Question

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)?

Answer 1

any options?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

yeah but what do I derive?

Answer 6

You derive t^3-t^2+t

Answer 7

Okay so its gonna constant POSITIVE ACCELERATION so its always positive

Answer 8

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

Answer 9

Yeah ! That's correct

Answer 10

Now solve 6t-2>0 for t

Answer 11

t > -2?

Answer 12

Not really...

Answer 13

oh I see

Answer 14

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

Answer 15

A

Answer 16

cuz it's positive....

Answer 17

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 !

Answer 18

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.

Answer 19

so it is C?

Answer 20

Yes

Answer 21

ty

Answer 22

yw