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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: b^3 on August 06, 2011, 01:43:04 pm

Title: The Sign/direction of Velocity
Post by: b^3 on August 06, 2011, 01:43:04 pm: Ok I can do the question but cannot figure out why only the positive solution is given.

Essential Specialist Maths - Ch13 - Dynamics - Chapter Review - Short Answer - Q11
Here is the Question:
A particle of mass 3 kg, moves in a straight line and, at time t, its displacement from a fixed origin is a and its speed is v. If the resultant force is 3 + 6x, and v = 2 when x = 0, find v when x = 2.

Now this is my working.
$m=3kg$
$F=3+6x, v=2, x=0$
$a=\frac{1}{3}(3+6x)$
$a=1+2x$
$\frac{d(\frac{1}{2}v^2)}{dx}=1+2x$
$\frac{1}{2}v^2=\int 1+2xdx$
$\frac{1}{2}v^2=x^2 +x+c$
$\frac{1}{2}(4)=0+c$
$c=2$
$\frac{1}{2}v^2=x^2 +x+2$

So now x=2 $\frac{1}{2}v^2=4+2+2$
$v^2=16$
$v=\pm 4$ m/s

Now the answer given is just +4m/s, I cannot work out why it is only the positive solution, the particle is at x=2 at two points, and there is no restrictions that I can see.

Anyone able to explain this or is the book wrong?
Title: Re: The Sign/direction of Velocity
Post by: pi on August 06, 2011, 01:46:23 pm: I'm not up to this yet, but when x = 2, a = 1+4 = 5. Therefore, the particle must be in the positive direction of motion at x = 2, as 'a' is a positive value. Hence v = 4, showing the positive direction of motion described by 'a'.

Again, I'm not up to this stuff yet, so I could be completely off. I'm going off my petty VCE physics knowledge on this one...

Moderator action: removed real name, sorry for the inconvenience

EDIT: 3.6k^th post 8)
Title: Re: The Sign/direction of Velocity
Post by: b^3 on August 06, 2011, 01:51:45 pm: Quote from: pi on August 06, 2011, 01:46:23 pm
I'm not up to this yet, but when x = 2, a = 1+4 = 5. Therefore, the particle must be in the positive direction of motion at x = 2, as 'a' is a positive value. Hence v = 4, showing the positive direction of motion described by 'a'.

Again, I'm not up to this stuff yet, so I could be completely off. I'm going off my petty VCE physics knowledge on this one...

EDIT: 3.6k^th post 8)
Congrats on 3.6k posts.
Thats what I was thinking originally but acceleration can be negative, even when it is moving in a positive direction in a positive quadrant, the particle would just be deccelerating from a positive velocity.

Moderator action: removed real name, sorry for the inconvenience
Title: Re: The Sign/direction of Velocity
Post by: pi on August 06, 2011, 02:00:51 pm: Hmmm. Can you have neg velocity when the force is positive?
Title: Re: The Sign/direction of Velocity
Post by: b^3 on August 06, 2011, 02:08:25 pm: Quote from: pi on August 06, 2011, 02:00:51 pm
Hmmm. Can you have neg velocity when the force is positive?
Yes. Say you start with a velocity of -20m/s on a mass of say 10kg. Say a force is acting of magnitude 50N, i.e. F=+50N. then accel would be +50/10=+5m/s^2. Initially the object is still travelling with a negative velocity, accelerating in a postive direction, while the force acting on it is postive. The particle will 'slow down' in the negative direction, until the velocity is zero, where it starts accelrating in a postive direction. In the example here this is at t=v-u/a=0+20/5=t=4s

Moderator action: removed real name, sorry for the inconvenience
Title: Re: The Sign/direction of Velocity
Post by: jane1234 on August 06, 2011, 02:08:58 pm: Quote from: b^3 on August 06, 2011, 01:43:04 pm
Ok I can do the question but cannot figure out why only the positive solution is given.

Essential Specialist Maths - Ch13 - Dynamics - Chapter Review - Short Answer - Q11
Here is the Question:
A particle of mass 3 kg, moves in a straight line and, at time t, its displacement from a fixed origin is a and its speed is v. If the resultant force is 3 + 6x, and v = 2 when x = 0, find v when x = 2.

Speed is the magnitude of velocity, thus it's always positive as it has no direction.
Could be wrong, but that's how I see it. :)
Title: Re: The Sign/direction of Velocity
Post by: b^3 on August 06, 2011, 02:10:18 pm: AH HA BINGO! THATS IT! Thankyou, I got to learn to read the dam question now don't I. Dam so simple. Again Thanks.

EDIT: Just found the worked solutions book and they did it wrong, leave it as plus or minus.
Title: Re: The Sign/direction of Velocity
Post by: jane1234 on August 06, 2011, 02:11:13 pm: Quote from: b^3 on August 06, 2011, 02:10:18 pm
AH HA BINGO! THATS IT! Thankyou, I got to learn to read the dam question now don't I. Dam so simple. Again Thanks.

Haha no worries. Don't worry, I always over-complicate things too. :P
Title: Re: The Sign/direction of Velocity
Post by: b^3 on August 06, 2011, 02:13:30 pm: I'm going to highlight the word "speed" and put that many arrows to it now that it won't be funny.
Title: Re: The Sign/direction of Velocity
Post by: jane1234 on August 06, 2011, 02:15:30 pm: Quote from: b^3 on August 06, 2011, 02:13:30 pm
I'm going to highlight the word "speed" and put that many arrows to it now that it won't be funny.

Hang on, but you said the worked solutions left it as +-? Are the worked solutions wrong then?
Title: Re: The Sign/direction of Velocity
Post by: b^3 on August 06, 2011, 02:18:54 pm: Yeh they must be wrong, the found v like I did, they did not find speed.
Title: Re: The Sign/direction of Velocity
Post by: Mao on August 06, 2011, 03:29:58 pm: The answer is +4 only. Consider this:

- Acceleration is positive for x>0
- The mass is initially moving in a positive direction. Imagine after some small amount of time $\delta t$ , it would have moved by a distance of $v\delta t = \delta x > 0$ , and since the mass is now at a positive x coordinate, its acceleration will also be positive, increasing velocity in the positive direction, increasing displacement in the positive direction, and so forth (positive loop). As time goes on, this mass will shoot towards infinity and never look back. At no point will acceleration oppose positive velocity at x>0.
- Thus, when the mass is at x=2, the only way to reach this point is by a unidirectional motion from x=0. Its velocity must be positive.

The equation of motion for this object is:

$x(t) = -\frac{1}{2} + \frac{\cosh (\sqrt{2}\, t)}{2} + \sqrt{2}\sinh(\sqrt{2}\,t)$ (defining t=0 when x=0)

The time when it reaches x=2 is at $t=\sqrt{2}\coth^{-1} \left(\frac{3}{\sqrt{2}}\right)$
Title: Re: The Sign/direction of Velocity
Post by: yawho on August 06, 2011, 04:49:30 pm: How do you know moving +/- direction initially?
Title: Re: The Sign/direction of Velocity
Post by: Mao on August 06, 2011, 05:12:13 pm: Quote from: yawho on August 06, 2011, 04:49:30 pm
How do you know moving +/- direction initially?

It cannot be negative. If it was, it will never reach x=2.

It can be show that with a large enough initial velocity in the negative direction, the force will not be able to turn the mass towards the positive side. This is the case for an initial velocity of v=-2, and the mass accelerates to negative infinity.

I have calculated the critical threshold to be $v_c=-\frac{\sqrt{2}}{2}$ . For initial velocity $v>v_c$ , the object will accelerate towards infinity. For initial velocity $v<v_c$ , the object will accelerate to negative infinity. For $v=v_c$ , the object will reach $x=-0.5$ and remain stationary.

The equation of motion with a variable initial velocity is $x = \frac{1}{2}\left( -1 + \cosh(\sqrt{2}\, t) + \sqrt{2}\, v \, \sinh(\sqrt{2}\, t) \right)$ , where v is the initial velocity. Try plotting on your calculator for different values of v.
Title: Re: The Sign/direction of Velocity
Post by: yawho on August 06, 2011, 09:33:09 pm: So this is not a simple question after all.