Be sure to read the information on testing.  All students should have received emails during the registration period, with an attachment containing links to information, including a clearly indicated link to testing information.  That link is provided here, for students who might have missed it.

http://vhcc2.vhcc.edu/dsmith/geninfo/testing_information.htm

You aren't required to submit any of the problems provided below.  However you should work them, at least in your head, and take note of all the hints.

You are invited to submit solutions, questions, etc. on all problems, or on just the ones you aren't sure of.

If you do submit, insert your answers in the usual manner.

Question:  Problem A
A coasting cart has velocities 4.5 m/s, 10 m/s and 2.25 m/s at clock times t = 0 s, 9 s and 13 s. Approximately how far does it move and at what average rate does it accelerate over each of the two time intervals (i.e., 0- 9 seconds and 9 - 13 seconds)?

Show how you could use a graph of velocity vs. time to obtain your results.

Directly reason out your results using the concept of rate.

If the automobile is coasting, and if air resistance is not a significant factor, then over which time interval do you think the average slope of the road is greatest?
 

Your solution: 

Confidence Assessment:

Given Solution: 

STUDENT ANSWER: The average acceleration for the cart from time 0s to 9s, the average velocity is

(4.5+10)m/s/2=7.25 m/s so displacement is

7.25 m/s * 9s =65.25m.

The displacement for the cart between clock time 9s and 13s is (2.25+10)m/s / 2 * 4s=24.5m

The acceleration of the cart between clock time 0 - 9s is

a = `dv / `dt ? (10-4.5)m/s / (9 s) =.6m/s^2.

The acceleration of the cart between clock time 9-13s is

(2.25-10( m/s / (4s) = -1.93m/^2.

You could use a graph of velocity vs. time to obtain the results. By placing the information on a graph of v vs. t you could see the rise and the run between the points. For example looking at 0,4.5 and 9,10 you could see that it had a run of 9 s and a rise of 5.5 m/s and then to find the acceleration you would have

ave accel = slope = 5.5 m/s / (9 s) = .61m/s^2.

If the automobile is coasting then the slope of the road is the greatest when the magnitude of the acceleration is greatest. This occurs on the second slope, where the magnitude of the acceleration is 1.93 m/s^2.

INSTRUCTOR NOTES FOR ALL STUDENTS:

Be sure you use units correctly in every calculation, and be sure you are using the definitions of velocity (rate of change of position, so vAve = `ds / `dt) and acceleration (rate of change of velocity, so a = `dv / `dt).

STUDENT RESPONSE:

1.8cm 2.722297s
4.2cm 2.518101s
6.7cm 2.6606s

'ds=78.

These results show that the smallest slope the time to coast is the slowest
The middle ramp has the fastest time down the ramp
These results show that acceleration is fastest down the middle ramp (4.2cm)

INSTRUCTOR COMMENT ON STUDENT RESPONSE:

** To support the idea that acceleration is linearly dependent on slope you first have to calculate the accelerations, as you did in the Video Experiments.

You then need to plot a graph of acceleration vs. ramp slope and see if it forms a straight line. **

 

Self-critique (if necessary):

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Question:  Problem B
A ball reaches a ramp of length 50 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 3.8 seconds. Its velocity at the end of the ramp is determined to be 6.31579 cm/s. What is its acceleration on the ramp?
 

Your solution: 

Confidence Assessment:

Given Solution: 

STUDENT RESPONSE: .'ds=50cm

vf=6.31579cm/s

'dt = 3.8s

The average velocity on the ramp is 50cm/3.8s =13.16m/s

The initial velocity will be v0 = 2 'ds /dt, v0 = 2 *50 /3.8, v0 = 26.31579 -vf which is 6.31579, so the v0 is 20.00

** You got it but this step isn?t correctly stated. v0 = 2 `ds/dt is not a correct statement. v0 = 2 vAve ? vf, or v0 = 2 `ds/`dt ? v0 (which is the same thing since vAve = `ds/`dt) is the correct equation.

The reasoning is that ave velocity is the average of initial and final velocities, since acceleration is uniform. So you have

vAve = (v0 + vf) / 2..

You know vAve from your previous calculation and vf is given. So you solve this equation for v0 and then substitute these values and simplify.

You get v0 = 2 vAve ? vf, then substitute.

This is actually what you did; just be careful you state it this way. If you don?t state it right there?s a chance you might not do it right. **

The accleration is 6.31579-20/3.8 = -3.601m/s/s

If your v0 was correct this would be right **
 

Self-critique (if necessary):

Self-critique rating:

Question:  Problem C
A projectile leaves the edge of a table at 18 cm/s, then falls freely a distance of 95 cm to the floor. How far does it travel in the horizontal direction during the fall?
 

Your solution: 

Confidence Assessment:

Given Solution: 

STUDENT RESPONSE:

For the vertical motion:

'ds=95cm

a=980cm/s/s

v0=0

First, we take the equation 'ds = v0'dt + .5(980) 'dt^2

95cm = .5 (980) 'dt^2
'dt = sqrt (95cm/.5(980 cm/s^2)) = sqrt (.194 s^2)
So, 'dt = sqrt (.194 s^2)
'dt=.44s

this is correct but see the note below

Next take the 18cm/s and multiply by the time duration .44s = 7.92cm/s

Note:  Generally this isn't the equation we would use with the given information to solve vertical motion, since v0 isn't always zero.

The fourth equation can be solved for vf, obtaining vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (0 cm/s)^2 + 2 * 980 cm/s^2 * 95 cm) - +- sqrt( 190 000 cm^2 / s^2) = +- 430 cm^2 / s^2.  Knowing that the final velocity in this situation is in the same direction as the displacement, we conclude that vf = + 430 cm/s and discard the negative solution.

From the initial and final velocities we find that the average velocity is about 215 cm/s, so that the time interval is about

`dt = 95 cm / (215 cm/s) = .44 s.

INSTRUCTOR COMMENT TO ALL STUDENTS:

Remember the basic principle used in this solution. You find the time of fall for the vertical motion, then find the horizontal displacement using the time interval and the initial horizontal velocity.

The basic principle is that vertical and horizontal motion are independent.  Vertical quantities don't appear in the analysis of horizontal motion, and horizontal quantities do not appear in the analysis of vertical motion.   Only the time interval is the same for the vertical and horizontal motion.

For an ideal projectile we also assume that the only force present is the gravitational force, which acts only in the vertical direction.  We regard air resistance and other forces that might actually be present, acting in the horizontal direction, as negligible.  So the horizontal acceleration is zero, and horizontal motion is at constant velocity.

Self-critique (if necessary):

Self-critique rating:

Question:  Problem D
What are the average acceleration and final velocity of an object which accelerates uniformly from rest, ending up with velocity 13.9 cm/sec after traveling a distance of 40 cm from start to finish?
 

Your solution: 

Confidence Assessment:

Given Solution: 

STUDENT RESPONSE: 'ds=40cm

v0=0

vf=13.9cm/s

If vf is 13.9cm/s then the using the equation a 'ds = vf^2 - v0^2)/2 will be , a * 40 cm = ( (13.9 cm/s)^2 ? 0^2) /2

40 * a = 193.21 cm^2/s^2 / 2
a = 96.61 cm^2/s^2 / (40 cm)
a = 2.415cm/s^2

** If vf^2 = v0^2 + 2 a `ds then a `ds = (vf^2 - v0^2) / 2, as you say, so a = (vf^2 - v0^2) / (2 `ds).

This is what you did (good job) but be careful that you state it this way. **

Then use the equation 'ds = v0 'dt + .5 a 'dt^2

** I don't recommend that you use this equation to solve for `dt, since the equation is quadratic in `dt.

At this stage you know all variables except `dt. The equation vf = v0 + a `dt would be much simpler.

Or you could just figure average velocity and divide into displacement. **

** You could have used the fact that vAve = (13.9 cm/s + 0) / 2 = 6.95 cm/s to obtain `dt:

Since vAve = `ds / `dt, we have `dt = `ds / vAve = 40 cm / (6.95 cm/s) = 5.8 sec, approx. **
 

Self-critique (if necessary):

Self-critique rating:

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

Below are some additional sample questions with hints included on some.  Some, and perhaps all problems don't include solutions.

Question: Problem Number 1

Give an example of a situation in which you are given v0, a and Ds, and reason out all possible conclusions that could be drawn from these three quantities, assuming uniform acceleration. Accompany your explanation with graphs and flow diagrams. Show how to generalize your result to obtain the symbolic expressions for vf and a.

Your solution: 

Confidence Assessment:

Given Solution: 

No detailed solution is at this point provided for this question.  However your description should include a description of the object and its situation (e.g., a car coasting up a hill (or down a hill), a ball rolling up or down an incline, a runner accelerating down a track.  In some of these situations acceleration is not likely to be uniform, but we can still solve the problem on the assumption that it is.

You should pick nonzero values of the three quantities (e.g., v0 = 10 m/s, a = 2 m/s^2, `ds = 20 m), then solve so that you know all the quantities (v0, vf, a, `dt, `ds , which of course gives you vAve and `dv), and state what your solution means for the situation you have described.

Self-critique (if necessary):

Self-critique rating:

Problem Number 2

If a cart coasts distances of  8.22996, .9279166, 7.202417 and 6.310089 cm, starting from rest each time, and requires respective times of 9.5 sec, 1 sec, 8.25 sec and 6.75 sec, is the hypothesis that acceleration is independent of position or velocity supported or not?

Your solution: 

Confidence Assessment:

Given Solution: 

No detailed solution is at this point provided for this question.

For each displacement the time required to accelerate thru that displacement, from rest, is given.  So it is straightforward to find the acceleration for each.

You would find:

• the acceleration corresponding to displacement 8.2 cm, from rest, in 9.5 sec

• the acceleration corresponding to displacement .92 cm, from rest, in 1 sec

• the acceleration corresponding to displacement 7.2 cm, from rest, in 8.25 sec

• the acceleration corresponding to displacement 6.3 cm, from rest, in 6.75 sec

To the extent that all these accelerations are the same, within reasonable uncertainty, the hypothesis would be supported.

Self-critique (if necessary):

Self-critique rating:

Problem Number 3

Determine the acceleration of an object which, after accelerating through a distance of 53 cm in 3.4 sec, is moving at 8.17647 cm/s.

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is at this point provided for this question.

You are given `ds, `dt and vf. 

You can easily reason this out:

• find the average velocity using the definition of average velocity
• sketch the v vs. trapezoid and determine initial velocity
• having determine initial velocity use the definition of acceleration to find the acceleration

You could also use the equations.

Either reasoning from definitions or equations, or any combination thereof, would be acceptable, since the method of solution wasn't prescribed by the problem.

Self-critique (if necessary):

Self-critique rating:

Problem Number 4

An object is given an unknown initial velocity up a ramp on which its acceleration is known to have magnitude 6 cm/s^2.  It reaches a maximum distance of 40.33 cm up the ramp before rolling back down. 

• What is its initial velocity? 

Hint:  You know three of the five quantities v0, vf, a, `dt and `ds for the motion up the ramp.  You will need to use the equations of motion to solve this one; you can't reason it out from the definitions without using simultaneous equations.

• How many seconds after the initial impetus does the object pass a point 5.8 cm up the ramp from its initial position (give all possible answers)? 

Hint:  The interval for this question is different from the interval for the preceding question.  However the initial velocity and the acceleration are both the same.

Your solution: 

Confidence Assessment:

Given Solution: 

We are given `ds and the magnitude of a.

The ramp is of unspecified length, meaning that in the absence of information to the contrary, the ball will travel until it comes to rest.  We conclude that vf = 0.

To solve, we need to specify our interval and our positive direction. 

• The interval runs from the initial instant until the ball comes to rest at the 40.3 cm position.
• Let's choose up the ramp as positive.

Solving the motion on this interval:

The initial velocity, being up the ramp, is therefore positive.  The final velocity is zero, so we conclude that the change in velocity, and therefore the acceleration, must be negative.

Our information is therefore:

• `ds = 40.3 cm
• a = - 6 cm/s^2
• vf = 0.

We can easily solve the fourth equation for v0, obtaining positive and negative values.  Knowing that v0 is in the direction we chose as positive, we discard the negative solution.

The initial velocity will be about v0 = +22 cm/s.

Our solution will also give us the time interval, which will be about 3.7 seconds

Thus the given conditions entail a ball with initial velocity 22 cm/s, coming to rest after accelerating for 3.7 s at an acceleration of -6 cm/s^2, having traveled about 40 cm during this interval.

We now answer the second question, which involves a different interval.

This interval will run from the initial position to the 5.8 cm position.  This interval shares its initial instant with the previous interval, so our initial velocity will be the same as the initial velocity we found for that interval.

To be consistent with our previous work we can choose 'up the ramp' as the positive direction.

Motion on this interval shares the initial velocity and acceleration with the previous interval:

Our previous solution found the initial velocity to be v0 = +22 cm/s, and the acceleration is again -6 cm/s^2.

We solve the situation v0 = +22 cm/s, a = -6 cm/s^2 and `ds = 5.8 cm, finding the final velocity and the time interval.

An approximate solution is vf = +-20.3 cm/s.

vf = +20.3 cm/s would indicate that the ball reaches the 5.8 cm position still moving up the ramp at 20.3 cm/s. 

• Its average velocity will be (22 cm/s + 20.3 cm/s) / 2 = 21.2 cm/s, and it will travel the 5.8 cm in `dt = 5.8 cm / (21.2 cm/s) = .25 sec. approx.

vf = -20.3 cm/s indicates that the ball reaches the 5.8 cm position moving down the ramp at 20.3 cm/s.  We might be tempted to discard this solution, but it makes perfect sense if we consider the following:

• After reaching the 5.8 cm position while traveling up the ramp, the ball could eventually come to rest, then accelerate back down, with no change in the -6 cm/s^2 acceleration.
• When the ball again reaches the 5.8 cm position it will be moving down the ramp, consistent with the negative velocity.

• So if the conditions of the problem persist long enough, the 5.8 cm displacement interval can be interpreted in two different ways.  According to the second interpretation the final event is reaching the 5.8 cm position while moving down the incline.
• The average velocity between these two events is vAve = (22 cm/s + (-20.3 cm/s) ) / 2 = .85 cm/s.
• The time required for this motion is `dt = `ds / vAve = 5.8 cm / (.85 cm/s) = 7 sec, approx..
• During this time the ball travels up the ramp until it comes to rest about 40 cm from the initial point, which as we saw earlier requires about 3.6 seconds, before accelerating back down the incline to the 5.8 cm point, which requires an additional 3.4 seconds.
• Note that it wasn't necessary to analyze separately the interval from the stopping position to the 5.8 cm position.  We simply combine the present result `dt = 7 sec with our previous result for motion from the initial position to the point of rest, and we draw the conclusion.  Had we analyzed this interval separately, our result would be consistent with the results we have so far.

Self-critique (if necessary):

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Problem Number 5

An object accelerates uniformly from rest, traveling a distance of 40 cm from start to finish and ending up with a velocity of 14.2 cm/sec.  What are the average acceleration and final velocity of the object?

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is at this point provided for this question.

We know v0, `ds and vf.  We can easily analyze this motion using the definitions.  Or we can easily use the equations of motion.

Recommended procedure:  Both solutions are short and easy, so solve both ways to confirm the solution.  If the two ways don't give you the same results, then figure out what you did wrong.  One practical reason for doing so:  if you're wrong, you'll be graded more leniently if you do it both ways and at least try to reconcile.

Self-critique (if necessary):

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Problem Number 6

Over a period of 10 seconds, an object increases its velocity at a uniform rate from 3 m/s to 27 m/s

• What is its acceleration and how far does it travel?
• Graph velocity vs. clock time for this object and explain what the slope of the graph means and why, and also what the area means and why.

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is at this point provided for this question.

This is the standard situation in which you are given v0, vf and `dt.  Reasoning is straightforward, as are the graphs and their interpretation.

Self-critique (if necessary):

Self-critique rating:

Problem Number 7

Using the equations which govern uniformly accelerated motion determine vf, v0, a, Ds and Dt for an object which starts at velocity 10 cm /s and accelerates at .333 cm/s/s through a distance of 103.5 cm.

Reason out the quantities v0, vf, Dv, vAve, a, Ds and Dt, using the basic definitions rather than the equations of motion: If an object’s velocity changes at a uniform rate from 10 cm/s to 13 cm/s as it travels 103.5 cm, then how long does it take to cover the distance?

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is at this point provided for this question.

You will need to use the equation of motion for the first question.

You are instructed to use reasoning on the second.  This means reason it out from the definitions and the v vs. t trapezoid.

Self-critique (if necessary):

Self-critique rating:

Problem Number 8

If the velocity of an object changes from 5 cm / sec to 15 cm / sec in 9 seconds, then at what average rate is the velocity changing?

A ball rolling from rest down a constant incline requires 3.9 seconds to roll the 49 centimeter length of the incline.

• What is its average velocity?

• What therefore is the final velocity of this ball?
• What average rate is the velocity of the ball therefore changing?

An automobile accelerates uniformly down a constant incline, starting from rest. It requires 13 seconds to cover a distance of 110 meters. At what average rate is the velocity of the automobile therefore changing?

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is at this point provided for this question.

On that last question, students frequently give the average rate of change of position with respect to clock time.  You will need to calculate that, but that's just the average velocity.  You would still have to find the average rate of change of velocity with respect to clock time. 

Self-critique (if necessary):

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Problem Number 10

How far does an object travel and what is its acceleration if its velocity increases at a uniform rate from 5 m/s to 20 m/s in 6 seconds?

Explain the meaning of the slope and area of the velocity vs. time for this object during this time interval.

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is provided for this question.

Self-critique (if necessary):

Self-critique rating:

Problem Number 11

Determine the acceleration of an object whose velocity is initially 17 cm/s and which accelerates uniformly through a distance of 70 cm in 5.9 seconds.

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is provided for this question.

Self-critique (if necessary):

Self-critique rating:

Problem Number 12

A projectile leaves the edge of a table and, while traveling horizontally at a constant 31 cm/s, falls freely a distance of 58 cm to the floor. If its vertical acceleration is 980 cm/s², how long does it take to fall and how far does it travel in the horizontal direction during the fall?

Hint:  For an ideal projectile, the horizontal velocity is constant.

Analyze the interval which begins with the ball leaving the tabletop and ends with the ball striking the floor.

The vertical motion is completely independent of the horizontal motion.

Which of the given quantities apply to the vertical motion?

Which apply to the horizontal motion?

Since the object is traveling in the horizontal direction at the instant it leaves the edge of the table, its initial vertical velocity is zero.

So for the vertical motion, v0 = 0.  Which of the other vertical quantities vf, `ds, a and `dt are you given?  Don't include any horizontal quantities in your list.

Use the equations of uniformly accelerated motion to find the remaining vertical quantities.  Your solution will include `dt.

`dt applies to both the vertical and the horizontal motion. 

Your solution: 

Confidence Assessment:

Given Solution: 

Vertical and horizontal motion are both analyzed for the interval starting at the edge of the table and ending at first contact with the floor.  Before and after this interval the acceleration changes, so we can only assume uniform acceleration for the interval between these events.

The vertical motion is characterized by

• v0 = 0
• `ds = 58 cm
• a = 980 cm/s^2

from which we easily enough find that `dt is about .35 sec.

The time interval runs from the initial to the final event and is therefore the same for the vertical and horizontal motion. 

So the horizontal motion is characterized by

• a = 0
• vAve = 31 cm/s
• `dt = .35 s

from which we easily find that `ds = vAve * `dt = 31 cm/s * .35 s = 11 cm, approx..

Note the following:

• 980 cm/s^2 is the vertical acceleration and will not appear in the analysis of the horizontal motion
• 31 cm/s is the horizontal velocity and will not appear in the analysis of the vertical motion
• 58 cm is a vertical displacement and will not appear in the analysis of the horizontal motion
• initial velocity in the vertical direction is zero; initial velocity in the horizontal direction is not zero
• during the specified interval the horizontal velocity does not change; the initial horizontal velocity is 31 cm/s, the final horizontal velocity is 31 cm/s; neither of these quantities is zero.  One or both might be zero sometime before the specified interval, and the same will likely be true sometime after the specified interval, but during this uniform-acceleration interval neither horizontal velocity is zero
• the final vertical velocity isn't zero either; the object will likely end up later with vertical velocity zero, but not during the specified interval

Self-critique (if necessary):

Self-critique rating:

Problem Number 13

How well does the following data set support the hypothesis that a graph of average acceleration vs. slope will be linear?

• A straight ramp 75 cm long is inclined at various slopes.
• The time required for a cart to coast 75 cm down the ramp, starting from rest, is 2.515303 seconds on the first incline, 2.394328 seconds on the next, and 2.556382 seconds on the last.
• The differences in elevation between one end and the other, for the different slopes, are 1.9, 4 and 5.9 cm.

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is at this point provided for this question.

• 2.515303 seconds on the first incline, 2.394328 seconds on the next, and 2.556382 seconds on the last.
• The differences in elevation between one end and the other, for the different slopes, are 1.9, 4 and 5.9 cm.

On this problem you would figure out the acceleration for each of three trials, each from rest.

• For example the first trial would involved a 75 cm displacement from rest in 2.51 seconds.  You easily analyze that to get the acceleration.
• Then you move on to the next trial, a 75 cm displacement from rest in 2.40 seconds.  You get a second acceleration.
• Then you get an acceleration for the third trial.

You figure out the three ramp slopes (the first would be 1.9 cm / (75 cm) = .025, approx..).

You construct a graph with acceleration on the vertical and ramp slope on the horizontal axis. 

You plot three points, one corresponding to each trial.

You see if the points can be reasonably well fit by a straight line, and draw your conclusions.

Nothing in the statement of this problem asks you if your results are consistent with what you think they should be.  It doesn't matter whether the graph looks the way you think it should; it looks the way it does if you do your calculations and your graph correctly.  The only question is whether the graph is consistent with a straight-line model.

Self-critique (if necessary):

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Problem Number 14

An object is given a velocity of 14 cm/s up a ramp on which its acceleration is known to have magnitude 6 cm/s^2. 

• At what later times does it pass a point 5 cm up the ramp from its initial position? 
• What is its maximum distance up the ramp from its initial point?

Hint:  These questions apply to different intervals.  One interval ends at the 5 cm position.  The other ends when the ball reaches its maximum distance up the ramp.  You therefore need to set up and solve two different problems, one for each question.

Hint:  You have to choose a positive direction and stick with it.

The acceleration has magnitude 6 cm/s^2.  This means that the acceleration could be +6 cm/s^2 or -6 cm/s^2.  Depending on what direction you choose as positive, one of the two accelerations will make sense and the other won't.

Your solution: 

Confidence Assessment:

Given Solution: 

No solution is at this point provided for this question.

The given information is a little different but the situation is otherwise completely analogous to the previous problem, on which the acceleration was also of magnitude 6 cm/s^2.

Self-critique (if necessary):

Self-critique rating:

Problem Number 15

At clock time t = 7 sec, a ball rolling straight down a hill is moving at 6 m/s and is 47 m from the top of the hill.  It accelerates uniformly at a rate of .6 m/s/s until clock time t = 17 sec.  What is its velocity at this point and what is its average velocity during this time?  How far is it from the starting point at t = 17 sec?

Hint:  Be very sure you clearly identify the interval for which you are given information.

Your solution: 

Confidence Assessment:

Given Solution: 

You don't know anything about what happened at t = 0.  You don't know where the ball was, how fast it was moving, or if it even existed at t = 0.

All you know is the state of motion at t = 7 sec, and at t = 17 sec, and that acceleration is uniform (which among other things justifies the use of the equations of uniformly accelerated motion, and the fact that the v vs. t graph on this interval is a trapezoid).

Your analysis must therefore be confined to the interval between t = 7 sec and t = 17 sec.