Applications And Problem Solving

Applications And Problem Solving-50
\[\begin25t\ 20\left( \right)\end\] At this point a quick sketch of the situation is probably in order so we can see just what is going on.In the sketch we will assume that the two cars have traveled long enough so that they are 300 miles apart. That means that we can use the Pythagorean Theorem to say, \[ = \] This is a quadratic equation, but it is going to need some fairly heavy simplification before we can solve it so let’s do that.Since statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies, still apply.

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So, let’s convert to decimals and see what the solutions actually are.

\[x = \frac = 7.2892\hspace\hspacex = \frac = - 10.2892\] So, we have one positive and one negative.

Also, we are going to assume that you can do the quadratic formula work and so we won’t be showing that work.

We will give the results of the quadratic formula, we just won’t be showing the work.

Let \(t\) be the amount of time it takes the first machine (Machine A) to stuff a batch of envelopes by itself.

That means that it will take the second machine (Machine B) \(t 1\) hours to stuff a batch of envelopes by itself.Now, we also know that area of a rectangle is length times width and so we know that, \[x\left( \right) = 75\] Now, this is a quadratic equation so let’s first write it in standard form.\[\begin 3x & = 75\ 3x - 75 & = 0\end\] Using the quadratic formula gives, \[x = \frac\] Now, at this point, we’ve got to deal with the fact that there are two solutions here and we only want a single answer.Also, as we will see, we will need to get decimal answer to these and so as a general rule here we will round all answers to 4 decimal places.So, we’ll let \(x\) be the length of the field and so we know that \(x 3\) will be the width of the field.Now, from a physical standpoint we can see that we should expect to NOT get complex solutions to these problems. Two hours later the second car starts driving east at 20 mph.Upon solving the quadratic equation we should get either two real distinct solutions or a double root. How long after the first car starts traveling does it take for the two cars to be 300 miles apart?\[t = \frac = 10.09998\hspace\hspace\hspace\,\,\,\,\,\,\,\,\,\,\,t = \frac = - 8.539011\] As with the previous example the negative answer just doesn’t make any sense.So, it looks like the car A traveled for 10.09998 hours when they were finally 300 miles apart.From the stand point of needing the dimensions of a field the negative solution doesn’t make any sense so we will ignore it. The width is 3 feet longer than this and so is 10.2892 feet. In this case this is more of a function of the problem.Notice that the width is almost the second solution to the quadratic equation. For a more complicated set up this will NOT happen.


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