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

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

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

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

ceb105Best ResponseYou've already chosen the best response.1
ds/dt=6t^248t+90 thus d^2s/dt^2=12t48 => aceleration is greater than zero when t>4
 one year ago

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

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

ceb105Best ResponseYou've already chosen the best response.1
So yea it is when t>4 that the acceleration is postivei .e. speed increases
 one year ago

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

ceb105Best ResponseYou've already chosen the best response.1
Well its the right one...
 one year ago

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

ceb105Best ResponseYou've already chosen the best response.1
is it possible that this is wrong?
 one year ago

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

ceb105Best ResponseYou've already chosen the best response.1
Okay well 3 and 5 are the critical points of the velocity
 one year ago

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

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

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

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

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

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

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

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

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

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

ksaimouliBest ResponseYou've already chosen the best response.0
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
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
yes 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
 one year ago
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