MedVision ad

Section I - Multiple Choice (3 Viewers)

superSAIyan2

Member
Joined
Apr 18, 2012
Messages
320
Gender
Male
HSC
2013
Okay, but the question didn't say that force was being applied on the particle?

I'm looking it a different way
Im assuming you dont do physics. But if an object accelerates, there must be a non zero force acting on it

Edit: saw your sig, i assumed correct:)
 

hjed

Member
Joined
Nov 10, 2011
Messages
211
Gender
Undisclosed
HSC
2013
Sadly it's a very ambiguous question, looks like the majority is going with D.

Both ways could be argued pretty fairly.
 

NFmusic1995

Member
Joined
Mar 30, 2013
Messages
274
Gender
Male
HSC
2013
its 100% D, its moving to the left, but it is accelerating positively, therefore it will slow down and eventually go to the positive side, simple physics.
 

Samir97

Member
Joined
Sep 26, 2013
Messages
109
Gender
Male
HSC
2014
1. D
2. B
3. A
4. A
5. C
6. B
7. B
8. C
9. C
10. B
Acceleration is in the opposite direction of the speed, hence it will eventually turn the other way and go with the acceleration, so it needs to stop eventually to turn around, so it must slow down to be able to come to a halt, hence speed is decreasing and the answer is D
 

hjed

Member
Joined
Nov 10, 2011
Messages
211
Gender
Undisclosed
HSC
2013
its 100% D, its moving to the left, but it is accelerating positively, therefore it will slow down and eventually go to the positive side, simple physics.
It depends if you look at speed or velocity.
Its speed could be taken as -3, positive acceleration then indicates that the speed is increasing (towards positive infinity). This is how I read the question and it is a valid interpretation.
However, you can also take it a speed = |velocity| (which I guess is valid, but you can have negative speed). In this case speed is decreasing. This is how a lot of others have read it, so it is likely the answer that will be marked correct. :(
 

TurtleyyEpic

New Member
Joined
Aug 19, 2013
Messages
12
Gender
Male
HSC
2013
It depends if you look at speed or velocity.
Its speed could be taken as -3, positive acceleration then indicates that the speed is increasing (towards positive infinity). This is how I read the question and it is a valid interpretation.
However, you can also take it a speed = |velocity| (which I guess is valid, but you can have negative speed). In this case speed is decreasing. This is how a lot of others have read it, so it is likely the answer that will be marked correct. :(
As Samir explained, the answer is D. Just wanted to add that speed is a scalar quantity that disregards direction - therefore it can't be negative as you say
 

Firmin

New Member
Joined
Feb 25, 2012
Messages
12
Gender
Male
HSC
2013
It depends if you look at speed or velocity.
Its speed could be taken as -3, positive acceleration then indicates that the speed is increasing (towards positive infinity). This is how I read the question and it is a valid interpretation.
However, you can also take it a speed = |velocity| (which I guess is valid, but you can have negative speed). In this case speed is decreasing. This is how a lot of others have read it, so it is likely the answer that will be marked correct. :(
Speed is scalar and does not have direction as you pointed out above. Your interpretation is not valid as acceleration can only be worked with velocity and then converted to speed (valid due to this being linear motion).
 

aDimitri

i'm the cook
Joined
Aug 22, 2013
Messages
505
Location
Blue Mountains
Gender
Male
HSC
2014
speed is the magnitude (the absolute value) or velocity. over the next brief period of time the velocity will increase but it's magnitude will decrease as it prepares to turn around. 10.D is correct
the rest of the answers are as follows:
1. D
2. B
3. A
4. A
5. C
6. D
7. B
8. B
9. C
10. D
 

Samir97

Member
Joined
Sep 26, 2013
Messages
109
Gender
Male
HSC
2014
If a > 0, the particle is travelling at an increasing speed.
If v < 0, the particle is travelling towards the left (or towards the origin).

For questions like: for what values of x is f(x) is increasing, isn't it when f'(x) > 0 ?

look at v as f(x) and a as f'(x)

Since a [f'(x)] is > 0, v [f(x)] is increasing. So the speed is increasing.

Therefore, the particle is moving towards the origin at an increasing rate (or speed, whatever it was).
If a>0 that doesnt mean its soeed is increasing, it means the speed is moving towards the positive direction, and if v<0 then the particle will follow the direction of acceleration eventually, so it needs to slow down, stop, then speed up in the other direction, hence its speed decreases at that instant, not increases
 

Users Who Are Viewing This Thread (Users: 0, Guests: 3)

Top