Calculus Problem: Dealing with velocity i think
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Calculus Problem: Dealing with velocity i think

[From: ] [author: ] [Date: 12-04-22] [Hit: ]
Then average the two to get a new estimate.please help!!! i dont understand what to do! If answer please be descriptive so i can kinda make something out of it to finish the rest of my problems!......
for time, t, in hours, 0 <(with the line under) t < (with the line under) , a bug bug is crawling at a velocity, v, un meters/hours given by

v= 4/9+t

use triangle t= .2 to estimate the distance that the bugs crawls during this hour. Find over estimate and an underestimate. Then average the two to get a new estimate.


please help!!! i don't understand what to do! If answer please be descriptive so i can kinda make something out of it to finish the rest of my problems!! thanks in advance :)

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Its okay Lila! I always forget what that triangle thingy is too :) as long as you know what it means :D

so... lets begin


so i want you to get a mental picture. Draw a straight line. Number it from 0-1, now you will have 6 points which are 0, .2, .4, .6, .8, and 1. this is your t :)

so v= 4/9+t so plug in your numbers t for each one since it is an underestimate its from 0-.8

this is how it should look!

(4/9+0 + 4/9.2 +4/9.4 +4/9.6 +4/9.8)= get your answer and multiply by delta t which is .2
your answer should be .417028891m

do the same exact thing for the over estimate. The overestimate is from 0.1-1

add those numbers up and multiply by .2

you should get .42591778m for overestimation.

and to find the average add up them both and divide by two :)

I know this helped :DD

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I'm not exactly sure what you mean by 'use' the triangle. But if you are trying to find the distance traveled you must integrate velocity. this will give you the displacement of the object (in this case the bug)

If you want to find the total distance traveled, you must ingrate the absolute value of velocity

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Triangle symbol is called delta. You really should know this by now.

Let d = distance
v = ∆d/∆t
∆d = v * ∆t
∆d = (4/9+t) * ∆t

∆t = 0.2, 0 ≤ t ≤ 0.2

t = 0 -----> ∆d = (4/9 + 0) * 0.2 = 4/45
t = 0.2 -----> ∆d = (4/9 + 0.2) * 0.2 = 29/225
avg = (4/45 + 29/225) / 2 = 49/450 = 0.108888889

However, if all of 9+t is in denominator, i.e. v = 4/(9+t) we get

t = 0 -----> ∆d = (4/(9+0)) * 0.2 = 4/45
t = 0.2 -----> ∆d = (4/(9+0.2)) * 0.2 = 2/23
avg = (4/45 + 2/23) / 2 = 91/1035 = 0.087922705

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v=0.6444
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