# -16(2a) -> How to isolate variable?

-16(2a) --> How to isolate variable?

## Homework Statement

-16(2a)
How do I isolate the variable? This is probably an EXTREMLY simple question. Its times like this that makes me wish I paid attention in highschool algebra.

## The Attempt at a Solution

Don't really know what to do without violating some basic arithmetic law.

I tried setting it equal to 0 but??

-32a = 0

My professor got the answer of $$a=\frac{5\sqrt{5}}{2}$$

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gb7nash
Homework Helper

Is there an equation associated with this?

In any case, you can simplify a little more. Use the associative law. For any real numbers x,y,z:

x(yz) = (xy)z

What's x? y? z?

Is there an equation associated with this?

In any case, you can simplify a little more. Use the associative law. For any real numbers x,y,z:

x(yz) = (xy)z

What's x? y? z?
I did that, maybe it wasn't very noticiable in my post. Regardless,

-16(2a) = -32a

Edit: No equation, just part of a word problem in my calc class. This is the very last step, just don't know how to solve for a. I have the problem copied from my professor so it isn't an error.

gb7nash
Homework Helper

With the information given, there's no way to properly answer your question. What's the original problem?

With the information given, there's no way to properly answer your question. What's the original problem?
Given the position function, $$s(t)= -16t^2 + 500$$

If a construction worker drops a wrench from a height of 500
feet, when will the wrench hit the ground? At what velocity
will the wrench impact the ground?

$$lim_{t\to a} \frac{s(a) - s(t)}{a-t} = \frac{-16at^2 + 500 - (16t^2 + 500 )} {a-t}$$
$$= \frac{-16a^2+16t^2}{a-t}$$
$$= \frac{-16(a^2 - t)}{a-t} = \frac{-16 (a+t)(a-t)}{a-t}$$
$$= lim_{t\to a} = -16(a+t)$$
$$= - 16(2a)$$

Professor said a = t, and set (a+t) is the same thing as 2a.

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eumyang
Homework Helper

Given the position function, $$s(t)= -16t^2 + 500$$

If a construction worker drops a wrench from a height of 500
feet, when will the wrench hit the ground?
(1) Did you answer this question? If not, when the wrench hits the ground, what does s(t) have to equal? You need to then solve for t.

$$lim_{t\to a} \frac{s(a) - s(t)}{a-t} = \frac{-16at^2 + 500 - (16t^2 + 500 )} {a-t}$$
I learned the formula as this:
$$s'(a) = lim_{t\to a} \frac{s(t) - s(a)}{t-a}$$

Professor said a = t, and set a = 2.
You sure? The a = 2 part isn't right. What you should plug in is the value of t you get from (1) above.

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gb7nash
Homework Helper

Like eumyang stated, there's two parts to this question. First, you need to find the time at which the wrench hits the ground. How do you do this?

Professor said a = t, and set a = 2.
Ignoring the latex typos, your methodology for solving for the velocity is fine. However, you need to approach the time at which the wrench hits the ground. If you properly solve for this time, you won't get 2.

Like eumyang stated, there's two parts to this question. First, you need to find the time at which the wrench hits the ground. How do you do this?

Ignoring the latex typos, your methodology for solving for the velocity is fine. However, you need to approach the time at which the wrench hits the ground. If you properly solve for this time, you won't get 2.

I'm sorry, that was a typo, what I really meant was $$\lim_{t\to a} (t+a) = 2a$$

So when I said a = 2, I meant since t approaches a its the same as 2a. I've been studying and practicing for a while, sometimes you get clumsy.

(1) Did you answer this question? If not, when the wrench hits the ground, what does s(t) have to equal? You need to then solve for t.

I learned the formula as this:
$$s'(a) = lim_{t\to a} \frac{s(t) - s(a)}{t-a}$$

You sure? The a = 2 part isn't right. What you should plug in is the value of t you get from (1) above.
I'm sorry that was a typo, read the above post.

Mark44
Mentor

There are quite a few mistakes here.
Given the position function, $$s(t)= -16t^2 + 500$$

If a construction worker drops a wrench from a height of 500
feet, when will the wrench hit the ground? At what velocity
will the wrench impact the ground?
The first question is the same as asking at what t is s(t) = 0? When you answer that question, you should get the same answer as you posted, which is (5$\sqrt{5}$)/2.

It looks like you skipped that question and went on to try to answer the second question.
$$lim_{t\to a} \frac{s(a) - s(t)}{a-t} = \frac{-16at^2 + 500 - (16t^2 + 500 )} {a-t}$$
s(a) $\neq$ -16at2 + 500.
$$= \frac{-16a^2+16t^2}{a-t}$$
$$= \frac{-16(a^2 - t)}{a-t} = \frac{-16 (a+t)(a-t)}{a-t}$$
$$= lim_{t\to a} = -16(a+t)$$
$$= - 16(2a)$$
Since you actually take the limit in the last expression above, each step preceding it should indicate that you haven't yet taken the limit. IOW, Each expression except the very last should be preceded with $\lim_{t \to a}...$.

Actually, since what you're really doing is finding s'(t), each limit shoud have been $\lim_{a \to t}...$.

BTW, as already noted by other folks in this thread there are a couple of errors in the work above.
Professor said a = t, and set (a+t) is the same thing as 2a.
What you have found here is s'(t), or the velocity at time t for the dropped wrench. You should have found that v(t) = -32t. If you had answered the first question, you could have used that value to find the velocity when the wrench hit the ground.

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Oh my... I was so fixated on figuring out how my professor solved for a, it crossed my mind to flip a page back for the other portion.:

$$-16t^2+500=0$$
$$-16t^2= -500$$
$$\sqrt{t^2}= \frac{\sqrt{125}}{\sqrt{4}}$$

I probably need a break lol..

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