#$&* course phy 201 9/26 2:20 am Be sure to read the information on testing. All students should have received emails during the registration period, with an attachment containing links to information, including a clearly indicated link to testing information. That link is provided here, for students who might have missed it.http://vhcc2.vhcc.edu/dsmith/geninfo/testing_information.htm
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Given Solution: 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). STUDENT RESPONSE: 1.8cm 2.722297s4.2cm 2.518101s 6.7cm 2.6606s 'ds=78. These results show that the smallest slope the time to coast is the slowestThe 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Problem BA 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Difference in time: 3.8 seconds; displacement: 50cm; vF: 6.31579 cm/scm/s; average velocity: 13.1579 cm/s; difference from vAve to vF: 6.84211 cm/s; v0: 20 cm/s; Change in velocity: -13.6842 cm/s; average acceleration: -3.6 cm/s^2; confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Problem CA 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vF^2 = v0^2 + 2*a*ds; vF^2 = 0^2 + * 95; vF = 43.15 m/s; 431.5 cm/s; average velocity: 215.75 cm/s; time passed: 0.44 seconds; distance travelled: 18 cm/s * 0.44 seconds: 7.92 centimeters;
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Given Solution: STUDENT RESPONSE: For the vertical motion: '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 cm/s^2)) = sqrt (.194 s^2)So, 'dt = sqrt (.194 s^2)'dt=.44s this is correct but see the note below Next take the 18cm/s and multiply by the time duration .44s = 7.92cm/s Note: Generally this isn't the equation we would use with the given information to solve vertical motion, since v0 isn't always zero. The fourth equation can be solved for vf, obtaining vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (0 cm/s)^2 + 2 * 980 cm/s^2 * 95 cm) - +- sqrt( 190 000 cm^2 / s^2) = +- 430 cm^2 / s^2. Knowing that the final velocity in this situation is in the same direction as the displacement, we conclude that vf = + 430 cm/s and discard the negative solution. From the initial and final velocities we find that the average velocity is about 215 cm/s, so that the time interval is about `dt = 95 cm / (215 cm/s) = .44 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. The basic principle is that vertical and horizontal motion are independent. Vertical quantities don't appear in the analysis of horizontal motion, and horizontal quantities do not appear in the analysis of vertical motion. Only the time interval is the same for the vertical and horizontal motion. For an ideal projectile we also assume that the only force present is the gravitational force, which acts only in the vertical direction. We regard air resistance and other forces that might actually be present, acting in the horizontal direction, as negligible. So the horizontal acceleration is zero, and horizontal motion is at constant velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK Question: Problem DWhat 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V0: 0 cm/s; vF: 13.9 cm/s; ds: 40 cm; dt: 2.877 seconds; change in velocity: 13.9 cm/s; average acceleration: 4.83 cm/s^2; confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 / 2a = 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 that you state it this way. ** Then use the equation 'ds = v0 'dt + .5 a 'dt^2 ** I don't recommend that you 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. ** ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK