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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Studyinghard on January 04, 2010, 09:49:51 pm

Title: Acceleration
Post by: Studyinghard on January 04, 2010, 09:49:51 pm: A car that is accelerating uniformly travels 18 m in the fi fth second of its motion.
a What is the acceleration of the car?
b How far does the car travel in 10.0 s?

Just wtf'ed me for some reason.
Title: Re: Acceleration
Post by: appianway on January 04, 2010, 09:56:58 pm: If it travels 15 metres in its 5th second of it's motion, it means the difference between the time of the 4th and 5th second was 18 m. Use the relevant kinematic equations, find the difference and solve. Then you can use the acceleration for part b.
Title: Re: Acceleration
Post by: Studyinghard on January 04, 2010, 10:02:00 pm: Quote from: appianway on January 04, 2010, 09:56:58 pm
If it travels 15 metres in its 5th second of it's motion, it means the difference between the time of the 4th and 5th second was 18 m. Use the relevant kinematic equations, find the difference and solve. Then you can use the acceleration for part b.

Sorry, I didnt do any 1/2 and I am doing this all on my own and I have absolutely no idea what the kinematic equations are.

well once it comes up I am sure it will hit me but I am blank as hell.
Title: Re: Acceleration
Post by: appianway on January 04, 2010, 10:06:54 pm: Do you have a textbook?

d=ut + 0.5 * at^2
v^2= u^2 + 2ad
v(av) = d/t = (u + v)/2

Things like that. They're pretty much common sense - you can derive them all from each other.
Title: Re: Acceleration
Post by: Aden on January 04, 2010, 10:18:38 pm: These ones:

$x = \frac{1}{2} (u+v)t$

$v = u +at$

$x = ut + \frac{1}{2}at^2$

$x = vt - \frac{1}{2}at^2$

$v^2 = u^2 +2ax$
Title: Re: Acceleration
Post by: Studyinghard on January 04, 2010, 10:52:07 pm: thanks for help. got another question but il post in the other thread.
Title: Re: Acceleration
Post by: kenhung123 on January 05, 2010, 12:10:14 am: Is there something missing here?
Title: Re: Acceleration
Post by: Gloamglozer on January 05, 2010, 12:14:55 am: If it's from the textbook, yes, the writers stuffed up.
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 12:16:43 am: http://hyperphysics.phy-astr.gsu.edu/HBASE/cf.html#cf2

that should help
Title: Re: Acceleration
Post by: kenhung123 on January 05, 2010, 12:27:41 am: Whats it supposed to be?
I am not sure how $(\frac{2\pi }{T})^2*\frac{1}{r}=\frac{4\pi^2 r}{T^2}$
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 12:39:51 am: so u begin with $a=\frac{v^{2}}{r}$

$a=r\times \omega ^{2}$ and u know $\omega =\frac{2\pi}{\tau }$
sub this back in:

$a=r\times (\frac{2\pi}{\tau })^{2}$
simplify

$a=\frac{4\pi^{2}r}{\tau ^{2}}$
Title: Re: Acceleration
Post by: kenhung123 on January 05, 2010, 12:44:54 am: Isnt a=v^2/r?
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 12:47:10 am: yes, im just showing you how they derive $a=\frac{4\pi^{2}r}{\tau ^{2}}$

dont really need to know how its derived, just know how to use the formula :P
Title: Re: Acceleration
Post by: kenhung123 on January 05, 2010, 01:24:43 am: How is there constant speed when the net force is 0? What is driving the speed to be constant?
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 01:31:17 am: newtons first law, object will remain at rest or at constant speed unless an unbalanced force acts on it ie. as there are no unbalanced forces ( net force: 0 N) it continues at constant speed.
Title: Re: Acceleration
Post by: /0 on January 05, 2010, 01:54:05 am: No superflya, in uniform circular motion there is always the centripetal force, which is directed towards the centre of the circle. The force changes the 'direction' of the speed, and hence the velocity, but it doesn't actually affect the speed.

It doesn't affect the speed because the centripetal force is at all times perpendicular to the velocity. In a sense, it 'tugs' the velocity vector to change direction, but it doesn't change the size of the vector (which is the speed).

Take special care of the word uniform. Uniform circular motion occurs when the only force is towards the centre of the circle.
If you vertically swing a ball tied to a string , it is not undergoing uniform circular motion because it is being affected by the gravitational field, and thus it does not move at constant speed.

The constancy of speed can be proven using mathematics, but isn't that elementary if you haven't learnt about vectors.

From the unit circle in trigonometry, $\sin{\theta} = \frac{y}{1}$ , $\cos{\theta} = \frac{x}{1}$

Therefore, two equations describing the circle are $x = \cos{\theta}$ , $y = \sin{\theta}$

Now, to turn this into a dynamic system, where the object is moving around the circle, we can replace $\theta$ with 't' for time.

$x = \cos{t}$ , $y=\sin{t}$

The period of the motion around the circle in this case will be $2\pi$ (the period of sin and cos), but that's not very important.

If we differentiate both with respect to time:

$\frac{dx}{dt} = -\sin{t}$

$\frac{dy}{dt} = \cos{t}$

These are the horizontal and vertical velocities of the object. Imagine these as being the magnitudes of vectors, pointing horizontally and vertically respectively.

The resultant sum of these vectors is the velocity,

$\mathbf{v} = \frac{dx}{dt}\mathbf{i}+\frac{dy}{dt}\mathbf{j}$

And the length of the velocity vector is the speed!

$speed =|\mathbf{v}|= \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} = \sqrt{(-\sin{t})^2+(\cos{t})^2} = 1=constant$ .

So the speed of an object moving around a circle is constant. No matter what radius or period we choose, the result still stands.
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 02:06:12 am: my bad soz I'm a tad drained. thanks for clearing it up /0 (;
Title: Re: Acceleration
Post by: kenhung123 on January 05, 2010, 02:09:11 am: Quote from: superflya on January 05, 2010, 01:31:17 am
newtons first law, object will remain at rest or at constant speed unless an unbalanced force acts on it ie. as there are no unbalanced forces ( net force: 0 N) it continues at constant speed.
Yea but for example and object is at rest, it is then lifted vertically up at 2m/s constant speed. So what makes it move at 2m/s since it was first at rest and the net force is 0?
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 02:09:53 am: so basically velocity is changing due to directional changes however the speed is still the same
Title: Re: Acceleration
Post by: /0 on January 05, 2010, 02:10:40 am: Quote from: superflya on January 05, 2010, 02:09:53 am
so basically velocity is changing due to directional changes however the speed is still the same

yep
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 02:14:40 am: sweet.
Title: Re: Acceleration
Post by: kenhung123 on January 05, 2010, 02:27:33 am: Superflya do you know what drives the object from rest to move at constant speed when the net force is zero?
Title: Re: Acceleration
Post by: Akirus on January 05, 2010, 02:32:19 am: There would first have to be a net force to move it, which can then be maintained by a net force of 0, I believe.
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 04:56:53 am: not too sure bout ur question however as the velocity is constantly changing, the Mass Is accelerating. Hence according to newtons second law, the mass of the stone must have a force applied to it that is producing the acceleration. this force is the centrepetal force acting towards the centre of the circle. Also note that there is a reaction force associated with the centrepetal force. This reaction force doesn't act on the body moving in a circular path, it acts directly away from the centre of the circle. Hope that helps :)
Title: Re: Acceleration
Post by: superflya on January 05, 2010, 05:06:49 am: I believe aqirus' theory sounds pretty reasonable as there has to be an initial force to lift it off the ground to it's desired height then only can the net force = 0.