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ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0\[s(t)=2t^324t^2+90t+7\]

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0for what values t is the speed of particle increasing

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1\[s'(t) \implies velocity \space \space s''(t) \implies acceleration\]

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0ya i got a(t)=0 i got 4

ceb105
 one year ago
Best ResponseYou've already chosen the best response.1ds/dt=6t^248t+90 thus d^2s/dt^2=12t48 => aceleration is greater than zero when t>4

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1well first get the turning point, you mean s'(t)=0

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0no i mean s''(t) gives acceleration =0 will give us information about velocity

ceb105
 one year ago
Best ResponseYou've already chosen the best response.1So yea it is when t>4 that the acceleration is postivei .e. speed increases

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0no that not the answer i am looking for

ceb105
 one year ago
Best ResponseYou've already chosen the best response.1Well its the right one...

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0but the answer is 3<t<4 and t>5 ( i was shocked !!!!

ceb105
 one year ago
Best ResponseYou've already chosen the best response.1is it possible that this is wrong?

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1dw:1360515901447:dw graph increasing the speed inceases when the grapg rises so x<a and x>b

ceb105
 one year ago
Best ResponseYou've already chosen the best response.1Okay well 3 and 5 are the critical points of the velocity

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0But why i need the critical points of velocity when they ask for velocity

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0This is no calculator question so i cannot graph

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1they did not say the rate of change in velocity nor speed just velocity increasing

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1yeah i know what i am saying is\[s'(t)=0\]

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0then how did they get 4

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1so you get 3 and 5 right

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1speed =velocity \[6t^28t+15>0\] i would like to explain grapghically 4 a sec

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0its actually velocity=I speedI

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0velocity is vector where as speed is scalar

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1similar quest http://math.stackexchange.com/questions/116405/thepositionofaparticlemovingalongalineisgivenby2t324t290t7

ksaimouli
 one year ago
Best ResponseYou've already chosen the best response.0i 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

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1yes so we found \[t^28t+15>0 \implies t<3 ,t>5\] but also for  velocity we put \[(t^28t+15)\] its positive for \[(2t8)>0 \implies t<4\] (3,4) and x>5
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