Question

what am i doing wrong?

Answer 1

\[a(t)=\frac{ (2t+2)(t+1)^2-(t^2+2t-8)(2t+2) }{ (t+1)^4 }\]

Answer 2

\[a(t)=\frac{ (2t+2)(t^2+2t+1-(t^2+2t-8) }{ (t+1)^4 }\]

Answer 3

\[a(t)=\frac{ (2t+2)(9) }{ (t+1)^4 }\]

Answer 4

\[a(t)=\frac{ 2(t+1)(9) }{ (t+1)^4 }\]

Answer 5

\[a(t)=\frac{ 18 }{ (t+1)^3}\]

Answer 6

its supposed to be -14/(t+1)^3

Answer 7

btw the v(t) equation is..
\[v(t)=\frac{ t^2+2t-8 }{ (t+1)^2 }\]

Answer 8

The first thing you're doing wrong, I'm sorry to say, is that you have not provided any instructions whatsoever for this problem. What are you supposed to do?

Answer 9

just simplify the equation. I'm pretty sure i set it up right but i keep simplifying it incorrectly

Answer 10

You did every single step correctly, I checked it over twice. Maybe the original equation was different?

Answer 11

I will post what my teacher did

Answer 12

You haven't expanded the rightmost (2t+2)

Answer 13

He didn't need to expand it, he distributed it with the leftmost (2t+2)

Answer 14

the original and v(t) are on the left and the a(t) is on the right

Answer 15

I'm pretty sure your teacher just did it wrong. Happens sometimes. She didn't show her work to simplify the equation you had up there.

Answer 16

I'm sorry but still feel that y ou have not recognized nor shared the instructions for this problem. You mention a(t) and v(t), which usually represent acceleration and velocity respectively. Were you supposed to obtain one from the other? To solve a problem, it is essential that you yourself know precisely what your goal is.

Answer 17

They did it a different way and it makes sense, but I'm just wondering why my method doesn't work?

Answer 18

See my comment above: You seem unsure regarding what you are supposed to do.

Answer 19

Yeah I'm doing particle motion but I'm just trying to find my acceleration equation which i keep messing up

Answer 20

Post the ENTIRE question so we can understand it and help.

Answer 21

After i can get that i can answer the rest of the questions. s(t) is the original (position/time), v(t) is deriv (velocity, slope of the first), and a(t) is the 2nd deriv (slope of the velocity)

Answer 22

A particle is moving along a horizontal line with position function as given. Find the following
a) time intervals when particle is moving right and moving left. I did this
b) times when particle is stopped. i did this one too (t=2)
c) time intervals when the particle is speeding up and slowing down.
\[s(t)=t+\frac{ 9 }{ t+1}+1\]

Answer 23

And how are you supposed to find a(t)? In the example you've posted, the three different quantities s(t), v(t) and a(t) show up. They are related to one another through differentiation (or, in reverse, through integration.) Next time, kindly mention that you are "doing particle motion."

You don't have to take my advice, but I suggest you review this post of yours and see how much time has been wasted because your instructions were not shared with the rest of us and you did not say "particle motion" up front.

Would you please repost this question with proper instructions if you still need and want help with it.