3

I don't understand how people can know how to solve math word problems.

I went to high school in the worst state in the USA for education. My math classes didn't have word problems at all, so I can't understand them now.

Something like this:

"A proton (mass $m = 1.67 \times 10^{-27}$ kg) is being accelerated along a straight line at $3.6 \times 10^{15}$ m/s$^2$ in a machine. If the proton has an initial speed of $2.4 \times 10^7$ m/s and travels 3.5 cm, what then is (a) its speed and (b) the increase in its kinetic energy?"

What does this even mean? How do you break this down into steps? How do you figure out which formulas apply to which steps? How can you tell if it's the right answer or not?

Sanderr
  • 523
user416503
  • 137
  • 1
    That's a physics problem, not a mathematics one. – hmakholm left over Monica Mar 21 '17 at 08:13
  • 3
    I'd call that one a physics question. It requires you to know physical concepts like kinetic energy, which you won't learn in a mathematics course. Of course physics questions often involve mathematics. In particular, it requires you to know the definition of work and kinetic energy. – celtschk Mar 21 '17 at 08:14
  • 4
    If you have this big trouble understanding word problems, then I personally don't think that a single written answer in a forum will be enough to help you. Learning to make sense of word problems is a process that takes time, and you need to start smaller than this. I would recommend that you get a few lessons from a private tutor, if that is even an option. – Arthur Mar 21 '17 at 08:15
  • 6
    You make sense of it the same way you do reading this sentence. – Zelos Malum Mar 21 '17 at 08:17
  • 3
    I'm not sure what kind of an answer you're looking for. To get an example, I guess you can solve the word problem "I'm 28 years old now. How old will I be this time next year?" How would you answer the questions in your last paragraph for that problem? – JiK Mar 21 '17 at 13:49
  • 1
    @user416503 , "I went to high school in the worst state in the USA for education", i don't know your religious or phyloshopical thoughts, but remember that who invented the wheel never went to school. So, cheer up, and welcome to this lovely world! If you want, read the biographies of https://en.wikipedia.org/wiki/Michael_Faraday who received little formal education and https://en.wikipedia.org/wiki/Srinivasa_Ramanujan who had almost no formal training in pure mathematics, amongst others. – cgiovanardi Mar 21 '17 at 22:19

3 Answers3

3

You can retranscript as you read

  1. mass $\color{green}m$ is given;

  2. acceleration $\color{green}a$ is given;

  3. initial speed $\color{green}{v_0}$ is given;

  4. traversed space $\color{green}{\Delta s}$ is given;

  5. final speed $v_t$ is asked;

  6. increase in kinetic energy $\Delta E=E_t-E_0$ is asked.

You should understand that this is related to a uniformly accelerated motion, for which you have two formulas at disposal (in the forms below or similar):

  • speed increase $\Delta v=v_t-\color{green}{v_0}=\color{green}a\Delta t$,

  • space increase $\color{green}{\Delta s}=s_t-s_0=\dfrac{\color{green}a\Delta^2t}2+\color{green}{v_0}\Delta t$,

and the formula for kinetic energy is

  • $E=\dfrac{\color{green}mv^2}2$.

Looking at the given data, you observe that $\Delta t$ is missing, but you can draw it from the space increase equation, which is of the quadratic type. There will be two roots, one of which should be discarded.

From $\Delta t$, you draw the increase in speed, hence $v_t$, and the increase in energy easily follows.

  • I would call it $(\Delta t)^2$ rather than $\Delta^2t$. It is, after all, the square of the length of one time period, not the change of length from one time period to the next. – Arthur Mar 21 '17 at 08:23
  • @Arthur: yep, my $\Delta^2t$ is to be read as $(\Delta t)^2$, not as a second order finite difference. –  Mar 21 '17 at 08:25
  • 5
    So why don't you write it like that? To me, it seems as strange as insisting that $\frac{d^2f}{dx^2}$ is the square of the derivative of $f$. I am aware of notation like $\sin^2x$, but that doesn't mean one has to perpetuate an ambiguity. – Arthur Mar 21 '17 at 08:27
  • Actually for the difference in energy, I'd use $F=ma$ and $\Delta E = F,\Delta s$. Anyway, calculating it both ways is a good way to check the answer: If the results differ, there must be an error somewhere. – celtschk Mar 21 '17 at 08:28
  • @celtschk: there is no strong reason to prefer one formula over the other, except maybe that the direct computation of the kinetic energy is closer to the problem statement (otherwise you are silently invoking the principle of energy conservation and the result would be wrong for a dissipative system or with other energy terms). –  Mar 21 '17 at 08:39
  • Given that dissipation is also a force and affects the acceleration, my version does not depend on energy conservation. And I didn't say that your way is wrong (I indeed explicitly stated that doing it both ways and making sure that the results agree is a good way to check calculations). However my way has the advantage that it calculates the energy difference directly from the data given, and therefore would not be affected by possible errors in the previous calculation of the final speed (this includes both mistakes and rounding errors). – celtschk Mar 21 '17 at 08:46
2

I assume you are aware of all the concepts that are used in your question, and you are just interested in how to convert everything into an answer to the posed question. Maybe these steps will help you for a general problem like the one you gave:

  1. First extract all the information given in the text. For example, you know that there is an object (a proton) with a certain mass. Also write down the unknown quantity that you are asked to figure out.
  2. Sometimes it is useful to make a sketch of the situation to make it clearer what's going on.
  3. Write down the relevant physical formulas you know. For example, since the question is about velocities and kinetic energy, write down the formulas for speed, acceleration, kinetic energy, etc. that you know.
  4. Use the formulas and given quantities you wrote down in steps 1 and 3 to figure out the unknown quantity that is asked. You might not need every formula you wrote down (depending on how many you wrote down of course), and you might need to combine multiple (one after each other) to arrive at the correct answer. This can feel like a puzzle sometimes, and can be rewarding when you solve it!

I hope this helps you for general word problems. For this specific problem, look at Yves' answer.

Sanderr
  • 523
1

I would start by assigning a pronumeral to each value in the verbal description and then writing out the value of each pronumeral as a list. Then, relate the variables with an equation and proceed to solve for unknowns.

For instance, here we are dealing with initial and final velocities. A high school convention I remember was to put $u = $ initial velocity and $v$ = final velocity, but you can use anything you want that makes sense to you e.g. $v_0$, $v_i$, $v_f$ etc.

Then, write out all of the information in the expression in a list:

E) $m = 1.67 * 10^{-27} $ kg = mass

$a = 3.6 * 10^{15}$ m s$^{-2}$ = acceleration

etc.

And finally, write out the equation that relates the variables in question e.g.

$$v^2 = u^2 + 2as$$

and substituting your defined pronumerals allows for an algebra setup in only one variable (or sometimes two equations in two variables which requires simultaneous equations), for instance

$$5^2 = u^2 + 2(-10)(1)$$

$$25 = u^2 - 20$$ etc. which can be solved for $u$.

Side note: remember to tag questions like yours with "soft question" as you're asking for help around the topic rather than with a specific question.

M.Diggerson
  • 222