The position of a particle moving along a line is given by

- ksaimouli

The position of a particle moving along a line is given by

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- ksaimouli

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

- ksaimouli

for what values t is the speed of particle increasing

- anonymous

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

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## More answers

- ksaimouli

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

- ksaimouli

as critical point

- anonymous

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

- anonymous

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

- ksaimouli

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

- anonymous

yes so you got 4 right

- ksaimouli

yes

- anonymous

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

- anonymous

yes

- ksaimouli

no that not the answer i am looking for

- anonymous

Well its the right one...

- ksaimouli

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

- anonymous

is it possible that this is wrong?

- ksaimouli

@hartnn

- anonymous

Okay well 3 and 5 are the critical points of the velocity

- anonymous

|dw:1360516136831:dw|

- anonymous

|dw:1360516160914:dw|

- ksaimouli

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

- ksaimouli

This is no calculator question so i cannot graph

- anonymous

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

- anonymous

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

- ksaimouli

i e 0,3,5

- ksaimouli

then how did they get 4

- anonymous

so you get 3 and 5 right

- ksaimouli

yes

- anonymous

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

- anonymous

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

- ksaimouli

its actually velocity=I speedI

- ksaimouli

velocity is vector where as speed is scalar

- 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

i agree with person 3

- 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

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

did the link help

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