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Next: Vectors Up: Motion with constant acceleration Previous: Solution

Relations between other quantities

The last two examples, in sections 3.3.5 and 3.3.4 posed a different kind of problem then we considered before. We started off in section 3.3.2 by asking

The last two examples asked the question

Which is an equally legitimate question. Do we want to go through the same rigmarole as in the last example? It'd be better to find the general formula relating the above quantities a, tex2html_wrap_inline1490 , tex2html_wrap_inline1466 , v, and x. We almost had that formula but we blew it, I'm sorry to say. Yes, I'm going to get preachy again and mention yet another rule that you should always follow, (well maybe not always, but most of the time).

Do not substitute numerical values into formulas until you've got to the last stage of the problem. Do everything in terms of symbols, not numbers.

I'll illustrate that for you with the problem at hand. Eq. 3.17 gave the general expression relating t to v, tex2html_wrap_inline1466 and a. We screwed up getting a nice general formula by foolishly substituting in for the numerical values of quantities, giving us tex2html_wrap_inline1608 . That wasn't necessary at all. We could have waited to do the substitution. Let's see what happens if we do that now.

Following the same logic we used in that example, the next thing to do was to substitute t into eq. 3.14. Instead of substituting in 1s let's substitute what's after the second equals sign in eq. 3.17, and keep all other quantities as symbols.

equation250

This can easily be simplified. It's just a little algebra. You finally get

  equation254

There we are. A general formula relating a, tex2html_wrap_inline1490 , tex2html_wrap_inline1466 , v, and x! That's what we wanted. With this formula, you can solve a lot of problems similar to examples 3.3.4 and 3.3.5 quite simply.

Another reason why doing things in symbols is so important is that it allows you to check you answers to see if they make sense. Look at eq. 3.20. Check what happens if a=0. Well then you see that the speed of the particle doesn't change. That sounds correct. How about if tex2html_wrap_inline1626 ? Yes the initial and final speeds are the same in that case also. You know in these limits, the answer is right. Also you can easily check that the units work out. This gives you some confidence that the formula is indeed correct. If you had instead plugged in the time as we did in the previous example (3.3.5), you couldn't check all these limits and would be far more likely to make some mistake.

It takes a while to get used to solving problems with symbols instead of numbers. It seems far too abstract for most people at first. You can get over this problem by using the same approach we did here. First solve the problem the more comfortable way by plugging in numbers as you go along. But after that go back, and go through the same steps but this time keep all your symbols. It's the same logic in both cases. You're just replacing a lot of multiplication and addition with algebraic manipulations.


next up previous
Next: Vectors Up: Motion with constant acceleration Previous: Solution

Joshua Deutsch
Mon Jan 6 00:05:26 PST 1997