Q9b - Acceleration-time graph (1 Viewer)

[daniel_fank]

That's the same reasoning as Intuition.
But if a particle is stationary and then decelerates, does it necessarily go backwards?

if something is at a standstill and deccelerates (which implies accelerating backwards) then it does move backwards. i think i had a brain fart :\ did you understand my reasoning though?

daniel

 

[G'day-john]

um, not sure if im too late for any1 to care, but after dinner my friend who does f 4 unit maths said he will try settle the q9b dispute, i will post them then.

.::AlL iN aLl YoUr JuSt AnOtHeR bRiCk In ThE wAlL::.

 

[undo]

yep i had t=4 as well.

 

[~ ReNcH ~]

if something is at a standstill and deccelerates (which implies accelerating backwards) then it does move backwards. i think i had a brain fart :\ did you understand my reasoning though?

daniel

Doesn't decelerating imply that you have first accelerated and then slowed down though? But then again, that's only in reality and this is maths...which is only theoretical.

 

[undo]

nope. you can decelerate from 0 velocity. you get a negative velocity ie. backwards.

 

[~ ReNcH ~]

Ok.
I think I got that one wrong then.
I wonder if I can still get 1 mark...

 

[JayWalker]

(i) The particle doesn't slow down at t=1, acceleration is still +ve but decreasing which means that the particle is still getting faster, but just at a slower rate.

(ii) Look at Jay's reasoning, which is what I used for this part i.e. t=5s.

Glad my reasoning could be of some help..
The max velocity would be when the graph crossed the time axis, i.e when a = 0, i think that occured at t = 2 i think i cant remember, i have the test at home, and will be home in about 2 hours, if you have any questions just pm me

 

[JayWalker]

That's the same reasoning as Intuition.
But if a particle is stationary and then decelerates, does it necessarily go backwards?

If a particle is stationary and it 'decelerates', then it is essentially accelerating in the -ve direction..

Rench either refer to my previous post or read this,

It starts accelerating at a constant rate, meaning the velocity is getting more and more +ve faster until the line crosses the t axis, and a goes from + to -, HOWEVER,
The velocity is still positive until t = 5 where the areas cancel out, this is when the velocity is zero, after that, its going backwards..

Just think of it in a car, and acceleration is how much you put your foot on the gas..

I wouldn't bother john, you do this in physics also, trust me its right....

 

[O-B-1 Ken-O-B]

Woah wait a minute...

Part one: I put down 3<=t<=5; because the velocity is the greatest when acceleration is at is minimum, since a=0 is not exactly the minimum value but a=-3 is the minimum value for the acceleration function.

Part two: 0<=t<=1; Remember that the question stated that the particle is initially at rest (ie: t=0)? Relating this information to the graph at t=0, the acceleration is also at its maximum (a=3). Therefore when the acceleration is at is maximum, and when the velocity is at rest (v=0), that means that the particle stopped and is changing direction, therefore the particle is at it's furthest from the origin.

Note: The relationship between, acceleration (x**), velocity (x*) and displacement (x) (a = amplitude):
(x=a, x*=0, x**=max) (x=0, x*=max, x**=minimum).

 

[lucyinthehole]

damn. i got 1 right but not 2 maybe they'll still like my reason, and think that i meant t=4.... *shifty*

 

[monkey187]

I dont think you can decelerate from 0, de celerating or accelerating implies changing speed it does not mean moving a direction does it?

 

[skypryn]

yeah pretty sure it's t=4. just add area under and above the curve. at t=4 they cancel ie. particle is stationary. then it accelerates towards the origin, making it closer at t=4 than at t=5

 

[rara]

i got 9(b)i t=2
and 9(b)ii t=3

I did it by drawing the velocity and displacement graphs.
The velocity graph is max at 2
The displacement graoh is at max at 3

Dont know if this is right

 

[garry]

Your reasoning is looking more correct to me, but the only thing I'm thinking about is the fact that acceleration is negative. If a particle decelerates from a stationary position, does velocity necessarily become negative and hence make it go backwards?

yeah i got t=4 as well, my reason was it cancels out the acceleration given from 0-2 so the particle should be stop at 4 and start accelerate back to origin

 

[bevstarrunner]

you can have negative acceleration...velocity and acceleration are vectors, hence have both size and direction...it just means accelerating normally in the other direction

I also got t=2 then t=4

 

[bahodl]

i got 9(b)i t=2
and 9(b)ii t=3

I did it by drawing the velocity and displacement graphs.
The velocity graph is max at 2
The displacement graoh is at max at 3

Dont know if this is right

i also drew the velocity and displacement graphs but came up with t=4

 

[Charon.]

yeah that sound like wat i put. I hope u know wat ur talking about

 

[JayWalker]

you can have negative acceleration...velocity and acceleration are vectors, hence have both size and direction...it just means accelerating normally in the other direction

I also got t=2 then t=4

Yeah that sounds right

 

[bevstarrunner]

correct me if i am wrong, but are they right..??
(not to scale of course, but the general shapes??)

 

[~ ReNcH ~]

Velocity-time looks right. Not sure about displacement coz I haven't drawn it up.

BTW. For you guys who got that question, did you also get out all of Q10? Coz IMO Q9 was the hardest, so I'm assuming guys who got Q9 got Q10 as well.