PHY 201 QA06

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course PHY 201

9:15 PM on 2/11/13

006.  Using equations with uniformly accelerated motion. 

 

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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.

 

 

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Your solution: 

a = (vf - v0) / 'dt; vf = 30 m/s, v0 = 10 m/s, 'dt = 15 s

a = (30 m/s - 10 m/s) / 15 s = 20 m/s / 15 s = 1.33 m/s^2 

 

 

confidence rating #$&*:#8232; 

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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.

 

 

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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?

 

 

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Your solution: 

 

 Find the difference between the final and starting velocities and divide it by the time interval. The same steps as using the equation, however, I can do this by not having to write the equation to solve for acceleration.

 

 

confidence rating #$&*:8232; 

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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.

 

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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.

 

 

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Your solution: 

 

v0 = ('ds / 'dt) * 2 - vf; 'ds = 80 m, 'dt = 10 s, vf = 6 m/s

v0 = (80 m / 10 s) * 2 - 6 m/s = 10 m/s

 

 

confidence rating #$&*:#8232; 

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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.

 

 

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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.

 

 

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Your solution: 

 

 vAve = 'ds / 'dt = 80 m / 10 s = 8 m/s.

vAve is somewhere between v0 and vf. If we know that vf = 6 m/s and vAve = 8 m/s, then v0 must be around 10 m/s because there is a 2 m/s difference between vAve and vf.

 

 

confidence rating #$&*:8232; 

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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.

 

 

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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.

 

 

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Your solution: 

 

v0 = ('ds - 0.5 a 'dt^2) / /dt

v0 = (80 m - 0.5(-2 m/s^2)(10^2 s) / 10 s = 18 m/s

 

 

confidence rating #$&*:8232; 

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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.

 

 

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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.

 

 

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Your solution: 

 

 a = 'dv / 'dt

'dv = a * 'dt = -2 m/s^2 * 10 s = - 20 m/s

'dv = 18 - 20 m/s = - 2 m/s

vAve = (18 + (-2 m/s) / 2 = 8 m/s

8 m/s traveled for 10 seconds would be 8 m/s * 10 s = 80 m.

 

 

confidence rating #$&*:8232; 

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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.

 

 

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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.

 

 

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Your solution: 

 

v0^2 = vf^2 - 2 a 'ds

v0 = sqrt((20 m/s)2 - 2(2 m/s^2)(80 m)) = 8.9 m/s

 

 

confidence rating #$&*:8232; 

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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.). 

 

 

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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.

 

 

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Your solution: 

 

vAve = (-8.9 m/s + 20 m/s) / 2 = 5.5 m/s

'dt = 'ds / vAve = 80 m / 5.5 m/s = 14.5 s

a = 'dv / 'dt = (20 - (-8.9) / 14.5 = 28.9 m/s / 14. 5 s = 1.99 m/s^s = 2.0 m/s^2

 

 

confidence rating #$&*:8232; 

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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.

 

 

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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?

 

 

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Your solution: 

 

 

confidence rating #$&*:32; 

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Given Solution: 

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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.

 

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Your solution: 

 

 v0 = 10 m/s

vf = 20 m/s

'ds = 60 m

vAve = (10 m/s + 20 m/s) / 2 = 15 m/s

'dt = 'ds / vAve = 60 m / 15 m/s = 4 s

aAve = 'dv / 'dt = (20 m/s - 10 m/s) / 4 s = 2.5 m/s^2

@&

Your quantities are correctly identified, and you've done an excellent job of reasoning out the remaining quantities. In other words you've used the definitions of average velocity and acceleration to obtain your results.

However you want to relate this to the four equations of motion, which are derived from the definitions.

The three quantities v0, vf, `ds are contained in the first equation

`ds = (vf + v0) / 2 * `dt

as well as the fourth

vf^2 = v0^2 + 2 a `ds.

The first equation also contained `dt, so you should solve that equation for `dt and substitute the three known quantities. You should get the same result for `dt as the one you reasoned out.

You should also solve the fourth equation for the variable a and compare with your reasoned-out result.

It would be a good idea to then send me a copy of the problem, your original solution, my note and your added solutions of the equations.

*@

 

 

confidence rating #$&*:

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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.

 

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Your solution: 

v0 = 10 cm/s

'ds = 40 cm

aAve = 5 m/s^2

'ds = v0 'dt + 0.5 a 'dt^2, find 'dt

I am having a brain fart on how to solve for 'dt using this equation. I am sure its very basic, but can't think of it right now, just a little rusty.

@&

That's the equation you want to avoid having to solve for `dt.

If you have v0, a and `ds, you have an alternative in the fourth equation

vf^2 = v0^2 + 2 a `ds.

You can solve this for vf, then use either of the first two equations to find `dt.

I suggest you submit this as well.

*@

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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.

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Your solution: 

 s1 = 10 cm

v0 = 5 cm/s

s2 = -30 cm

'dt = 8 s

vAve = 'ds / 'dt = (-30 cm - 10 cm) / 8 s = -5 cm/s

vAve = (v0 + vf) / 2 => 2(vAve) = v0 + vf => vf = 2(vAve) - v0 = 2(-5 cm/s) - 5 cm/s = -15 cm/s

aAve = 'dv / 'dt = (vf - v0) / 'dt = (-15 cm/s - 5 cm/s) / 8 s = -20 cm/s / 8 s = -2.5 cm/s^2

 

 

confidence rating #$&*:

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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.

 

STUDENT COMMENTS

 

Made relative sense, but still unsure in doubt to my answer that the object was increasing after it was moving in the right direction. Being negative, would it have started towards the south at that acceleration, and moving north would it have diminished its negativity? It seems this naturally.

 

INSTRUCTOR RESPONSE

 

Accelerating to the north (not 'moving' to the north; it does end up moving to the north, but the object starts out moving to the south), the speed in the southward direction would have diminished, as you say. 
Eventually it comes to rest, just for an instant, somewhere south of its starting point. Then the northward acceleration will give it an increasing northward velocity.

 

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&#Good responses. See my notes and let me know if you have questions. &#