qa 6

#$&*

course phy 201

006. Using equations with uniformly accelerated motion.

*********************************************

Question: `q001. Note that there are 12 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

30 = 10 + a x 15

20 = 15a subtract 10

a = 1.33 m/s/s divide by 15

@&

Good, but you aren't using units in your calculation. You are putting units on your final result, but you aren't including them throughout your solutions and don't appear to be doing the unit calculations.

*@

confidence rating #$&*:32; 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Plug and chug the values. There is no need to create another equation.

@&

There is every reason to be able to work this out without using the equations.

The equations can be used to get the answer without understanding any of the relationships, definitions, or meanings. Without this understanding you are unlikely to understand the physics.

*@

confidence rating #$&*:32; 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

80 = (6 + v0) / 2 x 10

2/10 x 80 = 6 + v0 multiply both sides by 2/10

16 = 6 + v0 subtract 6

v0 = 10 m/s

confidence rating #$&*:32; 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Since 80m/ 10 s = 8m/s that is the halfway point between 6m/s and 10m/s. Knowing that would allow us to know the initial is 10m/s.

confidence rating #$&*:32; 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

80 = v0 x 10 + .5(-2) x 10^2

80 = 10v0 - 100

180 = 10v0

v0 = 18m/s

confidence rating #$&*:32; 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The final velocity would -2m/s since 18-20 (from -2 x 10) = -2. From this we find the average velocity of 8m/s and multiply this by ‘dt to get 80m.

confidence rating #$&*:32; 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

20^2 = v0^2 + 2 x 2 x 80

400 = v0^2 + 320

80 = v0^2

square root = 8.9 m/s

confidence rating #$&*:32; 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

20 - 8.9 / 2 = 5.55

20 + 8.9 = 28.9

80 / 5.55 = 14.4

28.9 / 14.4 = 2m/s/s

confidence rating #$&*:32; 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

In starts South

The speed increases.

Eventually the acceleration of 2m/s/s creates a positive velocity and moves north.

confidence rating #$&*:32; 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

*********************************************

Question: `q010. An object speeds up from 10 m/s to 20 m/s, accelerating uniformly and traveling 60 meters during this interval.

Specify which of the quantities v_0, v_f, aAve, `ds and `dt are given, and specify the value of each.

Specify which of the four equations of uniformly accelerated motion include the given three quantities. There is at least one such equation, and there might be two.

For each of the equations you specified, identify the quantity for which the value is not given. Then symbolically solve the equation for each of these quantities, showing the steps of your algebra.

Substitute the three given quantities into your solution, and simplify.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

V0 = 10m/s

Vf = 20m/s

‘ds = 60m

aAve = (20-10)/ 2 = 5m/s/s

‘dt = 60 / (20+10) / 2 = 4s

@&

You haven't indicated which of the four equations of uniformly accelerated motion you are using.

Your acceleration is incorrect, as you would be able to tell by using the units. (20 m/s - 10 m/s) / 2 = 5 m/s, not 5 m/s/s.

Your calculation for `dt does appear to follow correctly from one of the equations, but you were asked to show the equation and solve it symbolically. I suspect you did this, from the form of your solution, but you haven't shown it.

You need to revise you solution to this problem accordingly.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q011. An cart initially moving at 10 cm/s travels 40 cm while accelerating at 5 m/s^2. Using the equations of uniformly accelerated motion determine the time required and the the cart's final velocity.

Hint: You will need to start out with either the third or the fourth equation of uniformly accelerated motion. You are advised that it's easier to start out with the fourth equation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

vf = v0 + a * `dt.

Vf = 10 + 5 x 4

@&

You aren't given `dt. Without first finding `dt you can't use it in this equation.

*@

Vf = 30cm/s

@&

Accelerating fro 10 cm/s to 30 cm/s at 5 m/s^2 would require 4 seconds.

However the average velocity would be 20 cm/s, and in 4 seconds the object would travel 80 cm.

So your solution is inconsistent with the given information.

This needs to be revised.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `q012. An object starts at position x = +10 cm with a velocity of +5 cm/s, and accelerates uniformly, ending up at position x = -30 cm after a time interval of 8 seconds. What is its velocity at this point, and what was its acceleration during this interval?

Principles of Physics students should not spend over 5 minutes on this problem, General College Physics students should not spend over 10 minutes.

University Physics students are expected to be able to solve this problem, but if it hasn't been solved within 15 minutes, should submit their best thinking and await the instructor's notes.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

40/8 = -5cm/s

@&

40/8 is 5, not -5.

However the displacement is -40 cm, so the calculation

-40 cm / (8 s) = -5 cm/s

does give you the average velocity on the interval.

However that isn't the velocity at the final point.

You also need to calculate the acceleration.

*@

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

@&

You're doing a number of things right, but you do not appear to be using units throughout your calculations and it doesn't appear that you are consistently using the equations of uniformly accelerated motion when asked to do so.

You're in a position to nail these things down. See if you can do so in a revision.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

&#

*@