course phy 121 I realize this wasn't required to send in, but since I have been gone for two weeks, I felt it was best to do it. I did the practice quiz in two days (had to stop for thunderstorm). I hope I have sent the two pieces correctly. ?????{?}?????assignment #001
......!!!!!!!!...................................
17:15:08 Problem Number 1 A coasting cart has velocities 4.5 m/s, 10 m/s and 2.25 m/s at clock times t = 0 s, 9 s and 13 s. Approximately how far does it move and at what average rate does it accelerate over each of the two time intervals (i.e., 0- 9 seconds and 9 - 13 seconds)? Show how you could use a graph of velocity vs. time to obtain your results. Directly reason out your results using the concept of rate. If the automobile is coasting, and if air resistance is not a significant factor, then over which time interval do you think the average slope of the road is greatest?
......!!!!!!!!...................................
RESPONSE --> I use a graph of velocity vs. time and plot out the points (0, 4.5) (9,10) (13, 2.25). To find the aAve I find the slope of the line from 0 - 9 secs and the slope of the line from 9-13 seconds by dividing rise by run. accel for 0-9 secs = (10-4.5)/ 9 = .61m/s/s accel for 9-13 secs = (2.25 - 10)/ 4 = -1.94m/s/s To find the distance traveled I find the area under each slope or use ds = vAve * dt and add the two together to get the total distance traveled. ds for 0-9 sec = (10+4.5)/2 * 9 = 65.25 ds for 9-13 sec = (2.25 + 10)/2 * 4 = 25 Total ds = 65.25 + 25 = 90.25m
.................................................
......!!!!!!!!...................................
17:18:46 STUDENT ANSWER: The average acceleration for the cart from time 0s to 9s, the average velocity is (4.5+10)m/s/2=7.25 m/s so displacement is 7.25 m/s * 9s =65.25m. The displacement for the cart between clock time 9s and 13s is (2.25+10)m/s / 2 * 4s=24.5m The acceleration of the cart between clock time 0 - 9s is a = `dv / `dt ? (10-4.5)m/s / (9 s) =.6m/s^2. The acceleration of the cart between clock time 9-13s is (2.25-10( m/s / (4s) = -1.93m/^2. You could use a graph of velocity vs. time to obtain the results. By placing the information on a graph of v vs. t you could see the rise and the run between the points. For example looking at 0,4.5 and 9,10 you could see that it had a run of 9 s and a rise of 5.5 m/s and then to find the acceleration you would have ave accel = slope = 5.5 m/s / (9 s) = .61m/s^2. If the automobile is coasting then the slope of the road is the greatest when the magnitude of the acceleration is greatest. This occurs on the second slope, where the magnitude of the acceleration is 1.93 m/s^2. INSTRUCTOR NOTES FOR ALL STUDENTS: Be sure you use units correctly in every calculation, and be sure you are using the definitions of velocity (rate of change of position, so vAve = `ds / `dt) and acceleration (rate of change of velocity, so a = `dv / `dt).
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. Thanks for reminding me to use a graph. The visual representation is a lot easier to use makes reasoning out my answers a lot easier than just trying to memorize the equations
.................................................
......!!!!!!!!...................................
17:47:01 Problem Number 2 We set up a straight ramp and measure the time required to coast down the ramp, from rest, at various slopes. The differences in elevation between one end and the other, for the different slopes, are 1.8, 4.2 and 6.7 cm. The time required for a cart to coast 78 cm down the ramp, starting from rest, is 2.722297 seconds on the first incline, 2.518101 seconds on the next, and 2.6606 seconds on the last. How well do these results support the hypothesis that acceleration is linearly dependent on slope?
......!!!!!!!!...................................
RESPONSE --> I've tried several different things and I am not getting that these support the hypothesis that acceleration is linearly dependent on slope. My final attempt went like this... First I found the slope of each ramp by dividing the difference in elevation by the length of the ramp s1 = 1.8/78 = .02 s2 = 4.2/ 78 = .05 s3 = 6.7/78 = .08 Then I found the acceleration of each trial by using the equation ds = v0 * dt + .5a * dt^2 trial 1: 78 = 0 + .5a * 2.722297^2 (78/7.410901)*2 = a = 21.05cm/s/s trial 2: 78 = 0 + .5a * 2.518101^2 (78/6.340833)*2 = a = 24.60cm/s/s trial 3: 78 = 0 + .5a * 2.6606^2 (78/7.07879)*2 = a = 22.04 cm/s/s Then plot acceleration vs. slope and they do not form a straight line. Of course this is easy to see just by looking at the times required because they do not incrementally go down as the ramp slope increases. The last one actually goes up.
.................................................
......!!!!!!!!...................................
17:48:21 STUDENT RESPONSE: 1.8cm 2.722297s 4.2cm 2.518101s 6.7cm 2.6606s 'ds=78. These results show that the smallest slope the time to coast is the slowest The middle ramp has the fastest time down the ramp These results show that acceleration is fastest down the middle ramp (4.2cm) INSTRUCTOR COMMENT ON STUDENT RESPONSE: ** To support the idea that acceleration is linearly dependent on slope you first have to calculate the accelerations, as you did in the Video Experiments. You then need to plot a graph of acceleration vs. ramp slope and see if it forms a straight line. **
......!!!!!!!!...................................
RESPONSE --> Just as I suspected. I spent WAY too much time on this. I did demonstrate that I know how to do it though.
.................................................
......!!!!!!!!...................................
18:08:54 Problem Number 3 A ball reaches a ramp of length 50 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 3.8 seconds. Its velocity at the end of the ramp is determined to be 6.31579 cm/s. What is its acceleration on the ramp?
......!!!!!!!!...................................
RESPONSE --> v0 = ? ds = 50cm dt = 3.8s vf = 6.3178cm/s a = ? ds = (vf+v0)/2 * dt 50cm = (6.3178cm/s + v0)/2 * 3.8s (50cm/3.8s)*2 = 6.3178 + v0 26.316 - 6.3158 = v0 = 20 cm/s aAve = dv/dt aAve = (6.3158cm/s - 20 cm/s) / 3.8s aAve = -13.6842 / 3.8 aAve = -3.6011cm/s/s Problem Number 3 A ball reaches a ramp of length 50 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 3.8 seconds. Its velocity at the end of the ramp is determined to be 6.31579 cm/s. What is its acceleration on the ramp?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
07:21:38 STUDENT RESPONSE: .'ds=50cm vf=6.31579cm/s 'dt = 3.8s The average velocity on the ramp is 50cm/3.8s =13.16m/s The initial velocity will be v0 = 2 'ds /dt, v0 = 2 *50 /3.8, v0 = 26.31579 -vf which is 6.31579, so the v0 is 20.00 ** You got it but this step isn?t correctly stated. v0 = 2 `ds/dt is not a correct statement. v0 = 2 vAve ? vf, or v0 = 2 `ds/`dt ? v0 (which is the same thing since vAve = `ds/`dt) is the correct equation. The reasoning is that ave velocity is the average of initial and final velocities, since acceleration is uniform. So you have vAve = (v0 + vf) / 2.. You know vAve from your previous calculation and vf is given. So you solve this equation for v0 and then substitute these values and simplify. You get v0 = 2 vAve ? vf, then substitute. This is actually what you did; just be careful you state it this way. If you don?t state it right there?s a chance you might not do it right. ** The accleration is 6.31579-20/3.8 = -3.601m/s/s If your v0 was correct this would be right **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
07:31:16 Problem Number 4 A projectile leaves the edge of a table at 18 cm/s, then falls freely a distance of 95 cm to the floor. How far does it travel in the horizontal direction during the fall?
......!!!!!!!!...................................
RESPONSE --> Horiz v = 18 cm/s Vert ds = 95cm Hor ds = ? Vert v0 = 0 dt = ? Vert aAve = 980cm/s/s ds = v0*dt + .5a*dt^2 95cm = 0cm/s*dt + .5(980cm/s/s)*dt^2 95cm = 490cm/s/s *dt^2 .1939 = dt^2 .44s = dt Horiz ds = vAve *dt Horiz ds = 18cm/s * .44s = 7.92cm
.................................................
......!!!!!!!!...................................
07:31:33 STUDENT RESPONSE: .'ds=95cm a=980cm/s/s v0=0 First, we take the equation 'ds = v0'dt + .5(980) 'dt^2 95cm = .5 (980) 'dt^2 'dt = sqrt 95cm/.5(980)=.194 So, 'dt = sqrt .194 'dt=.44s Next take the 18cm/s and multiply by the time duration .44s = 7.92cm/s INSTRUCTOR COMMENT TO ALL STUDENTS: Remember the basic principle used in this solution. You find the time of fall for the vertical motion, then find the horizontal displacement using the time interval and the initial horizontal velocity.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
07:31:36 STUDENT RESPONSE: .'ds=95cm a=980cm/s/s v0=0 First, we take the equation 'ds = v0'dt + .5(980) 'dt^2 95cm = .5 (980) 'dt^2 'dt = sqrt 95cm/.5(980)=.194 So, 'dt = sqrt .194 'dt=.44s Next take the 18cm/s and multiply by the time duration .44s = 7.92cm/s INSTRUCTOR COMMENT TO ALL STUDENTS: Remember the basic principle used in this solution. You find the time of fall for the vertical motion, then find the horizontal displacement using the time interval and the initial horizontal velocity.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
07:46:10 Problem Number 5 What are the average acceleration and final velocity of an object which accelerates uniformly from rest, ending up with velocity 13.9 cm/sec after traveling a distance of 40 cm from start to finish?
......!!!!!!!!...................................
RESPONSE --> v0 = 0 vf = 13.9 cm/s ds = 40 cm ds = vAve * dt 40cm = 13.9cm/s / 2 * dt 40cm/6.95cm/s = dt 5.76s = dt aAve = dv/dt aAve = (13.9cm/s - 0)/5.76s aAve = 13.9cm/s / 5.76s aAve = 2.41cm/s/s
.................................................
......!!!!!!!!...................................
07:48:54 STUDENT RESPONSE: 'ds=40cm v0=0 vf=13.9cm/s If vf is 13.9cm/s then the using the equation a 'ds = vf^2 - v0^2)/2 will be , a * 40 cm = ( (13.9 cm/s)^2 ? 0^2) /2 40 * a = 193.21 cm^2/s^2 / 2 a = 96.61 cm^2/s^2 / (40 cm) a = 2.415cm/s^2 ** If vf^2 = v0^2 + 2 a `ds then a `ds = (vf^2 ? v0^2) / 2, as you say, so a = (vf^2 ? v0^2) / (2 `ds). This is what you did (good job) but be careful to state it this way. ** Then use the equation 'ds = v0 'dt + .5 a 'dt^2 ** I wouldn?t use this equation to solve for `dt, since the equation is quadratic in `dt. At this stage you know all variables except `dt. The equation vf = v0 + a `dt would be much simpler. Or you could just figure average velocity and divide into displacement. Either way would be fine. ** 40= +.5 2.415cm/s/s dt^2 dt = sqrt 40/ ** You could have used the fact that vAve = (13.9 cm/s + 0) / 2 = 6.95 cm/s to obtain `dt: Since vAve = `ds / `dt, we have `dt = `ds / vAve = 40 cm / (6.95 cm/s) = 5.8 sec, approx. **
......!!!!!!!!...................................
RESPONSE -->