Instantaneous Velocity Review
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Instantaneous Velocity Review

[From: ] [author: ] [Date: 12-04-18] [Hit: ]
so if you can help Ill certainly reward it best I can.Im reviewing some of the odd questions (admittedly, sounds like doing my homework, but wait!) limit concepts in prep for calculus, and I came across this one,......
(NOTE: I had to delete the other I asked. I'd rather award no best response than let it go to auto-spam. I also do give best answer quick, so if you can help I'll certainly reward it best I can.)

I'm reviewing some of the odd questions (admittedly, sounds like doing my homework, but wait!) limit concepts in prep for calculus, and I came across this one, and want to verify I have the concept. I'll skip straight to my answer now so you know it isn't you doing my work guys!

v(ins) ~ 78.4 ft/s
Rock hits ~ 2.45 s

Now here's the question:

A rock is dropped from a height of 96 feet. The function s(t) (in feet) is equal to 16t^2.
They want when it hits, and to make a conjecture of approximate instantaneous velocity.

So what I did was (omitting units, we know what they are):
96 = 16t^2
6 = t^2 (squareroot)
t ~ 2.45 seconds.

At which point I took the function as..
[s(t) - s(2.45)] / (t - 2.45)
Where s(t) was to be 2.450001.
At which point I received the answer 78.4 ft/s

I just want to verify it's either correctly done or not, and correct any errors.

Thanks!

(btw, try and go easy on me. This is all pretty new, we didn't get TOO far into calculus last quarter. Just trying to understand concepts.)

-
yes, I believe that you correctly did the problem and understand the concept

the idea that you are using is a tangent line approximation with a point extremely close to the time when the rock hits, the closer to 2.45s, the more exact it becomes

instantaneous velocity is the slope of position at any instant of time
also known as the derivative of position which you will learn later in calculus and use to find the exact instantaneous velocity
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