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Smart VLR users come

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#1
RickyIndian

what would the instantaneous velocity be at 7.50 seconds and how would i solve instantaneous velocity for other values?

https://imgur.com/a/CGma8pG

edit: solved, thanks for help

#2
aspect1
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Frags
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atleast 2

#3
var1ance
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search it online yw

#9
RickyIndian
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i searched up how to do it and it's making me draw a tangent line

it's online so i obv can't draw it

#37
kzeriar
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vlr.gg is online

#4
Congo1
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solve it yourself

#5
number1_cned_hater
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bros asking vlr.gg users math questions

#6
FNATIC_OMEGALUL
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bro taking physics

#7
s0ber
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dx/dt = V so just x coordinate/time coordinate of the points

#11
RickyIndian
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so -2/7.5 ?

#15
s0ber
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yea but it would +ve since it is moving forward (x coordinate increasing

#8
terribletyranny
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o7 no clue

#10
blueberry
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The instantaneous velocity on a position vs time graph can be found by finding the slope of the line at any given point.

#12
CryoZanderDerrekEnjoyer
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bro really asked a website with an aveage iq of sentinels lan appearances in the past 2 years

#13
EntryKJ
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Instantaneous velocity is dx/dt which is the slope of the graph.
The answer is the slope of the graph at the required instant (t = 7.5).

slope at 7.5 = (0 - -5)/(8-7) = 5/1 = 5

assuming x is -5 at t=7 and x is 0 at t=8.

It depends on being able to read the value of x precisely from the graph. Assuming x to be -2 at 7.5 is also fine, but you'll get a different answer.

#17
RickyIndian
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i included the graph that the question is referring to and the question is asking what the instantaneous velocity would be at t = 7.50 seconds

#19
EntryKJ
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mb, edited

#26
RickyIndian
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i see, thank you very much

#21
Zeah
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ofc indian will answer this with detail

#28
LightningLeaf
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ok bro

#14
bptee
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5

#16
BoF7ooM
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For instantaneous velocity at 7.5 seconds, find the point at which it is 7.5 second and find its gradient. Here from 7.5 seconds to 8 seconds it is a constant slope, so just find the gradient of that line.

#23
RickyIndian
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so at 7.5 seconds, the x value is -2 and at 8 seconds the x value is 0.

(0 - - 2) / (8 - 7.5)?

#27
EntryKJ
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yeah, it depends on the whether x value is actually -2. It's hard to make out from the graph.

I assumed x to be -5 at t=7, but I can't be sure.

#30
RickyIndian
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yup, that worked.

just so i fully understand, if i were to solve for t = 3.0 and t = 4.5, it would go like this

t = 3
(5-10) / (4-2) = -1.25

t = 4.5
(5-5) / (5-4) = 0

#31
EntryKJ
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It's correct (edited)

#32
RickyIndian
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so what should i change in the calculations?

#33
EntryKJ
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It depends on which instant you want to find the instantaneous velocity for. If I want to find the instantaneous velocity at 4.5, I would use t=4 and t = 5 since it's a straight line there, so the slope can be found easily. If I want to find the instantaneous velocity at 3, I will use 2 and 4.

#34
EntryKJ
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wait, mb I didn't read the calculations fully, you are correct for both. Sorry man, I should have read the equations.

#38
RickyIndian
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allg thanks for all your help

#29
BoF7ooM
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Yes. Correct.

#18
Dustlized
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it says that the image is nsfw
MaTh Is NsFW???????????

#20
cocoluna
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bro is in calculus

#22
Poh1238
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average american who needs help of dawg eater's huh?

#24
FLU1D_
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Idk of what instant it's asking the instantaneous velocity of but whatever it is just write the equation of line passing through that point and the slope of that line is your instantaneous velocity

#25
lipton13
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Velocity is distance over time. Instantaneous velocity is the velocity at THAT specified moment (in your case, 7.5 secs). So just take what the y-coordinate is at 7.5 secs. Take that y-coordiate/7.5 secs. You have the answer

p.s. (rusty maths so maybe wrong)

#35
H0r0sha
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v=dx/dt
since it is a (x vs t) graph , just find the slope of graph at the required time.
From img:
v= (0-(-2))(8-7.5) = 4
#36
RickyIndian
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answer is 5 actually

(0--5) / (8-7)

#39
H0r0sha
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Yeah, then x is not -2 at t=7.5, I think graph should have more lines to figure it out accurately.

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