course phy 201
......!!!!!!!!...................................
14:53:40 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 --> The first distance is 65.3m and then second one is 24.5m. The acceleartion for the first is 0.61m/s^2 and the second one is -1.9m/s^2. You could use a velocity vs time graph to obtain this information by have the time on the x axis and the velocity on the y axis. Then for each of the time intervals that you wanted to obtain. 0-9s and 9-13 you would put a line vertically for it till it touched the line that was graphed. Then you would start with 0s and see where the velocity which is 4.5m/s was at 0 seconds. then you would move over to 9s and see where the vertical line toughed the graphed line and see what velocity which is 10m/s. You would then find the average velocities of the two and then determine that it was 9s interval . You would then multiply the two to find the displacement which is 7.25m/s * 9s = 65.25m.. you would then do this again for the other interval. To find the acceleration you would do the same intervals but instead of finding the average velocity you would find the change in velocity which is when you subtract the initial from the final 10m/s - 4.5m/s = 5.5m/s and then divide it by the time interval which is 9s ==== 5.5m/s / 9s = .61m/s^2. then you would do this again for the other interval
.................................................
......!!!!!!!!...................................
14:55:00 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 --> right.
.................................................
......!!!!!!!!...................................
14:57:38 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 --> This information does not supprt the hypothesis because the last slope is the largest and it is not the fastest. The larger the slope is the fasters the time should be. This could be due to human error but in general when you increase the slope the time interval for the slope should also increase.
.................................................
......!!!!!!!!...................................
15:00:29 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 --> If you calculate the accelerations and then plot them you would find that the do not lie in a linear line that the last ramp does not go with the information.
.................................................
......!!!!!!!!...................................
15:19:30 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 --> First you have to find the velocity average vAve= 'ds/'dt vAve = 50cm/3.8s = 13.16cm/s now we need to find the initial velocity vAve = (vf+v0)/2 2*(vAve-vf) = v0 2(*13.16cm/s - 6.31579cm/s) = 13.69cm/s the change in velocity is vf-v0 6.31579-13.69 =-7.37cm/s then the accerleration of the ramp is change in velocity / time interval -7.37 cm/s / 3.8s = -1.93cm/s^2
.................................................
......!!!!!!!!...................................
15:23:01 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 --> to find the initial velocit I had the right equation but I subtracted the vf from the average velocity before i multiplied by the two.. the rest I did right just i still had the wrong initial so it changed my information.
.................................................
......!!!!!!!!...................................
15:27:56 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 --> The initial falling velocity is 0cm/s so we use the equation x=0.5at^2 a = 980cm/s^2 for gravity.... so we then solve the equation for the time interval t = srt(2*x/a) = srt(2*95cm/980cm/s^2) = 0.4403s You then can use this information to find the displacement. The initial horizontal velocity was 18cm/s and you would multiply this by the time interval you just obtained 0.4403s 18cm/s * 0.4403s = 7.93cm
.................................................
......!!!!!!!!...................................
15:28:08 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 --> right
.................................................
......!!!!!!!!...................................
15:35:04 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 --> The initial velocity is 0cm/s because it starts from rest the final velocity is given which is 13.9cm/s the displacement is 40cm Then to find the time interval you would first set up the equation vAve = 'ds/'dt so 'dt = ds/vAve vAve also = (vf+v0)/2 'dt = 'ds/ ((vf+v0)/2) 40cm/((13.9cm/s +0cm/s)/2) 'dt = 5.76s acceleration is 'dv/ 'dt (vf-vo)/'dt (13.9cm/s - 0cm/s)/5.76s = 2.4cm/s^2
.................................................
......!!!!!!!!...................................
15:35:42 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 --> right
.................................................
"