Exp Growth and Decay Q (1 Viewer)

[Shuter]

I can normally do them but this one I can't for some reason.

The formula for the acceleration of a particle is given by A = 5x - (e^2x) + 3, where x is the dispacement of the particle. If the acceleration of the paritcle is at a constant rate -9.8m/s/s, find the rate of change of it's displacement when the displacement is -0.3m.

 

[sub]

use a = d((v^2)/2)/dx... go on from there...i think thats the right method...

 

[Shuter]

That's what I thought, but that seemed too complex for what is meant to be a 2unit question.

 

[sub]

does it work tho? cos thats all that matters...

i dun think its 2u...2u exp groth and decay is just sunbstitution, so its atleast a 3u q...but yeah, do it and see if it works...

 

[mojako]

A = 5x - (e^2x) + 3, where x is the dispacement of the particle. If the acceleration of the paritcle is at a constant rate -9.8m/s/s

That doesn't make sense.
Given the A formula, A won't be constant as the particle moves.

Also this is not exponential growth and decay.

 

[CrashOveride]

Since we're talking about motion, i was wondering what's the best way to determine things like "will the particle ever turn back" or something. usually if its easy to do u cud just find x in terms of t and test as t-->inf, but what about other situations. for eg, there was a question few years back where u had to like graph the velocity^2 against time to determine it. id rather not graph stuff, so any better suggestions? [ this was question 6bii) 1998 ]

 

[Rorix]

if v=0 at any time t, the particle may turn (you need to use your head here, like if v= 1-sinx or something you know it won't)

 

[mathock]

yeah, if u have a general idea of what the graph looks like it makes it a lot easier

 

[jumb]

If you did physics this is alot easier

 

[Dougie]

yeah, it's only now we physics ppl get the adv.

 

[Shuter]

That doesn't make sense.
Given the A formula, A won't be constant as the particle moves.

Also this is not exponential growth and decay.

I know that's what I thought, it's question 10 from Ex6.2, Maths In Focus: Ext1, Book2.

And right you are, it's rates of change, not exp growth.

 

[becany]

whats the answer to this question? i got -2.51m..
ur given dA/dt = -9.8 and u wanna find dx/dt, which is dx/DA times dA/dt
and since ur given the eqn for A, dA/dx equals blah..flip it, then times it and sub in x = -0.3?

 

[akira276]

wut da?? it seys acceleration is a constant....in otha words dv/dt=-9.8 not da/dt....i dun fink there is such a thing as da/dt in da course...

 

[Shuter]

whats the answer to this question? i got -2.51m..
ur given dA/dt = -9.8 and u wanna find dx/dt, which is dx/DA times dA/dt
and since ur given the eqn for A, dA/dx equals blah..flip it, then times it and sub in x = -0.3?

Yes you're right it says -2.5m/s in the back of the book.

 

[jumb]

wut da?? it seys acceleration is a constant....in otha words dv/dt=-9.8 not da/dt....i dun fink there is such a thing as da/dt in da course...

da/dt is the speed of acceleration, ie, acceleration is changing. Where as it's not. Also use full stop and commas instead of ellipsis. They're annoying.

Edit: Also, in 3 unit, gravity is -10

 

[gordo]

wut da?? it seys acceleration is a constant....in otha words dv/dt=-9.8 not da/dt....i dun fink there is such a thing as da/dt in da course...

wat level english do u do?

 

[jumb]

wat level english do u do?

Advanced Sms talk.

It's where they butcher the english language worse then New Zealanders. It's all the rage these days.

 

[mojako]

da/dt is the speed of acceleration, ie, acceleration is changing. Where as it's not. Also use full stop and commas instead of ellipsis. They're annoying.

Edit: Also, in 3 unit, gravity is -10

what?

the original question (assuming typed correctly) said:
"If the acceleration of the paritcle is at a constant rate -9.8m/s/s"

if it's da/dt then it should say "If the acceleration of the paritcle is changing at" and the unit should be m/s/s/s

 

[jumb]

Meh, I just read the last few posts and misinterpreted the quetsion. My bad.

 

[Shuter]

what?

the original question (assuming typed correctly) said:
"If the acceleration of the paritcle is at a constant rate -9.8m/s/s"

if it's da/dt then it should say "If the acceleration of the paritcle is changing at" and the unit should be m/s/s/s

I typed everything exactly correctly. I tihnk the people who wrote the book stuffed up on this one.