course Phy 231 ߰J Uassignment #005
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18:42:57 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
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RESPONSE --> We calulate 'dv = a *'dt. Then we calculate vf = v0 + 'dv. We average v0 and vf to find average velocity. We calculate 'ds by multiplying vAve and 'dt.
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18:43:10 ** To find final velocity from the given quantities initial velocity, acceleration and `dt: Multiply `dt by accel to get `dv. Then add change in velocity `dv to init vel , and you have the final velocity**
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18:57:52 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> In the first level are the given quanitities v0, vf and 'dt In the second level is the quantity vAve with lines from v0 and vf. The third level has the quanitity 'ds, with lines from 'dt and vAve. The final level has the quantity a with lines from vAve and 'dt.
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19:06:44 ** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantites at a time. How we get each quantity may also be included. From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve. Then from vAve and `dt we get `ds, with the accompanying lines indicating from vAve and `dt to `ds, while from `dv and `dt we get acceleration, indicated similarly. **
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RESPONSE --> Of course, the change in velocity divided by the interval over which the change took place gives us the acceleration. Though the average velocity is calculated from the initial and final velocities, it is not unique to those particular inital and final velocities. That's why I need to use the change in velocity to calculate the acceleration.
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19:26:49 Principles of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly.
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RESPONSE --> Running 16 hours a day, a runner can travel 3961 km in approximately 23 days at an average speed of 10 km/hr. I found an estimate in km of the distance between New York City and Los Angeles. I divided that by 10 to obtain 396.1. Dividing that by 16 leaves our runner eight hours of rest each day and gives us the number of days she has to run.
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19:27:39 It is about 3000 miles from coast to coast. A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately. At 10 km / hr, the time required would be 5000 km / (10 km / hr) = 500 km / (km/hr) = 500 km * (hr / km) = 500 (km / km) * hr = 500 hr.
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RESPONSE --> The estimate I found was labelled in km, but must have been measured in miles
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19:35:49 All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result?
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RESPONSE --> In an average lifetime of 80 years, a human heart will beat an estimated 2,943,360,000 times. This assumes an average heart rate of 70 beats per minute and uses 525,600 as the number of minutes in a year: 70 bpm * 525600 min/yr * 80 yr = 2,943,360,000 beats per year.
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19:36:04 ** Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion, approximately. **
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19:44:16 University Physics Students Only: Problem 1.52 (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?
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RESPONSE --> The angle between these vectors is 165 degrees: magnitudes - vector A = sqrt(40) vector B = sqrt(13) Scalar product of vector A and vector B = -22. A negative scalar product indicates an angle between 90 and 180 degrees. arccos (-22 / sqrt(529)) = 164.8 degrees
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19:44:36 Add comments on any surprises or insights you experienced as a result of this assignment.
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19:44:41 ** I had to get a little help from a friend on vectors, but now I think I understand them. They are not as difficult to deal with as I thought. **
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yaHɮ Student Name: assignment #006
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19:53:35 `q001. Note that there are 9 questions in this assignment. Using the equation vf = v0 + a * `dt determine the acceleration of an object whose velocity increases at a uniform rate from 10 m/s to 30 m/s in 15 seconds. Begin by solving the equation for the acceleration a, then 'plug in' your initial and final velocities. Describe your work step y step.
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RESPONSE --> First, I subtracted v0 from each side to obtain vf - v0 = a * 'dt. Then I divided each side by 'dt to obtain (vf - v0) / 'dt = a Solving this with given values gives 1.33 m / sec^2
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19:53:47 The equation vf = v0 + a * `dt is solved for a by first adding -v0 to both sides to obtain vf - v0 = v0 + a * `dt - v0, which simplifies to vf - v0 = a * `dt. Both sides are then divided by `dt to obtain (vf - v0) / `dt = a. Reversing left-and right-hand sides we obtain the formula a = (vf - v0) / `dt. We then plug in our given values of initial and final velocities and the time interval. Since velocity increases from 10 m/s to 30 m/s, initial velocity is v0 = 10 m/s and final velocity is vf = 30 m/s. The time interval `dt is 15 seconds, so we have a = (30 m/s - 10 m/s) / (15 s) = 20 m/s / (15 s) = 1.33.. m/s^2.
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19:56:27 `q002. It wasn't necessary to use a equation to solve this problem. How could this problem had been reasoned out without the use of an equation?
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RESPONSE --> We can see that the change in velocity is the difference between the initial and final velocities, and divide this by the interval to obtain the change in velocity with respect to time. This is the definition of acceleration.
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19:56:46 Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second. We would then divided change in velocity by the time interval to get 20 meters/second / (15 sec) = 1.33 m/s^2.
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20:05:33 `q003. Use the equation `ds = (vf + v0) / 2 * `dt to determine the initial velocity of an object which accelerates uniformly through a distance of 80 meters in 10 seconds, ending up at a velocity of 6 meters / sec. begin by solving the equation for the desired quantity. Show every step of your solution.
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RESPONSE --> First, multiplying both sides by 2 / 'dt gives (2 * 'ds) / 'dt = vf + v0. Subtracting vf from each sides gives (2 * 'ds) / 'dt - vf = v0. Substituting the given values and solving gives v0 = 10 m/sec
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20:05:41 We begin by solving the equation for v0. Starting with `ds = (vf + v0) / 2 * `dt, we can first multiply both sides of the equation by 2 / `dt, which gives us `ds * 2 / `dt = (vf + v0) / 2 * `dt * 2 / `dt. The right-hand side can be rearranged to give (vf + v0) * `dt / `dt * 2 / 2; since `dt / `dt = 1 and 2 / 2 = 1 the right-hand side becomes just vf + v0. The equation therefore becomes 2 * `ds / `dt = vf + v0. Adding -vf to both sides we obtain v0 = 2 * `ds / `dt - vf. We now plug in `ds = 80 meters, `dt = 10 sec and vf = 6 m/s to get v0 = 2 * 80 meters / 10 sec - 6 m/s = 160 meters / 10 sec - 6 m/s = 16 m/s - 6 m/s = 10 m/s.
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20:17:14 `q004. We can reconcile the above solution with straightforward reasoning. How could the initial velocity have been reasoned out from the given information without the use of an equation? Hint: two of the quantities given in the problem can be combined to give another important quantity, which can then be combined with the third given quantity to reason out the final velocity.
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RESPONSE --> We can combine 'ds and 'dt to obtain average velocity of 80 cm/sec. Average velocity can also be expressed as (v0 + vf) / 2. Setting both expressions equal to each other and substituting the known value of vf, we can then multiply both sides by 2 to obtain: 16 m/sec = v0 + 6 m/sec. This is solved for v0 = 10 m/sec
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20:17:30 The average velocity of the object is the average rate at which its position changes, which is equal to the 80 meters change in position divided by the 10 s change in clock time, or 80 meters / 10 sec = 8 meters / sec. Since the 8 m/s average velocity is equal to the average of the unknown initial velocity and the 6 m/s final velocity, we ask what quantity when average with 6 m/s will give us 8 m/s. Knowing that the average must be halfway between the two numbers being averaged, we see that the initial velocity must be 10 m/s. That is, 8 m/s is halfway between 6 m/s and 10 m/s.
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20:44:43 `q005. Using the equation `ds = v0 `dt + .5 a `dt^2 determine the initial velocity of an object which accelerates uniformly at -2 m/s^2, starting at some unknown velocity, and is displaced 80 meters in 10 seconds. Begin by solving the equation for the unknown quantity and show every step.
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RESPONSE --> The equation solved for v0 : ('ds / 'dt) - (a 'dt/ 2 )= v0. First both sides are divided by zero, then a/2 is subtracted from each side. Substituting given values, the equation is solved for v0 = 14 cm/sec
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21:12:49 The unknown quantity is the initial velocity v0. To solve for v0 we start with `ds = v0 `dt + .5 a `dt^2. We first add -.5 a `dt^2 to both sides to obtain `ds - .5 a `dt^2 = v0 `dt. We then divide both sides by `dt to obtain (`ds - .5 a `dt^2) / `dt = v0. Then we substitute the given displacement `ds = 80 meters, acceleration a = -2 m/s^2 and time interval `dt = 10 seconds to obtain v0 = [ 80 meters - .5 * (-2 m/s^2) * (10 sec)^2 ] / (10 sec) = [ 80 meters - .5 * (-2 m/s^2) * 100 s^2 ] / (10 sec) = [ 80 m - (-100 m) ] / (10 sec) = 180 m / (10 s) = 18 m/s.
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RESPONSE --> It was my arithmetic, not my equation that was in error.
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21:32:34 `q006. Check the consistency of this result by verifying, by direct reasoning rather than equations, that an object whose initial velocity is 18 m/s and which accelerates for 10 seconds at an acceleration of -2 m/s^2 does indeed experience a displacement of 80 meters.
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RESPONSE --> If an object slows by 2 m/ sec every second, why doesn't it slow by 20 m / sec in 10 seconds?
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21:37:14 The change in the velocity of the object will be -2 m/s^2 * 10 s = -20 m/s. The object will therefore have a final velocity of 18 m/s - 20 m/s = -2 m/s. Its average velocity will be the average (18 m/s + (-2 m/s) ) / 2 = 8 m/s. An object which travels at an average velocity of 8 m/s for 10 sec will travel 80 meters.
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RESPONSE --> I confused the final and average velocity and had an object that traveled -20 m. Correcting that shows me the correct answer.
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21:45:04 `q007. Using the equation vf^2 = v0^2 + 2 a `ds determine the initial velocity of an object which attains a final velocity of 20 meters/second after accelerating uniformly at 2 meters/second^2 through a displacement of 80 meters. Begin by solving the equation for the unknown quantity and show every step.
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RESPONSE --> First, I subtracted 2 a 'ds from both sides to obtain : v0^2 = vf^2 - 2 a 'ds Then I substituted the known values to obtain:vo^2 = (400m^2 / sec^2) - (320m^2/ sec^2) This is solved v0 = 9 m/s
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21:46:40 To solve for the unknown initial velocity v0 we start with vf^2 = v0^2 + 2 a `ds. We first add -2 a `ds to both sides to obtain vf^2 - 2 a `ds = v0^2. We then reverse the right-and left-hand sides and take the square root of both sides, obtaining v0 = +- `sqrt( vf^2 - 2 a `ds). We then substitute the given quantities vf = 20 m/s, `ds = 80 m and a = 3 m/s^2 to obtain v0 = +- `sqrt( (20 m/s)^2 - 2 * 2 m/s^2 * 80 m) = +- `sqrt( 400 m^2 / s^2 - 320 m^2 / s^2) = +- `sqrt(80 m^2 / s^2) = +- 8.9 m/s (approx.).
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RESPONSE --> I rounded my answer to 1 significant digit because acceleration was given as ""2""
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22:14:33 `q008. We can verify that starting at +8.9 m/s an object which attains a final velocity of 20 m/s while displacing 80 meters must accelerate at 2 m/s^2. In this case the average velocity will be ( 8.9 m/s + 20 m/s) / 2 = 14.5 m/s (approx) and the change in velocity will be 20 m/s - 8.9 m/s = 11.1 m/s. At average velocity 14.5 meters/second the time required to displace the 80 meters will be 80 m / (14.5 sec) = 5.5 sec (approx). The velocity change of 11.1 meters/second in 5.5 sec implies an average acceleration of 11.1 m/s / (5.5 sec) = 2 m/s^2 (approx), consistent with our results. Verify that starting at -8.9 m/s the object will also have acceleration 2 meters/second^2 if it ends up at velocity 20 m/s while displacing 80 meters.
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RESPONSE --> I also failed to consider that by taking the square root, we could not be sure whether the initial velocity was positive or negative. An object with v0 = -8.9 m/s and vf = 20 m/s has 'dv = 28.9 m/s. The average of these velocities is 5.55 m/s. We can divide the displacement by the average velocity to find the time interval: 80 m/s / 5.55 m/s = 14.41 sec. A change in velocity of 28.9 m/s in 14.41 seconds corresponds to an acceleration of 2 m/sec. The difference between this situation and having an initial velocity of -8.9 m/sec is the length of time it takes to achieve a velocity of 20 m/s
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22:14:41 In this case the average velocity will be ( -8.9 m/s + 20 m/s) / 2 = 5.5 m/s (approx) and the change in velocity will be 20 m/s - (-8.9 m/s) = 28.9 m/s (approx). At average velocity 5.5 meters/second the time required to displace the 80 meters will be 80 m / (5.5 sec) = 14.5 sec (approx). The velocity change of 28.5 meters/second in 14.5 sec implies an average acceleration of 28.5 m/s / (14.5 sec) = 2 m/s^2 (approx), again consistent with our results.
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22:18:23 `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does?
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RESPONSE --> This particle starts out heading south at 8.9 m/s. It's southward speed slows at a rate of 2 m/sec until it is momentarily not in motion. It immediately begins moving north, accelerating at a rate of 2 m/sec. When it passes a point 80 meters north of its starting point, it is traveling at 20 m/s.
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22:19:11 The object ends up at position +80 meters, which is assumed to be 80 meters to the North. Its initial velocity is -8.9 m/s, with the - sign indicating that the initial velocity is in the direction opposite to the displacement of the object. So the object must start out moving to the South at 8.9 meters/second. Its acceleration is +2 m/s^2, which is in the opposite direction to its initial velocity. This means that the velocity of the object changes by +2 m/s every second. After 1 second the velocity of the object will therefore be -8.9 m/s + 2 m/s = -6.9 m/s. After another second the velocity will be -6.9 m/s + 2 m/s = -4.9 m/s. After another second the velocity will be -2.9 m/s, after another -.9 m/s, and after another -.9 m/s + 2 m/s = +1.1 m/s. The speed of the object must therefore decrease, starting at 8.9 m/s (remember speed is always positive because speed doesn't have direction) and decreasing to 6.9 m/s, then 4.9 m/s, etc. until it reaches 0 for an instant, and then starts increasing again. Since velocities after that instant become positive, the object will therefore start moving to the North immediately after coming to a stop, picking up speed at 2 m/s every second. This will continue until the object has attained a velocity of +20 meters/second and has displaced +80 meters from its initial position.{}{}It is important to understand that it is possible for velocity to be in one direction and acceleration in the other. In this case the initial velocity is negative while the acceleration is positive. If this continues long enough the velocity will reach zero, then will become positive.
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