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dynamics help
Posted: Mon Mar 09, 2009 11:27 am
by hemagoku
how to solve this ? i got exam after 2 days ,and can't solve this

the word solution means i solve it in there ,not mean there is a solution under there btw
Re: dynamics help
Posted: Mon Mar 09, 2009 12:36 pm
by chickenfeather
i. HINT: if you plot your velocity from 2s to 10s, you will notice that the particle is moving with a negative velocity "backwards" and then it changes direction and moves "forwards". Your distance will be the distance traveled while the particle is moving backwards plus the distance traveled while the particle is moving forwards.
ii. displacement is the integral of your velocity with respect to time from 2s to 10s. Your constant Xo is provided.
iii. these answers can be obtained based on information from parts i and ii.
Re: dynamics help
Posted: Mon Mar 09, 2009 12:41 pm
by hemagoku
chickenfeather wrote:i. HINT: if you plot your velocity from 2s to 10s, you will notice that the particle is moving with a negative velocity "backwards" and then it changes direction and moves "forwards". Your distance will be the distance traveled while the particle is moving backwards plus the distance traveled while the particle is moving forwards.
ii. displacement is the integral of your velocity with respect to time from 2s to 10s. Your constant Xo is provided.
iii. these answers can be obtained based on information from parts i and ii.
question ,how u know its going backwards first not forward first ? also to get distance in backwards and forward to add them ?
last on ,isn't avg speed is avg velocity ?
Re: dynamics help
Posted: Mon Mar 09, 2009 1:20 pm
by chickenfeather
hemagoku wrote:question ,how u know its going backwards first not forward first ? also to get distance in backwards and forward to add them ?
last on ,isn't avg speed is avg velocity ?
Since the velocity vector is one dimensional, I made the assumption that negative velocity would mean that the particle is moving "backwards" and positive velocity would mean that the particle is moving "forwards". This is just something I assumed. You can have it going the other way as well.
To get the backwards and forward distance, you would have to integrate the velocity in parts, with respect to the times that the particle is going either direction.
For example: if the particle was going backwards from 2s to 5s and then moving forward from 5s to 10s, you would have two integrals evaluated from 2s to 5s and from 5s to 10s. I think you would then have to take the absolute value to make sure your distances are positive.
Avg. speed is based on your distance.
Avg. velocity is based on your displacement.
Re: dynamics help
Posted: Mon Mar 09, 2009 1:29 pm
by MrJoey
It occurs to me to tell you, that if you're taking a class in the first place that has this question on the exam, you should either A. know it already, or B. know how to look it up in the text.
That being said, kudos to chickenfeather for giving hints and help on how to do it rather than just outright saying the answer.
Re: dynamics help
Posted: Mon Mar 09, 2009 2:44 pm
by hemagoku
am grateful ,just to check is displacement =1680 ?
Re: dynamics help
Posted: Mon Mar 09, 2009 8:03 pm
by SM-Count
Actually...

Edit: I'm working under the assumption they want displacement relative to the origin, which is mostly the case. When I say s'(x)=v(x) I mean s'(t)=v(t), typo. Btw, I got 708 m for displacement when I did it in my head... you might want to check that though.
Re: dynamics help
Posted: Mon Mar 09, 2009 9:15 pm
by hemagoku
SM-Count wrote:Actually...

Edit: I'm working under the assumption they want displacement relative to the origin, which is mostly the case. When I say s'(x)=v(x) I mean s'(t)=v(t), typo. Btw, I got 708 m for displacement when I did it in my head... you might want to check that though.
waaaa u rock !! .but i got displacement = 688.and distance = 992 ,i multiplied the 6 with whole the inside though maybe got confused and did something wrong
Re: dynamics help
Posted: Mon Mar 09, 2009 9:41 pm
by SM-Count
Don't forget, when you do displacement relative to the origin, you started with an extra 20 meters, therefore you need to add that to the 688. Also, I should warn you, I'd integrate with a calculator when doing integrals.
Re: dynamics help
Posted: Mon Mar 09, 2009 9:51 pm
by hemagoku
SM-Count wrote:Don't forget, when you do displacement relative to the origin, you started with an extra 20 meters, therefore you need to add that to the 688. Also, I should warn you, I'd integrate with a calculator when doing integrals.
they don't allow that

,btw i got a question ,in the modulus ,it means value is +ve ,but do i do modulus on just the final answer of the equation or each thing in it (like making it t2+2t+15) ? or is the same ?