course phy 201 Wednesday, September 30,2009 at 1006 am If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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):not necessary ------------------------------------------------ Self-critique rating: not nercessary ********************************************* 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.. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Write the equation 3x+7= 2 Subtract 7 from both sides 3x+7-7 = 2-7 3x= -5 Divide both sides by 3 (3x)/3 = -5/3 The 3/3 simplifies to 1 x= -5/3 ***The answer is x= -5/3 of -1 2/3 confidence rating: very confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 expresion x = -1.666... or as an approximate rounded expression, e.g., x = -1.67. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): not necessary ------------------------------------------------ Self-critique rating: not necessary ********************************************* 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.. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Write down the equation ax + b= c Subract b from both sides ax+b-b = c-b . ax= c-b Divide both sides by a x= (c-b)/a confidence rating: very condifent ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. 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): not necessary ------------------------------------------------ Self-critique rating: not necessary ********************************************* Question: `q004. Solve the equation vf = v0 + a `dt for `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Write the equation vf= v0 + a `dt Subtract v0 from both sides vf- v0 = v0 -v0 +a`dt Vf-v0 = a`dt Divide both sides by a (vf- v0)/ a = (a`dt)/a The a in the numerator and (vf - v0)/a = `dt Denominator on the left simplify to 1 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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) / `dt. 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): not necessary ------------------------------------------------ Self-critique rating: not necessary ********************************************* Question: `q005. Solve the equation `ds = v0 `dt + .5 a `dt^2 for a. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Write the equation `ds= v0 `dt + .5a `dt^2 Subtract the v0`dt from both sides `ds- v0`dt= v0`dt- v0`dt + .5a`dt^2 Simplify ‘ds- v0`dt= 0.5a`dt^2 Multiply both sides by 2 2(`ds-v0`dt)= a`dt^2 Divide both sides by `dt^2 2(`ds- v0`dt)/ `dt^2 = a`dt^2/ `dt^2 The `dt^2 on the right simplift to 1 2(1ds-v0`dt)/ `dy^2= a confidence rating: very cofident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): not nercessary ------------------------------------------------ Self-critique rating: not necessary ********************************************* Question: `q006. Solve the equation vf^2 = v0^2 + 2 a `ds for `ds. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Write the equation vf^2 = v0^2 + 2 a `ds Subtract v0^2 from both sides vf^2- v0^2 = v0^2- v0^2 + 2 a `ds Simplify vf^2-v0^2 = 2 a `ds Divide both sides by 2a (vf^2-v0^2)/ 2a = 2a`dt/2a The 2a/2a on the right simplifies to 1 (vf^2-v0^2)/ 2a = `ds confidence rating: very confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): not necessary ------------------------------------------------ Self-critique rating: not necessary ********************************************* Question: `q007. Solve the equation `ds = (v0 + vf) / 2 * `dt for `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Write the equation `ds = (v0 + vf) / 2 * `dt Multiply both sides by 2 2`ds= (v0+vf)/2 *2 *`dt The division and multiplication of the 2 on the right makes Them equal to 1 2`ds= (v0+vf)* ‘dt Divide both sides by (v0+vf) 2`ds/ (v0+vf)= (v0+vf)/(v0+vf)*`dt The (v0+vf)/ (v0+vf)= 1 2`ds/ (v0+vf)= `dt confidence rating: Very confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): not necessary ------------------------------------------------ Self-critique rating: not necessary ********************************************* Question: `q008. Solve the equation `ds = v0 `dt + .5 a `dt^2 for v0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Write the equation `ds = v0 `dt + .5 a `dt^2 Subract .5a`dt^2 from both sides `ds- 0.5a`dt^2 = v0 `dt + .5 a `dt^2- .5a`dt^2 Simplify `ds- 0.5a`dt^2 = v0 `dt Divide both sides by `dt (`ds- 0.5a`dt^2) /`dt= (v0 `dt)/`dt The `dt/`dt on right simplify to 1 (`ds- 0.5a`dt^2) /`dt= v0 confidence rating: Very confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): not necessary ------------------------------------------------ Self-critique rating: not necessary ********************************************* 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) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Write the equation `ds = v0 `dt + .5 a `dt^2 Subract `ds from both sides to make the equation =0 `ds- `ds = v0 `dt + .5 a `dt^2 -ds Simplify 0= v0 `dt + .5 a `dt^2 -ds Figure out what A, B and C would be for using A= .5a The quadratic equation B= v0 C= -`ds Write the quadratic equation `dt= -B+-sqrt (B^2- 4AC))/(2A) Plug in A, B and C `dt= (-v0+-sort(v0^2- 4(.5a)(-`ds)))/ (2* (.5a) Simplify ‘dt= (-v0+- sort (v0^2+ 2a`ds))/a confidence rating: fairly confident ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. This can be written, using the distributive law, as `dt = (-v0 / a) +- (sqrt( v0^2 + 2 a `ds) / a). 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): not necessary ------------------------------------------------ Self-critique rating: not necessary "