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course Phys201
6/20@9:13pm
Solving Uniform Acceleration Problems
Possible Combinations of Variables Direct Reasoning
Using Equations Problem
Possible Combinations of Variables
There are ten possible combinations of three of the the five variables v0, vf, a, Δt and Δs. These ten combinations are summarized in the table below:
1 v0 vf a
2 v0 vf dt
3 v0 vf ds
4 v0 a dt
5 v0 a ds *
6 v0 dt ds
7 vf a dt
8 vf a ds *
9 vf dt ds
10 a dt ds
If we know the three variables we can easily solve for the other two, using either direct reasoning or the equations of uniformly accelerated motion (the definitions of average velocity and acceleration, and the two equations derived from these by eliminating Δt and then eliminating vf).
Only two of these situations require equations for their solution; the rest can be solved by direct reasoning using the seven quantities v0, vf, a, Δt, Δs, Δv and vAve. These two situations, numbers 5 and 8 on the table, are indicated by the asterisks in the last column.
Direct Reasoning
We learn more physics by reasoning directly than by using equations. In direct reasoning we think about the meaning of each calculation and visualize each calculation.
When reasoning directly using v0, vf, `dv, vAve, `ds, `dt and a we use two known variables at a time to determine the value of an unknown variable, which then becomes known. Each step should be accompanied by visualization of the meaning of the calculation and by thinking of the meaning of the calculation. A 'flow diagram' is helpful here.
Using Equations
When using equations, we need to find the equation that contains the three known variables.
• We solve that equation for the remaining, unknown, variable in that equation.
• We obtain the value of the unknown variable by plugging in the values of the three known variables and simplifying.
• At this point we know the values of four of the five variables.
• Then any equation containing the fifth variable can be solved for this variable, and the values of the remaining variables plugged in to obtain the value of this final variable.
Problem
Do the following:
• Make up a problem for situation # 10, and solve it using direct reasoning.
• Accompany your solution with an explanation of the meaning of each step and with a flow diagram.
• Then solve the same problem using the equations of uniformly accelerated motion.
• Make up a problem for situation # 8, and solve it using the equations of uniformly accelerated motion.
A student releases a cylinder from a ramp that is 10 feet long at an acceleration of .1 ft/sec^2. It takes 5 sec for the cylinder to go down the ramp. Solve for the final and initial velocities.
Vf=total distance/total time
Vf=10/5
Vf= 2 ft/sec
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You have divided displacement by time interval. This does not give you vf.
It does give you an important quantity, though. Can you identify it?
Think in terms of the definitions of average velocity and average acceleration. One of those definitions is directly applicable to this question.
Note that you did not use units in your calculation, which would be 10 ft / (5 sec). You did put the correct units on your result.
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V=a `dt +v0
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The analysis of the uniformly accelerated motion on an interval, according to our notation, uses the quantities v0, vf, vAve and `dv.
It does not use the quantity v, which would stand for instantaeous velocity at some specified instant.
The equation you quote is valid, if v is understood to be the velocity at clock time t. However you will think more clearly about motion on an interval if you call this vf, the final velocity (i.e., the velocity at the end of the interval `dt).
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2=.1(5)+v0
v0=1.5 ft/sec
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Your result does not hold up when it is checked. Initial velocity 1.5 ft / sec, final velocity 2.0 ft / sec imply average velocity 1.75 ft / sec. In 5 sec that would imply a displacement of only 8.75 ft, not the 10 ft you assumed.
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`ds is the length of change in position so that would be the 10 ft of the ramp. `dt is the change in time which would be 5 sec. The acceleration is .1 ft/sec. The final velocity will be your change in position divided by the change in time. The initial velocity can be found using acceleration, time, and final velocity (v=a `dt+v0)
`dt `ds a
vf=`ds/`dt
v0=vf-a`dt
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You've done pretty well with yoru solution, but it doesn't quite hold up. This is due to one basic error at the beginning. Otherwise your application of equations would have led you to a good solution.
That's very good.
However you need to use direct reasoning on this problem, as specified.
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`` You have used the equations rather than direct reasoning to solve this problem.Direct reasoning is based on the definitions of velocity and acceleration, not on equations.
Direct reasoning gets you much closer to the meanings of the various steps in your solution than does the use of equations.
The equations are important, of course, because direct reasoning is not always possible.
`` To directly reason this situation:You know `ds, a and `dt.
`` From `ds and `dt we apply the definition of average velocity to get vAve. vAve is average rate of change of position with respect to clock time, so vAve = change in position / change in clock time = `ds / `dt. Given `ds and `dt, then, we simply calculate vAve = `ds / `dt.
`` From a and `dt you can find `dv, the change in velocity. (This is connected to the definition of average acceleration as the average rate of change of velocity with respect to clock time, which is (change in velocity) / (change in clock time). Thus a = `dv / `dt, so `dv = a `dt)
`` From vAve and `dv we can reason out both v0 and vf, both of which deviate from vAve by the same amount (since vAve is halfway between v0 and vf, and `dv is the difference between v0 and vf). Thus both v0 and vf deviate from vAve by half of `dv.To get v0 and vf, then, we respectively subtract and add `dv / 2 to the average velocity vAve.
We have now reasoned out the values of vAve, v0, vf and `dv.
When we reason out a problem of this nature, we will at the end know the values of all five of the quantities v0, vf, `ds, `dt and a, in addition to the values of vAve and `dv.
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****** There would be a line between `dt and `ds connecting to vf=`ds/`dt. Another line would connect a (acceleration) to v0=vf-a`dt. A line would connect `dt to v0=vf-a`dt. Another line connect vf=`ds/`dt to v0=vf-a`dt*******
The idea of uniformly accelerated motion is that the difference of average velocity to initial velocity equals the difference from final velocity to average velocity. I found final velocity to be 2ft/sec and initial velocity to be 1.5 ft/sec therefore, the average velocity is the 1.75 ft/sec and both differences are .25 ft/sec.
A student pushes a toy car down a 10 foot ramp. The car is going down the ramp at .1 ft/sec^2. The car reaches the end of the ramp at 5 ft/sec. Find `dt and v0.
vf^2=v0^2+2a`ds
v0= sq rt. vf^2-2a`ds
=5ft/sec(.1ft/sec^2)-2(.1ft/sec^2)(10 ft)
v0= sq rt 23
v0=4.80 ft/sec
v0+vf/2=vAve
4.80ft/sec^2+5/ft/sec2=vAve
vAve=4.9 ft/sec
vAve=`ds/`dt
`dt=`ds/vAve
`dt=10ft/4.9 ft/sec^2
`dt=2.04 ft/sec
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You have a good analysis here, based on the equations. This is the method specified for the problem.
However your final step has units which are very much incorrect.
ft / (ft / sec^2) does not give you ft / sec.
Also, ft / sec is not a unit of time.
To divide two fractions you invert and multiply. You need to review multiplcation and division of fractions.
The units you do get if you do your stated calculation using correct algebra of the units will not be units of time.
If you use the right units for vAve (see your previous calculation, where the units are right), you will get a calculation that ends up with units of time.
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You are doing very well with the equations.
You aren't doing quite as well with units and reasoning, which are related in that they are closer to the concept and understanding of uniformly accelerated motion than the equations.
However the equations are very important, so you're making good progress.
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You should be able to correct you equation-based solutions in a short time.
You will probably require more time to do the reasoning. However once you understand the reasoning, it is actually more efficient than the equations on those problems to which it applies.
&#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. &#
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