ksaimouli
  • ksaimouli
The position of a particle moving along a line is given by
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ksaimouli
  • ksaimouli
\[s(t)=2t^3-24t^2+90t+7\]
ksaimouli
  • ksaimouli
for what values t is the speed of particle increasing
anonymous
  • anonymous
\[s'(t) \implies velocity \space \space s''(t) \implies acceleration\]

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ksaimouli
  • ksaimouli
ya i got a(t)=0 i got 4
ksaimouli
  • ksaimouli
as critical point
anonymous
  • anonymous
ds/dt=6t^2-48t+90 thus d^2s/dt^2=12t-48 => aceleration is greater than zero when t>4
anonymous
  • anonymous
well first get the turning point, you mean s'(t)=0
ksaimouli
  • ksaimouli
no i mean s''(t) gives acceleration =0 will give us information about velocity
anonymous
  • anonymous
yes so you got 4 right
ksaimouli
  • ksaimouli
yes
anonymous
  • anonymous
So yea it is when t>4 that the acceleration is postivei .e. speed increases
anonymous
  • anonymous
yes
ksaimouli
  • ksaimouli
no that not the answer i am looking for
anonymous
  • anonymous
Well its the right one...
ksaimouli
  • ksaimouli
but the answer is 35 ( i was shocked !!!!
anonymous
  • anonymous
is it possible that this is wrong?
ksaimouli
  • ksaimouli
@hartnn
anonymous
  • anonymous
|dw:1360515901447:dw| graph increasing the speed inceases when the grapg rises so xb
anonymous
  • anonymous
Okay well 3 and 5 are the critical points of the velocity
anonymous
  • anonymous
|dw:1360516136831:dw|
anonymous
  • anonymous
|dw:1360516160914:dw|
ksaimouli
  • ksaimouli
But why i need the critical points of velocity when they ask for velocity
ksaimouli
  • ksaimouli
This is no calculator question so i cannot graph
anonymous
  • anonymous
they did not say the rate of change in velocity nor speed just velocity increasing
anonymous
  • anonymous
yeah i know what i am saying is\[s'(t)=0\]
ksaimouli
  • ksaimouli
i e 0,3,5
ksaimouli
  • ksaimouli
then how did they get 4
anonymous
  • anonymous
so you get 3 and 5 right
ksaimouli
  • ksaimouli
yes
anonymous
  • anonymous
speed =|velocity| \[|6t^2-8t+15|>0\] i would like to explain grapghically 4 a sec
anonymous
  • anonymous
i mean\[|t^2-8t+15|\]
ksaimouli
  • ksaimouli
its actually velocity=I speedI
ksaimouli
  • ksaimouli
velocity is vector where as speed is scalar
anonymous
  • anonymous
similar quest http://math.stackexchange.com/questions/116405/the-position-of-a-particle-moving-along-a-line-is-given-by-2t3-24t290t-7
anonymous
  • anonymous
i agree with person 3
ksaimouli
  • ksaimouli
i found out that they used velocity and acceleration derivatives and they set them =0 and found the answer but i have learned that to find graph of y' use y'' to find max or min
anonymous
  • anonymous
yes so we found \[|t^2-8t+15|>0 \implies t<3 ,t>5\] but also for - velocity we put \[-(t^2-8t+15)\] its positive for \[-(2t-8)>0 \implies t<4\] (3,4) and x>5
anonymous
  • anonymous
did the link help

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