Question

An object moves along a coordinate line, its position at each time t> 0 given by x (t). Find the acceleration at time t0.
t(0) = 3

Answer 1

Is there any more information to this question?

Answer 2

It definitely seems to be missing something :)

Answer 3

let me3 checl

Answer 4

Maybe a picture?

Answer 5

An object moves along a coordinate line, its position at each time t> 0 given by x (t). Find the acceleration at time t0.
https://assessment.casa.uh.edu/Assessment/Images//1019445.gif

Answer 6

AHH...much better :) One sec while I look it over

Answer 7

Yes...one sec...phone

Answer 8

Is this a calc-based course? Just curious

Answer 9

pre-calc i mean

Answer 10

I'm still with you....just checking over my work here

Answer 11

You're given:
\[x(t)=(t^2-3t)(t^2+3t)\]and
\[x(0)=3\]So I think I'd start off by factoring the equation:\[3=(t^2-3t)(t^2+3t)=t^4-9t^2\]\[t^4-9t^2-3=0\]Since velocity is the derivative of position and acceleration is the derivative of velocity, you need to get the second derivative of this equation:
\[V(t)=X(t)'=4t^3-18t\]\[A(t)=V(t)'=12t^2-18\]
Now that you have the formula for acceleration, you can plug in t = 0 to get -18m/s^2.

Hopefully that's one of your choices :)

Answer 12

lemme check

Answer 13

the answer choices are
a) -54, b) 54, c) 90, d)-14, e)756

Answer 14

did i mention that t(0) = 3

Answer 15

Ok...let me see what I did wrong

Answer 16

Ah...yes. I forgot to plug in 3 as you said. If you plug in 3 to the last equation you get 90

Answer 17

thank you so much.... :)

Answer 18

np :)