According to the order of operations 1/a * ax = 1/a * (c - b) can be written as follows
1/a * (c-b)/1 = 1 * (c - b) / a = (c - b) / a can be written as follows
x = (c - b) / a can be written as follows
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.
061. Doing the Algebra
Many students are doing very well with the algebra, and others need only a brush-up on the details. Some students are having a significant amount of trouble with the algebra. This exercise begins with some very easy problems and ends with a very challenging problem so it should be useful, as some level, to all students.
Note on the notation: You might want to run through the exercise Typewriter Notation, if you haven't already done so. This exercise might not have been assigned at this point, but if not it will be assigned soon; if you have time (of course, who does?) you should run through it as soon as possible. Among other things it includes links at the end that should be very helpful. You shouldn't spend more than 30-60 minutes on the Typewriter Notation exercise, but whether you do it now or later, be sure to read all the information.
It is assumed you have completed two years of algebra. There's actually nothing here you shouldn't have learned in first-year algebra, so this isn't absolutely necessary, but the additional practice and mathematical sophistication you develop in second-year algebra can be immensely helpful. If you didn't do well in second-year algebra you will probably need to spend a significant amount of time on this exercise.
Question: `q001. If you want to solve the equation 2 x + 4 = 9, what steps do you follow? Include all steps, showing what you add to both sides, what you multiply by both sides, etc..
Your solution:
Confidence Assessment:
Given Solution:
To solve 2 x + 4 = 9 we proceed as follows:
Starting with
2x + 4 = 9, we add -4 to both sides (or if you prefer subtract 4 from both sides) to get
2x + 4 - 4 = 9 - 4.
The right-hand side simplifies easily enough to 5.
Since 4 - 4 = 0, the left-hand side simplifies to 2x + 0.
Since 2x + 0 = 2x, the left-hand side becomes just 2x so we have
2x = 5. We now multiply both sides by 1/2 (of if you prefer divide both sides by 2) to get
1/2 * 2x = 1/2 * 5.
The right-hand side is 1/2 * 5, which is the same as 1/2 * 5/1.
1/2 * 5/1 = 1 * 5 / (2 * 1). = 5 / 2.
This could be expressed as 2.5, or even as 2 1/2. We generally avoid mixed numbers, so we'll use 5 / 2 or 2.5 to represent the right-hand side.
Since 1/2 * 2 = 1, the expression 1/2 * 2x simplifies to 1 * x.
Since 1 * x = x, the left-hand side simplifies to x.
Thus our equation becomes
x = 5/2.
This can also be written as
x = 2.5.
Self-critique (if necessary):
Self-critique rating:
Question: `q002. If you want to solve the equation 3 x + 7 = 2, what steps do you follow? Include all steps, showing what you add to both sides, what you multiply by both sides, etc..
Your solution:
Confidence Assessment:
Given Solution:
To solve 3 x + 7 = 2 we proceed as follows:
Starting with
3x + 7 = 2, we add -7 to both sides (or if you prefer subtract 7 from both sides) to get
3x + 7 - 7 = 2 - 7.
The right-hand side simplifies easily enough to -5.
Since 7 - 7 = 0, the left-hand side simplifies to 3x + 0.
Since 3x + 0 = 3x, the left-hand side becomes just 3x so we have
3x = -5. We now multiply both sides by 1/3 (of if you prefer divide both sides by 3) to get
1/3 * 3x = 1/3 * (-5).
The right-hand side is 1/3 * (-5), which is the same as 1/3 * (-5/1).
1/3 * (-5/1) = 1 * 5 / (2 * 1). = 5 / 2.
This could be expressed as -5/3 or even as -1 2/3. We generally avoid mixed numbers, so we'll use -5 / 3 to represent the right-hand side.
-5/3 can also be expressed as a decimal in the form -1.666....
-5/3 is not the same as -1.66, and is not the same as 1.66666. It is approximately equal to 1.67 (which represents a rounding error of about .2%), or 1.66667 (which represents a rounding error of about +.0002%). Depending on experimental uncertainty, it might be valid to express the solution as a rounded decimal, provided we include the work 'approximately' or 'approx').
So the right-hand side becomes -5/3, or -1 1/3, or even a rounded expression lik -1.667 (approx.).
Since 1/2 * 2 = 1, the left-hand side 1/3 * 3x simplifies to 1 * x.
Since 1 * x = x, the left-hand side simplifies to x.
Thus our equation becomes
x = -5/3,
which could be written as the exact expression
x = -1.666...
or as an approximate rounded expression, e.g.,
x = -1.67.
Self-critique (if necessary):
Self-critique rating:
Question: `q003. If you want to solve the equation a x + b = c, what steps do you follow? Include all steps, showing what you add to both sides, what you multiply by both sides, etc..
Your solution:
Confidence Assessment:
Given Solution:
To solve ax + b = c we proceed as follows:
Starting with
ax + b = c, we add -b to both sides (or if you prefer subtract b from both sides) to get
ax + b - b = c - b.
Since b - b = 0, the left-hand side simplifies to ax + 0.
Since ax + 0 = ax, the left-hand side becomes just ax so we have
ax = c - b. We now multiply both sides by 1/a (of if you prefer divide both sides by a) to get
1/a * ax = 1/a * (c - b).
The right-hand side is 1/a * (c - b), which is the same as 1/a * (c - b) / 1.
1/a * (c-b)/1 = 1 * (c - b) / a = (c - b) / a.
Since 1/a * a = 1, the left-hand side 1/a * ax simplifies to 1 * x.
Since 1 * x = x, the left-hand side simplifies to x.
Thus our equation becomes
x = (c - b) / a.
Note on standard mathematical notation for selected expressions:
According to the order of operations 1/a * ax = 1/a * (c - b) can be written as follows
1/a * (c-b)/1 = 1 * (c - b) / a = (c - b) / a can be written as follows
x = (c - b) / a can be written as follows
IMPORTANT NOTES:
The expression (c - b) / a means 'subtract b from c, then divide the result by a'.
This is very different from the expression c - b / a. If the expression is written this way, then the order of operations dictates that we first divide b by a, the subtract the result from c.
As a numerical example, (5 - 3) / 4 means 2 / 4 or 1/2.
However if we write this as 5 - 3 / 4, we have to divide 3 by 4 before subtracting. In decimals 3/4 = .75, so we would end up with 5 - .75 = 4.25, which is very different than the 1/2 we got when 5 - 3 was grouped.
Lesson: Be careful of grouping. when you write your expressions.
Note also that the distributive law of multiplication over addition applies to this expression.
(c - b) / a means the same as
(c - b) * (1 / a), which by the distributive law means
(c / a) - (b / a).
Repeating for emphasis:
As we saw before, (c - b) / a does not mean c - b / a or the equivalent expression c - (b / a).
Self-critique (if necessary):
Self-critique rating:
Question: `q004. Solve the equation vf = v0 + a `dt for `dt.
Your solution:
Confidence Assessment:
Given Solution:
Brief solution:
vf = v0 + a `dt so
vf - v0 = a `dt so
`dt = (vf - v0) / a.
However you should understand why this works. Otherwise chances are you're doing your mathematics by habit (or superstition) when you should understand all the details.
So here's the full solution:
Starting with
vf = v0 + a `dt we subtract v0 from both sides to get
vf - v0 = v0 + a `dt - v0.
Since v0 + a `dt = a `dt + v0 (order of addition doesn't matter) we can arrange (v0 + a `dt - v0) as (a `dt + v0 - v0); then side v0 - v0 = 0 we get (a `dt + 0), which is just (a `dt). So our equation becomes
vf - v0 = a `dt. We then multiply both sides by 1/a to get
1/a ( vf - v0) = 1/a * a `dt.
It might seem obvious that the right-hand side simplifies to just `dt, but it's useful to understand exactly why this is so:
The right-hand side means 1 / a * (a `dt) / 1, which multiples out to (1 * a `dt) / (a * 1), which is just (a `dt / a)
We can express this as the product of two fractions in a different way: you can easily verify that (a `dt / a) is the same as (a / a) * (`dt / 1).
Since a / a = 1, this becomes just 1 * `dt / 1, which is just `dt.
So the right-hand side simplifies to `dt.
The left-hand side is the same as 1 / a * (vf - v0) / 1 = (1 * (vf - v0) ) / (a * 1) = (vf - v0) / a.
So our equation becomes
(vf - v0) / a = `dt. We can of course reverse the right- and left-hand sides to express this as
`dt = (vf - v0) / a.
Note on standard mathematical notation for selected expressions:
According to the order of operations 1/a ( vf - v0) = 1/a * a `dt can be written as follows
According to the order of operations 1 / a * (vf - v0) / 1 = (1 * (vf - v0) ) / (a * 1) = (vf - v0) / a can be written as follows
According to the order of operations `dt = (vf - v0) / a can be written as follows
COMMON ERRONEOUS SOLUTION
Starting with
vf = v0 + a `dt, many students will begin by subtracting a from both sides to get
vf - a = v0 + `dt. However, this is wrong for many reasons.
Reason 1 why this is wrong:
In the first place vf - a is not a valid subtraction. Since vf and a are unlike terms, one with units of distance / time and the other with units of distance / time^2, they can' be subtracted. Similarly v0 and `dt do not have the same units, so they can't be added. Thus the equation vf - a = v0 + `dt doesn't make sense (e.g., what would it mean to subtract 90 cm/s^2 from 30 cm/s, or add 10 cm/s to 5 seconds? Neither makes any sense at all).
Reason 2 why this is wrong:
In the second place, if you subtract a from v0 + a `dt, you don't get v0 + `dt. You get
v0 + a `dt - a.
Now a `dt - a = a * (`dt - 1) ; if you don't believe this, apply the distributive law to the expression a * (`dt - 1). You get a * `dt - a * 1, which is equal to a `dt - a.
So
v0 + a `dt - a. = v0 + a * (`dt - 1), which is completely different that v0 + `dt.
An addition reason why this is wrong:
You can also verify that the units don't work. v0 and a `dt both result in units of distance / time, whereas a is in units of distance / time^2. So while the expression v0 + a `dt makes perfect sense, it makes no sense to subtract a from the expression v0 + a `dt.
Self-critique (if necessary):
Self-critique rating:
Question: `q005. Solve the equation `ds = v0 `dt + .5 a `dt^2 for a.
Your solution:
Confidence Assessment:
Given Solution:
Brief solution:
`ds = v0 `dt + .5 a `dt^2 so
.5 a `dt^2 = `ds - v0 `dt so
a = (`ds - v0 `dt) / (.5 `dt^2) or
a = 2 (`ds - v0 `dt) / `dt^2.
However you should understand why this works. Otherwise chances are you're doing your mathematics by habit (or superstition) when you should understand all the details.
So here's the full solution:
Starting with
`ds = v0 `dt + .5 a `dt^2 we begin by subtracting v0 `dt from both sides to get
`ds - v0 `dt = v0 `dt + .5 a `dt^2 - v0 `dt. Since v0 `dt - v0 `dt = 0, etc. (see preceding solution) we get
`ds - v0 `dt = .5 a `dt^2. If we multiply both sides by 2 we get get
2 * (`ds - v0 `dt) = 2 * (.5 a `dt^2), and since 2 * .5 = 1 we have
2 * (`ds - v0 `dt) = a `dt^2. Finally we multiply both sides by 1 / `dt^2 to get
1 / `dt^2 * 2 (`ds - v0 `dt) = 1 / `dt^2 * a `dt^2.
Using reasoning similar to that used in preceding solutions we simplify the right-hand side to just a, and the left-hand side to 2 (`ds - v0 `dt) / `dt^2. Our solution is thus
a = 2 ( `ds - v0 `dt) / `dt^2.
Note on standard mathematical notation for selected expressions:
According to the order of operations 1 / `dt^2 * 2 (`ds - v0 `dt) = 1 / `dt^2 * a `dt^2 can be written as follows
According to the order of operations a = 2 ( `ds - v0 `dt) / `dt^2 can be written as follows
Self-critique (if necessary):
Self-critique rating:
Question: `q006. Solve the equation vf^2 = v0^2 + 2 a `ds for `ds.
Your solution:
Confidence Assessment:
Given Solution:
Starting with
vf^2 = v0^2 + 2 a `ds we subtract v0^2 from both sides to get
vf^2 - v0^2 = 2 a `ds, then divide both sides by 2 a and switch right- and left-hand sides to get
`ds = (vf^2 - v0^2) / (2 a).
Self-critique (if necessary):
Self-critique rating:
Question: `q007. Solve the equation `ds = (v0 + vf) / 2 * `dt for `dt.
Your solution:
Confidence Assessment:
Given Solution:
Starting with
`ds = (vf + v0) / 2 * `dt, we multiply both sides by 2 / (vf + v0) to get
`ds * 2 / (vf + v0) = (vf + v0) / 2 * `dt * 2 / (vf + v0).
The left-hand sides is just 2 `ds / (vf + v0).
The numerators on the right-hand side are (vf + v0) ,`dt and 2 so the numerator of the product is 2 (vf + v0) `dt. The denominators are 2 and (vf + v0) so the denominator of the product is 2 ( vf + v0). Our right-hand side is therefore 2 (vf + v0) `dt / (2 ( vf + v0) ). Since 2/2 = 1 and (vf + v0) / (vf + v0) = 1, the right-hand side simplifies to just `dt, and our equation is
2 `ds / (vf + v0) = `dt, or switching right- and left-hand sides
`dt = 2 `ds / (vf + v0).
Note on standard mathematical notation for selected expressions:
According to the order of operations can be written as follows
According to the order of operations `ds = (vf + v0) / 2 * `dt can be written as follows
According to the order of operations `ds * 2 / (vf + v0) = (vf + v0) / 2 * `dt * 2 / (vf + v0) can be written as follows
According to the order of operations `dt = 2 `ds / (vf + v0) can be written as follows
STUDENT COMMENT:
It’s possible I have the right answer
although it doesn’t quite match. I cross multiplied and divided. I didn’t do it
the same way, but I did come out with the same solution, just flipped around. I
got ‘dt = (vf + v0)/ 2 ‘ds instead of what you got for the answer.
INSTRUCTOR RESPONSE:
You appear to have treated (2 * `dt) as the denominator.
Only the 2 is the denominator of the expression (vf + v0) / 2 `dt. (v0 + vf) / 2 * `dt means (vf + v0) divided by two, with the result multiplied by `dt.
The units of your solution are not consistent:
The units of (vf + v0) / 2 * `ds would be m/sec * m = m^2 / sec, while `dt is in units of sec. So `dt = (v0 + vf) / 2 * `ds does not have correct units.
If you intended (vf + v0) / (2 * `ds), then note that the parentheses are necessary. However, the units of this expression would be (m/sec) / (m) = (m / sec) * (1 / m) = m / (m * sec) = 1 / sec, not sec, and would not agree with the units of `dt.
Self-critique (if necessary):
Self-critique rating:
Question: `q008. Solve the equation `ds = v0 `dt + .5 a `dt^2 for v0.
Your solution:
Confidence Assessment:
Given Solution:
Starting with
`ds = v0 `dt + .5 a `dt^2, subtract .5 a `dt^2 from both sides to get
`ds - .5 a `dt^2 = v0 `dt then multiply both sides by 1/`dt to get
1 / `dt * (`ds - .5 a `dt^2) = v0 * `dt * (1 / `dt), which simplifies to
(`ds - .5 a `dt^2) / `dt = v0.
Self-critique (if necessary):
Self-critique rating:
Question: `q009. Solve the equation `ds = v0 `dt + .5 a `dt^2 for `dt.
(warning: if your algebra isn't really good you should skip this and just see the given solution, so you'll understand why we recommend that Phy 121 and Phy 201 student avoid this situation at this point)
Your solution:
Confidence Assessment:
Given Solution:
If you're a University Physics student you will soon need to understand all these details, so you might as well get used to it now.
If you're a Principles of Physics student you won't ever be tested on your ability to solve this equation for `dt and you can safely skip it, though if you have a good background in mathematics and might go into a field where mathematics is used, you might choose to study this solution.
If you're a General College Physics student you will eventually need to understand most of this and you should take a good look at this now, and at least be sure you remember the quadratic formula. However let yourself don't get bogged down in the finer details of the solution at this point of the course.
The equation `ds = v0 `dt + .5 a `dt^2 includes terms with both `dt and `dt^2. It's not possible to isolate `dt by adding or subtracting the same quantity to both sides of the equation or by multiplying or dividing both sides by the same quantity. So these methods won't lead to a solution.
You could complete the square to find a solution, but to do this with symbols is very challenging for most students at this stage of the course.
The easiest way to solve the equation is to recognize that it's quadratic in `dt and put it into the standard form of a quadratic equation, and apply the quadratic formula.
In standard for the equation is
.5 a `dt^2 + v0 `dt - `ds = 0,
which is of the form A x^2 + B x + C = 0, with x = `dt and A = .5 a, B = v0 and C = -`ds.
The solution is
x = (-B +- sqrt( B^2 - 4 A C) ) / (2 A), which after substitution is
`dt = (-v0 +- sqrt( v0^2 - 4 * (.5 a) * (-`ds) ) ) / (2 * (.5 a) ). Simplifying we have
`dt = (-v0 +- sqrt( v0^2 + 2 a `ds) ) / a
the last expression can be written in standard notation as follows
.
This can be written, using the distributive law, as
`dt = (-v0 / a) +- (sqrt( v0^2 + 2 a `ds) / a)
written in standard notation as follows
.
There are two solutions, one for the + of the +- symbol and one for the -.
The quantity v0^2 + 2 a `ds might be negative, so that sqrt( v0^2 + 2 a `ds) would not be a real number. The conclusion would be that t isn't a real number, so no solution exists. We would conclude that the situation described by our quantities `ds, a and v0 is impossible
One example of an impossible situation: Accelerate thru displacement 100 meters, starting with velocity 5 meters / second, accelerating at -10 m/s^2. This is unreasonable, since after only 1 second we would already be moving at -5 m/s, i.e., moving in the negative direction. We would be moving away from the 100 meter point, without having had time to reach it, and we would continue to accelerate away from it. We could never get there.
If we plug v0 = 5 m/s, a = -10 m/s^2 and `ds = 100 m into the expression v0^2 + 2 a `ds we will get (5 m/s)^2 + 2 * (-10 m/s^2) * (100 m) = 25 m^2 / s^2 - 2000 m^2 / s^2 = -1975 m^2 / s^2. Thus sqrt(v0^2 + 2 a `ds) = sqrt(-1975 m^2 / s^2) = sqrt(-1975) * sqrt(m^2 / s^2) = sqrt( -1975) m/s. However the square root of a negative number is not a real number, so there is no real-number solution to the problem.
If sqrt(v0^2 + 2 a `ds) is zero, then the +- makes no difference and we get only one solution.
If sqrt(v0^2 + 2 a `ds) is positive, then the + and the - give us different solutions. One or both of these solutions might or might not make sense for the given problem. Both solutions should be interpreted to see if they make sense.
Note on standard mathematical notation for selected expressions:
According to the order of operations can be written as follows
Self-critique (if necessary):
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