Asked by Anonymous
Consider a particle moving along the x-axis where
x(t)
is the position of the particle at time t,
x' (t)
is its velocity, and
x'' (t)
is its acceleration.
x(t) = t3 − 12t2 + 21t − 7, 0 ≤ t ≤ 10
(a) Find the velocity and acceleration of the particle.
x' (t) =
x'' (t) =
x(t)
is the position of the particle at time t,
x' (t)
is its velocity, and
x'' (t)
is its acceleration.
x(t) = t3 − 12t2 + 21t − 7, 0 ≤ t ≤ 10
(a) Find the velocity and acceleration of the particle.
x' (t) =
x'' (t) =
Answers
Answered by
Anonymous
Find the open t-intervals on which the particle is moving to the right. (Enter your answer using interval notation.)
Answered by
oobleck
moving to the right means x is increasing, so x' > 0
x = t^3 − 12t^2 + 21t − 7
x' = 3t^2 - 24t + 21 = 3(t^2-8t+7) = 3(t-1)(t-7)
x" = 6t - 24 = 6(t-4)
so, check where x' > 0 (hint: not between the roots, since it is a parabola which opens up)
Remember the domain.
x = t^3 − 12t^2 + 21t − 7
x' = 3t^2 - 24t + 21 = 3(t^2-8t+7) = 3(t-1)(t-7)
x" = 6t - 24 = 6(t-4)
so, check where x' > 0 (hint: not between the roots, since it is a parabola which opens up)
Remember the domain.
Answered by
Anonymous
(0,1)?
Answered by
oobleck
no, duh. it is negative between the roots, which are NOT at 0 and 1!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.