If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
006. Using equations with uniformly accelerated motion.
Question: `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.
Your solution:
Confidence rating:
Given Solution:
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.
Self-critique (if necessary):
Self-critique rating:
Question: `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?
Your solution:
Confidence rating:
Given Solution:
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.
STUDENT QUESTION (about reasoning vs. using the equation)
I understand but the steps taken to get
to the acceleration were the steps of the equation?????
INSTRUCTOR RESPONSE
The steps outlined here are the steps we could use to derive the equation. However it's possible to use the equation blindly, without understanding the reasoning behind it. In fact this is how most student use the equation, if not asked questions of this nature about the reasoning.
So, this question asks for the reasoning.
The first statement in the given solution is
'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.'
When using the equation you never explicitly find or reason out
the change in velocity, though of course the change in velocity is there
in the equation, represented by the term a * `dt. In other words, you do find
it, but you can use the equation without ever recognizing that
you have done so.
Similarly the step a = (30 m/s - 10 m/s) / 15 s in your equation-based solution
does correctly divide the change in velocity by the time interval, but you can
use the equation to do this without ever recognizing that you have done so.
The direct reasoning solution never mentions or uses the
equation, though of course direct reasoning can be used to derive
the equation.
This should help illustrate the difference between direct reasoning and using an
equation. Both skills are important.
Self-critique (if necessary):
Self-critique rating:
Question: `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.
Your solution:
Confidence rating:
Given Solution:
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.
Self-critique (if necessary):
Self-critique rating:
Question: `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.
Your solution:
Confidence rating:
Given Solution:
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.
Self-critique (if necessary):
Self-critique rating:
Question: `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.
Your solution:
Confidence rating:
Given Solution:
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.
Self-critique (if necessary):
Self-critique rating:
Question: `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.
Your solution:
Confidence rating:
Given Solution:
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.
Self-critique (if necessary):
Self-critique rating:
Question: `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.
Your solution:
Confidence rating:
Given Solution:
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.).
Self-critique (if necessary):
Self-critique rating:
Question: `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.
Your solution:
Confidence rating:
Given Solution:
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.
Self-critique (if necessary):
Self-critique rating:
Question: `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?
Your solution:
Confidence rating:
Given Solution:
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.
STUDENT QUESTION
I understood the negative velocity but was unsure how to
explain the rest. I am still rather confused by the last paragraph, expecially
where it says that it is possible for velocity to be in one direction and
acceleration in the other.
INSTRUCTOR RESPONSE
If you speed up the acceleration is in the direction of
motion.
If you slow down the acceleration is opposite the direction of motion.
To speed up a wagon you can get behind it and push in the direction of its
motion, giving it an acceleration in its direction of motion.
To slow it down you can get in front of it and push it against its direction of
motion (not advisable if it's a big wagon; think of stopping a child in a small
wagon), giving it an acceleration in the direction opposite its motion.
Self-critique (if necessary):
Self-critique rating:
You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook).
If the course is not specified for a problem, then students in all physics courses should do that problem.
Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics.
General College Physics students need not do questions or problems specified for University Physics.
University Physics students should do all questions and problems.
Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.)
General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students.
You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook.
Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it.
Questions related to q_a_
1. Using the equations of uniformly accelerated motion, find the initial velocity of a coasting automobile which accelerates at a uniform rate of 2 m/s^2 for 12 seconds, ending up with a velocity of 30 m/s. Also determine how far the automobile coasts during this interval.
Then explain how to reason this problem out using the definitions of velocity and acceleration, without reference to the equations of uniformly accelerated motion.
2. Using the equations of uniformly accelerated motion, find the initial velocity of a coasting automobile which accelerates at a uniform rate of 2 m/s^2 as it travels 125 meters, ending up with a velocity of 30 m/s. Also determine how long it takes the automobile to coast this distance.
Using the initial velocity you found, the given final velocity, and the time interval you found, use direct reasoning, in terms of the definitions of velocity and acceleration, to find the acceleration and displacement. Verify that your results agree with the given acceleration and displacement.
3. Show how to solve the equation `ds = v0 `dt + .5 a `dt^2 for v0.
4. The fourth equation of uniformly accelerated motion is vf^2 = v0^2 + 2 a `ds. Suppose that a = .5 m/s^2, `ds = 9 m and v0 = 4 m/s. Solve for vf. (recall that there are two solutions to the equation x^2 = c; the solutions are x = sqrt(c) and x = - sqrt(c); for example if the equation is x^2 = 9 then both x = 3 and x = -3 are clearly solutions)
As you see, there are two solutions for vf, one with a positive value and one with a negative value.
Using the positive value of vf:
Repeat the preceding, using the negative value of vf.
5. An automobile coasts at high speed down a hill, speeding up as it goes. It encounters significant air resistance, which causes it to speed up less quickly than it would in the absence of air resistance.
If the positive direction is down the hill, then
If the positive direction is up the hill, then
Questions related to Class Notes
1. A ball rolls a variety of distances down an incline, from rest, and from a variety of starting positions. The corresponding intervals are timed using the TIMER program. All time intervals last at least 1 second but none exceeds 2 seconds. The resulting acceleration values range from 43.2 cm/s^2 to 49.6 cm/s^2.
Questions/problems for General College Physics Students
1. My garden is 30 meters long and 15 meters wide. To protect against drought I want to build a pond containing a month's supply of water, equivalent to about 3 inches of rainfall. If the pond is 10 meters long and 6 meters wide, how deep does it have to be?
2. At 1200 liters/day per family how much would the level of a lake with surface area 50 km^2 fall in a year if supplying town of population 40000?
3. Estimate the number of gallons of gasoline used by all drivers in the United States in a month. Base your estimate on reasonable assumptions. State your assumptions and explain how they lead to your conclusion.
Questions/problems for University Physics Students
1. A sailor sails 2 km due east, then 3.5 km toward the southeast, then through an unknown displacement. He ends up 5.8 km to the east of his starting point.
2. We have two vectors, one of magnitude 3.6 directed at an angle of +70 deg relative to the positive x axis, and another of magnitude 2.4 directed at +210 deg from the positive x axis.