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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
So yea it is when t>4 that the acceleration is postivei .e. speed increases
no that not the answer i am looking for
Well its the right one...
but the answer is 3
5 ( i was shocked !!!!
is it possible that this is wrong?
Okay well 3 and 5 are the critical points of the velocity
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
speed =|velocity| \[|6t^2-8t+15|>0\] i would like to explain grapghically 4 a sec
its actually velocity=I speedI
velocity is vector where as speed is scalar
similar quest http://math.stackexchange.com/questions/116405/the-position-of-a-particle-moving-along-a-line-is-given-by-2t3-24t290t-7
i agree with person 3
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
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
did the link help