Off Topic"If you don't like your job, you don't strike. You just go in every day and do it really half-assed. That's the American way."
Welcome to civicforums.com!
Welcome to civicforums.com.
You are currently viewing our forum as a guest, which gives you limited access to view most discussions and access our other features. By joining our community, at no cost, you will have access to start new topics, reply to conversations, privately message other members (PM), respond to polls, upload content and access many other special features. Registration is free, fast and simple, so please join civicforums.com today!
Can someone please help on this problem? I cant figure it out.
1. A particle moves along the x-axis so that its velocity at any time t>0 is given by v(t)=3t^2-2t-1. The position x(t) is 5 for t=2
a. Write a polynomial expression for the position of the particle at any time t>0
b. For what values of t, 0<t<3, is the partilce's instantaneous velocity the same as its average velocity on the closed interval [0,3]
c. Find the total distance traveled by the particle from time t=0 until time t=3
To remove this ad, register today for free or log in if already registered!
Sponsored Links
To avoid seeing this ad in our forum please register at CivicForums.com
By joining our free community you will have access to post topics, respond to polls, upload content and access many other special features.
Quote
[hr]Originally posted by: stan536
Can someone please help on this problem? I cant figure it out.
1. A particle moves along the x-axis so that its velocity at any time t>0 is given by v(t)=3t^2-2t-1. The position x(t) is 5 for t=2
a. Write a polynomial expression for the position of the particle at any time t>0
b. For what values of t, 0<t<3, is the partilce's instantaneous velocity the same as its average velocity on the closed interval [0,3]
c. Find the total distance traveled by the particle from time t=0 until time t=3[hr]
a) d(t)= t^3-t^2-t+3
b) solve this: 6t-2=(1/3)(integral of v(t) from 0 to 3)
c) Integrate |t^3-t^2-t+3| from 0 to 3. Yeah, it's an absolute value.
Someone please double check me. I did that in my head.
Is the particle alone or with another particle? That would effect the probability of the polynominal expression of the x-axis divided by 4 then multiplied by 7.9853%.
In doing some numbers, it looks as though RyzRocket has the right idea, but I think the numbers and variables are suspect.... I'm going to rekindle my memory on this and I'll get back to you....
ok velocity is the derivative of position. therefore,
if v(t) = 3t^2-2t-1, the integral of that (which is the position formula) will be t^3-t^2-t + c (a constant).
since x(2) = 5, and x(t) = t^3-t^2-t+c, c=3
average velocity is v(final) - v(initial)/t(final) - t(initial), which is 20-(-1)/3-0 which equals 7.
instantaneous velocity you just plug in numbers for t, so solving for t where v(t) = 7, the instantaneous velocity is equal to the average velocity at t=2 (also at t=-4/3 but thats not in the domain)
total distance traveled by the particle from t=0 to t=3, you just plug in 3 to the position formula above, because initial position is 0. the answer to that part is 18.
check my math plz [IMG]i/expressions/face-icon-small-wink.gif[/IMG]
Quote
[hr]Originally posted by: dragoon
ok velocity is the derivative of position. therefore,
if v(t) = 3t^2-2t-1, the integral of that (which is the position formula) will be t^3-t^2-t + c (a constant).
since x(2) = 5, and x(t) = t^3-t^2-t+c, c=3
average velocity is v(final) - v(initial)/t(final) - t(initial), which is 20-(-1)/3-0 which equals 7.
instantaneous velocity you just plug in numbers for t, so solving for t where v(t) = 7, the instantaneous velocity is equal to the average velocity at t=2 (also at t=-4/3 but thats not in the domain)
total distance traveled by the particle from t=0 to t=3, you just plug in 3 to the position formula above, because initial position is 0. the answer to that part is 18.
check my math plz [IMG]i/expressions/face-icon-small-wink.gif[/IMG][hr]
Actually, average velocity is the change in displacement over the change in time, not the change in velocity over the change in time, so all of that is wrong after what you said there
for b, it is 18 meters / 3 seconds = 6 meters per second is the average velocity so plug that into the original equation to find then instantaneous equals average. I am in Diff. Eq. right now has a senior in H.S. and I am getting real tired of math already.
were reviewing for finals, our final is going to be in AP style format. My teacher makes every assignment worth alot, or at least what she thinks is important. I need this problem get a "B" in that class, I had like three other problems I figured out how to do them, but this one I just dont remember how to do.
