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\[s(t)=2t^3-24t^2+90t+7\]

for what values t is the speed of particle increasing

\[s'(t) \implies velocity \space \space s''(t) \implies acceleration\]

ya i got a(t)=0 i got 4

as critical point

ds/dt=6t^2-48t+90
thus
d^2s/dt^2=12t-48 => aceleration is greater than zero when t>4

well first get the turning point,
you mean s'(t)=0

no i mean s''(t) gives acceleration =0 will give us information about velocity

yes so you got 4 right

yes

So yea it is when t>4 that the acceleration is postivei .e. speed increases

yes

no that not the answer i am looking for

Well its the right one...

but the answer is 35 ( i was shocked !!!!

is it possible that this is wrong?

|dw:1360516136831:dw|

|dw:1360516160914:dw|

But why i need the critical points of velocity when they ask for velocity

This is no calculator question so i cannot graph

they did not say the rate of change in velocity nor speed just velocity increasing

yeah i know what i am saying is\[s'(t)=0\]

i e 0,3,5

then how did they get 4

so you get 3 and 5 right

yes

speed =|velocity|
\[|6t^2-8t+15|>0\]
i would like to explain grapghically 4 a sec

i mean\[|t^2-8t+15|\]

its actually velocity=I speedI

velocity is vector where as speed is scalar

i agree with person 3

did the link help