Q9b - Acceleration-time graph (1 Viewer)

[~ ReNcH ~]

What did you guys get for that question - it was frustrating me during the exam.
This is what I got:
(i) t=2s where a=0
(ii) t=5s because the particle never moves backwards, just slows down.

What did you guys get?
This and Q9c(iii) were the ones that troubled me most.

 

[Sweet_Lemon]

i stuffed part ii i think~~ got something like t=0....definitely wrong~~ =.=|||

 

[JayWalker]

What did you guys get for that question - it was frustrating me during the exam.
This is what I got:
(i) t=2s where a=0
(ii) t=5s because the particle never moves backwards, just slows down.

What did you guys get?
This and Q9c(iii) were the ones that troubled me most.

Answer is t = 5 because area underneath acceleration vs time graph is velocity. Therefore up until t = 5, velocity was +ve..

think about it, you can be decelerating but still going forward (like in a car breaking / slowing down)

 

[Wild Dan Hibiki]

dammit i got ii) wrong

 

[~ ReNcH ~]

Answer is t = 5 because area underneath acceleration vs time graph is velocity. Therefore up until t = 5, velocity was +ve..

think about it, you can be decelerating but still going forward (like in a car breaking / slowing down)

That was for (ii)? In that case my reasoning and my response were correct - yay.
How about (i) - or was t=5s for (i)?

My reasoning in full:
(i) t=2s:
For 0<t<=2, the acceleration is +ve which means the velocity is increasing.
For 2<t<=5, the acceleration is -ve which means the velocity is decreasing.
Which means that the maximum velocity occurs @ t=2.

(ii) t=5s
Basically the reasoning that was given by Jay.

 

[smegthehead]

9 b) i)

its when t = 1, as after this point the particle begins to slow down and after t = 2, starts to go backwards

b) ii)

its when t = 2.

if you look at the symmetry of the graph, the particle ends up right back at the origin at t = 4. At t = 5, the particle is only the same distance away from the origin as it was at t = 1. hence t=2 is the furthest displacement from the origin as this is where it actually stops

 

[~ ReNcH ~]

9 b) i)

its when t = 1, as after this point the particle begins to slow down and after t = 2, starts to go backwards

b) ii)

its when t = 2.

if you look at the symmetry of the graph, the particle ends up right back at the origin at t = 4. At t = 5, the particle is only the same distance away from the origin as it was at t = 1. hence t=2 is the furthest displacement from the origin as this is where it actually stops

(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.

 

[intuition]

Sorry, it doesnt go backwards after t = 2.
It just SLOWS down, but it's still moving in a positive direction

The max velocity occurs when acc = 0, hence when t = 2

I believe the max distance from origin is 4 sec
At this instant, the positive and negative accelerations cancel each other out and RIGHT after this instant is when the particle starts going backward (because it still carries negative acceleration for another sec up to t = 5)

9 b) i)

its when t = 1, as after this point the particle begins to slow down and after t = 2, starts to go backwards

b) ii)

its when t = 2.

if you look at the symmetry of the graph, the particle ends up right back at the origin at t = 4. At t = 5, the particle is only the same distance away from the origin as it was at t = 1. hence t=2 is the furthest displacement from the origin as this is where it actually stops

 

[bloodysunday]

9 b) i)

its when t = 1, as after this point the particle begins to slow down and after t = 2, starts to go backwards

b) ii)

its when t = 2.

if you look at the symmetry of the graph, the particle ends up right back at the origin at t = 4. At t = 5, the particle is only the same distance away from the origin as it was at t = 1. hence t=2 is the furthest displacement from the origin as this is where it actually stops

Yeah thats what I said, except I was kinda half guessing :s

 

[mizz_smee]

i always thought a negative acceletation meant it was going backwards
oh well screwed that question
i thought t=5 was too simple

 

[smegthehead]

fuck, i just realised, i got negative acceleration mixed up with negative velocity

 

[Acid]

for (i) I got t = 2 because if you draw the velocity from that it looks like a maximum. Looks like I got that one right.

For (ii) I had nfi so I just guessed that it was t = 1

 

[~ ReNcH ~]

Sorry, it doesnt go backwards after t = 2.
It just SLOWS down, but it's still moving in a positive direction

The max velocity occurs when acc = 0, hence when t = 2

I believe the max distance from origin is 4 sec
At this instant, the positive and negative accelerations cancel each other out and RIGHT after this instant is when the particle starts going backward (because it still carries negative acceleration for another sec up to t = 5)

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?

 

[Smurfwow]

I put t=2 and t=5, which seems to be the consensus.

My question is, will i still get marks if i have a really shitty reason. I have no idea why i put this reason, but it was "because it's had a good two seconds of deceleration" (thats verbatim )

 

[~ ReNcH ~]

I put t=2 and t=5, which seems to be the consensus.

My question is, will i still get marks if i have a really shitty reason. I have no idea why i put this reason, but it was "because it's had a good two seconds of deceleration" (thats verbatim )

t=2 and t=5. That's what I put, but even though it's the consensus I'm still thinking that Intuition's reasoning is more "logical", in which case we're wrong.
But fingers crossed

 

[sparkl3z]

yer i got t = 2
and t = 5
but i didnt state my reasons properly lol what was it out of, 2?? i cant be bothered to check, anyway i'll probably get half of whatever the mark was.

 

[~ ReNcH ~]

Or 0, if your (ii) was wrong.

Does anyone here have a conclusive reason for why the answer to (ii) is t=5?

 

[Will_Sparky]

Mine was t=2 for part i), simply because the acc/t graph has crosses the axis there, so the graph of V has a SP here. I think that I said something like t=3 for part ii) but come to think of it, i'm wrong. Damn! I think my reasoning was something like this is wher the V/time graph crosses the axis, and thus is a MAX SP for the displacement time graph. But yeah...

 

[daniel_fank]

i wrote t = 4. my reasoning is as follows:

the particle is accelerating (to the right) until t = 2, where it reaches its maximum velocity. it then deccelerates until t = 4, where it becomes stationary (as the interval t = 0 to t = 2 is equal and opposite in magnitude to the interval t = 2 to t = 4). the particle then accelerates in the opposite direction (to the left) which means t = 4 is where the particle is furtherest from the origin

hope that helps

daniel

 

[~ ReNcH ~]

i wrote t = 4. my reasoning is as follows:

the particle is accelerating (to the right) until t = 2, where it reaches its maximum velocity. it then deccelerates until t = 4, where it becomes stationary (as the interval t = 0 to t = 2 is equal and opposite in magnitude to the interval t = 2 to t = 4). the particle then accelerates in the opposite direction (to the left) which means t = 4 is where the particle is furtherest from the origin

hope that helps

daniel

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