OpenStudy (anonymous):

CALC QUESTION: (Will medal) An object moves along the x-axis with velocity of v(t) = t(sqrt(2t+3)) -3t For which 't' is the object moving to the right?

1 year ago
OpenStudy (anonymous):

Will I have to integrate?

1 year ago
OpenStudy (anonymous):

$v(t) = t \sqrt{2t+3} - 3t$

1 year ago
OpenStudy (anonymous):

$\int\limits_{?}^{?}t \sqrt{2t+3} - 3t?????$

1 year ago
OpenStudy (misty1212):

HI!!

1 year ago
OpenStudy (anonymous):

Hi!

1 year ago
OpenStudy (misty1212):

no $$v$$ is the velocity, same as speed except it can be positive or negative you are being asked where (for what value of $$t$$ ) is it positive

1 year ago
OpenStudy (anonymous):

Oh so I use the acceleration I found (second derivative)? Where would I go from there?

1 year ago
OpenStudy (misty1212):

you are thinking too hard

1 year ago
OpenStudy (misty1212):

if the velocity is positive, it is going to the right

1 year ago
OpenStudy (anonymous):

Should I plug in 0 to the velocity function?

1 year ago
OpenStudy (misty1212):

$t\sqrt{2t+3}-3t>0$ solve for $$t$$

1 year ago
OpenStudy (anonymous):

Oh ok!

1 year ago
OpenStudy (anonymous):

I got t > 3. Thank you! (Hope this is right) ((I need to be more confident).

1 year ago
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