Question

need help with partical motion

Answer 1

A particle moves along the x-axis so that its position is given by \[x(t) = t(t-1)^{3}\]
1) at what time is the partical at rest?
2) during what interval is the particle moving right? Justify your answer.
3) find the position of the particle when the acceleration is 0

Answer 2

1. take the derivative of the function and set it equal to zero. That will be your time at particle when v(t) = 0.

2. The particle is moving to the right when v(t) > 0

3. find the 2nd derivative and and set it equal to zero.

Answer 3

this is probably a stupid question, but how do i find the derrivative of the function? im confused because its x(t) = t(t-1)^3 instead of something like = x^4 - 6x^2 + 9

Answer 4

Are you familar with product rule of derivatives?

Answer 5

kinda. do i factor the t into (t-1)^3 ?

Answer 6

\[(f g)' = f'g+ fg'\]
where: \[f = t\] and \[g = (t-1)\]
It might also be helpful if you know the chain rule (in case you don't know): \[\frac{ d }{ dx }(f(g(x))) = f'(g(x))g'(x)\]
where f(x) = \[f(x)=(t-1)^3 = (a)^3 \times g(x) = (t-1)\]

Answer 7

\[3(t-1)^{3-1} \times (t-1)\]

Answer 8

so would i use the chain rule or the product rule? or both?

Answer 9

You use both. Let me set it up for you so you can see what you're doing. You can figure out everything else because it's just algebra:
\[[t \times \frac{ d }{ dt } (t-1)^3]+[(t-1)^3 \times \frac{ d }{ dt } t] = [t \times 3(t-1)^{3-1}]+[(t-1)^3 \times 1]=0\]

Answer 10

I'm sorry. I know this should be super easy, but I'm having a hard time grasping it. so do i then take the derrivative of that?

Answer 11

No. That IS the derivative. I just used product rule along with chain rule. All you have to do is simplify everything.

Answer 12

Simplify: \[[t \times 3(t-1)^{3-1}]+[(t-1)^3 \times 1]=0\] that's just basic algebra and you can do that.