OpenStudy (anonymous):

If an object’s position is described by d(t)=t^3-t^2+t , where d is distance (meters) and t is time (seconds). Which of the following describe the behavior of d ‘’(t)?

3 years ago
OpenStudy (anonymous):

3 years ago
OpenStudy (anonymous):

any options?

3 years ago
OpenStudy (anonymous):

d'' should represent the acceleration of the particle and given that graph i would say it has constant acceleration

3 years ago
OpenStudy (anonymous):

d ‘’(t) is always positive. d ‘’(t) is always negative. d ‘’(t) is positive when d(t) t>1/3 is concave up and d’’(t) is negative when d(t) is concave down for t<1/3 d ‘’(t) is positive when d(t) t<1/3 is concave up and d’’(t) is negative when d(t) is concave down for t>1/3

3 years ago
OpenStudy (math&ing001):

You just have to find the second derivative of d(t). Do you know how to derivate ?

3 years ago
OpenStudy (anonymous):

yeah but what do I derive?

3 years ago
OpenStudy (math&ing001):

You derive t^3-t^2+t

3 years ago
OpenStudy (anonymous):

Okay so its gonna constant POSITIVE ACCELERATION so its always positive

3 years ago
OpenStudy (anonymous):

' = 3t^2 - 2t + 1 '' = 6t -2

3 years ago
OpenStudy (math&ing001):

Yeah ! That's correct

3 years ago
OpenStudy (math&ing001):

Now solve 6t-2>0 for t

3 years ago
OpenStudy (anonymous):

t > -2?

3 years ago
OpenStudy (math&ing001):

Not really...|dw:1388614952569:dw|

3 years ago
OpenStudy (anonymous):

oh I see

3 years ago
OpenStudy (math&ing001):

So what do you think would be the solution from the 4 choices ?

3 years ago
OpenStudy (anonymous):

A

3 years ago
OpenStudy (anonymous):

cuz it's positive....

3 years ago
OpenStudy (math&ing001):

Well if you plug in 0 on the acceleration equation you found before you find d''(0)=6*0-2=-2 so it's not always positive !

3 years ago
OpenStudy (math&ing001):

And @fortune_paul It's true that acceleration is a vector which always has a non negative magnitude. However it may act in one of two directions along its line of action. So we represent that by using the positive or negative sign.

3 years ago
OpenStudy (anonymous):

so it is C?

3 years ago
OpenStudy (math&ing001):

Yes

3 years ago
OpenStudy (anonymous):

ty

3 years ago
OpenStudy (math&ing001):

yw

3 years ago
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