given by Problem 29. |
| 63. Initially, the back of the train is (95 +
180) m = 275 m from the worker. To get the velocity of
the back of the train when it arrives at the location of the
worker, we use: v22 = v12 + 2a(x2 - x1) , where v1 = 0 and (x2 - x1) = 275 m. Find v2 . However, we need the acceleration a. You can find it using additional information: The front of the train has speed 25 m/s after traveling 180 m. Find a using, vf2 = vi2 + 2a(xf - xi), where = 180 m, vi = 0 and vf = 25 m/s. Find the acceleration and plug into the first equation to find the final speed of the train rear. |
| 68. The fugitive and empty box car will have the
same displacement (xf - xi) when
he jumps on.
(b) You can answer part (b) immediately. You can find the
distance that both the box car and fugitive travel using: THE ABOVE HINT IS WRONG: Here is the correct method: Meanwhile in this first time interval the train has moved a distance d = (6 m/s)(2 s) = 12 (m). Thus the train is 4 (m) in front of the fugitive. To find the time the fugitive takes to catch up to the train, write 4 + (6 m/s)t' = (8 m/s)t', where t' = the second time interval for the fugitive to catch up. See problem 12. Solving for t', we get t' = 2 seconds. Thus the total time is t + t' = 4 seconds. (b) In the total time interval of 4 seconds, the train and fugitive have moved (6 m/s) (4 s) = 24 m. |
| 73. The auto and train front will have the same
displacement (xf - xi) when the
auto passes.
Write the equation for the position of each moving object: Then plug t into the equation for the auto to find its position x. If the auto and train are traveling in opposite
directions, we have the two set of equations: |
| 74. vf2 = vi2 + 2a(xf - xi), where = 3.5 m, vi = 0 and vf = 44 m/s. Find the acceleration. |
| More hints posted later. Check back soon. |