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ksaimouli
 2 years ago
The position of a particle moving along a line is given by
ksaimouli
 2 years ago
The position of a particle moving along a line is given by

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

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

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

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

ceb105
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1well first get the turning point, you mean s'(t)=0

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

ceb105
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0no that not the answer i am looking for

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

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

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

Jonask
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Okay well 3 and 5 are the critical points of the velocity

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

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

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

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

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

Jonask
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0its actually velocity=I speedI

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

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

ksaimouli
 2 years 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
 2 years 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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