course Phy121
......!!!!!!!!...................................
14:40:46 `q001. Note that there are 9 questions in this assignment. If the acceleration of an object is uniform, then the following statements apply: 1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Suppose that an object increases its velocity at a uniform rate, from an initial velocity of 5 m/s to a final velocity of 25 m/s during a time interval of 4 seconds. By how much does the velocity of the object change? What is the average acceleration of the object? What is the average velocity of the object?
......!!!!!!!!...................................
RESPONSE --> 25-5=20 25-5=20 =20/4= 5 25+5=30 30/2=15 confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:41:36 The velocity of the object changes from 5 meters/second to 25 meters/second so the change in velocity is 20 meters/second. The average acceleration is therefore (20 meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the average of its initial and final velocities, as asserted above, and is therefore equal to (5 meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers are averaged by adding them and dividing by 2).
......!!!!!!!!...................................
RESPONSE --> i understand self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:43:28 `q002. How far does the object of the preceding problem travel in the 4 seconds?
......!!!!!!!!...................................
RESPONSE --> x=.5(vi+vf)t == x=60 confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:43:39 The displacement `ds of the object is the product vAve `dt of its average velocity and the time interval, so this object travels 15 m/s * 4 s = 60 meters during the 4-second time interval.
......!!!!!!!!...................................
RESPONSE --> cool self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:46:33 `q003. Explain in commonsense terms how we determine the acceleration and distance traveled if we know the initial velocity v0, and final velocity vf and the time interval `dt.
......!!!!!!!!...................................
RESPONSE --> take the difference of velocities and divide them by the time to find acceleration. to find distance you would take difference in velocities and multiply by time. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:46:58 In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:50:07 `q004. Symbolize the situation by first giving the expression for the acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf and `dt.
......!!!!!!!!...................................
RESPONSE --> vf=vi+a*t (vf+vi)/2== vAve confidence assessment:
.................................................
......!!!!!!!!...................................
14:51:01 The acceleration is equal to the change in velocity divided by the time interval; since the change in velocity is vf - v0 we see that the acceleration is a = ( vf - v0 ) / `dt.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
14:51:26 `q005. The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2.
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:51:52 When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:59:45 `q006. This situation is identical to the previous, and the conditions implied by uniformly accelerated motion are repeated here for your review: If the acceleration of an object is uniform, then the following statements apply: 1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Describe a graph of velocity vs. clock time, assuming that the initial velocity occurs at clock time t = 0. At what clock time is the final velocity then attained? What are the coordinates of the point on the graph corresponding to the initial velocity (hint: the t coordinate is 0, as specified here; what is the v coordinate at this clock time? i.e., what is the velocity when t = 0?). What are the coordinates of the point corresponding to the final velocity?
......!!!!!!!!...................................
RESPONSE --> the graph will increase as the time and velocity increase. if the clock time ends at 4 sec then the final velocity attained vi =0 at t=0 t=4 vf=20 confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:00:18 The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5 m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4 seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the corresponding graph point is ( 4 s, 25 m/s).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 0
.................................................
......!!!!!!!!...................................
15:01:20 `q007. Is the graph increasing, decreasing or level between the two points, and if increasing or decreasing is the increase or decrease at a constant, increasing or decreasing rate?
......!!!!!!!!...................................
RESPONSE --> the graph is increasing between the two points and the increase is at a constant rate confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:01:40 Since the acceleration is uniform, the graph is a straight line. The graph therefore increases at a constant rate from the point (0, 5 m/s) to the point (4 s, 25 m/s).
......!!!!!!!!...................................
RESPONSE --> yea self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:04:30 `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope?
......!!!!!!!!...................................
RESPONSE --> (y1-y2)/(x1-x2)= m 20-0)/(4-0)= 5 confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:04:54 The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object.
......!!!!!!!!...................................
RESPONSE --> okk self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:13:42 The two altitudes are 5 meters/second and 25 meters/second, and their average is 15 meters/second. This represents the average velocity of the object on the time interval. The area of the trapezoid is equal to the product of the average altitude and the base, which is 15 m/s * 4 s = 60 meters. This represents the product of the average velocity and the time interval, which is the displacement during the time interval.
......!!!!!!!!...................................
RESPONSE --> i hit the wrong button but i understand the problem self critique assessment: 3
.................................................
course Phy121 іzOvIعassignment #005
......!!!!!!!!...................................
15:40:28 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> vf=vi+a*t x=vi*t+.5at^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:40:42 ** 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**
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
15:41:50 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
......!!!!!!!!...................................
RESPONSE --> vo goes to vf in amount of dt confidence assessment:
.................................................
......!!!!!!!!...................................
15:42:26 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
15:43:58 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.
......!!!!!!!!...................................
RESPONSE --> i need to know the distance i would plug it in the v=(x/t) confidence assessment:
.................................................
......!!!!!!!!...................................
15:44:18 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
15:48:02 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?
......!!!!!!!!...................................
RESPONSE --> 70 beats a min *60 an hour =4200 beats an hour = 2759400000 beats in a life time confidence assessment:
.................................................
......!!!!!!!!...................................
15:48:23 ** 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. **
......!!!!!!!!...................................
RESPONSE --> right on self critique assessment:
.................................................
......!!!!!!!!...................................
15:49:43 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?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
15:50:05 ** For the given vectors we have dot product =-2 * 2 + 6 * (-3) = -22 magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40) magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13) Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ] = arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
15:50:46 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> i learned alot self critique assessment:
.................................................
......!!!!!!!!...................................
15:50:56 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................