Does this group also help with physics questions??
i know its a form of mathematics and all but am i able to get help if need be? :)

- anonymous

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- katieb

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- anonymous

i can try

- anonymous

There's a physics board, and I can try also.

- Preetha

Post it in the Physics group.

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- anonymous

i did take physics so

- anonymous

whats the question

- anonymous

i have tip

- anonymous

post it on yahoo answers

- anonymous

after 3 hrs u usually get answers

- anonymous

okay thank you guys, i dont have a question just yet, i haven't started any work but if i have and im in need of help then i will ask/post it up.

- anonymous

will do

- anonymous

if you don't mind post it to this thread, so i can help. i took physics and got an A in it, so i should be able to help, but alas i have a calculus exam to study for, and if you post it in here i can look at it as you have questions.

- anonymous

okay thank you (:
are you any good with the setting out? because i mainly go wrong when it comes to setting out.

- anonymous

"Setting out"?

- anonymous

idk about that either? Sorry :(
i am guessing that is college level, i was in high for physics , so that course really did not cover that

- anonymous

no no high school level. umm you know when im solving or doing an equation i must write the formulas and show you i got a curtain answer within the equation. like it as to be in depth.

- anonymous

If you could literally just type the exact problem, I could solve it. Honestly, and I mean no offense to this, I have no idea what you're saying.

- anonymous

okay
A top sprinter reaches a velocity of 11.13m.s^-1 within 3.15s of start. Assuming that the acceleration is uniform, calculate its magnitude.

- TuringTest

acceleration is change in velocity over change in time:\[a=\frac{\Delta v}{\Delta t}\]how much does the velocity change in the problem?
how long does that take to happen?
plug in the numbers

- anonymous

Yup. v=v_0+at, or v-v_0=at, or (v-v_0)/t=a

- anonymous

okay, i understand but how do you figure out the magnitude?

- anonymous

The "magnitude" of something is just the absolute value. For instance, if I was traveling at a velocity of -x, the magnitude would just be x.
Now, the sprinter's acceleration is uniform, i.e., constant. So we can use simple division v/t=a.
We know his change in velocity, so plug that in. We know the amount of time it took for that change in velocity, so plug that in. Voila, you have the value for acceleration.

- anonymous

okay so it'll be
Vav= s/t where Vav=11.13m.s^-1
t=3.15s
and so s=Vav/t
s=11.13m.s^-1/ 3.15s
therefore s=3.53m.s^-2
is that right?

- anonymous

As long as you did the division correctly, yeah, that's right. Remember that your "s" is actually acceleration, though.

- anonymous

is acceleration the same as displacement?? like same concept?

- anonymous

No. Think about it this way. Movement is displacement. Displacement over time is velocity. Velocity over time is acceleration.

- anonymous

Therefore, movement over time squared is acceleration.

- anonymous

ohh okay i get it (:
thank you

- anonymous

i proberly sound really dumb by asking this but how do you find the uniform acceleration?

- anonymous

Alright, so uniform acceleration (I think) means constant acceleration. That means it doesn't change over time in the system you're given.
If there's constant acceleration, it must be the same as the average acceleration. After all, the average of the set (1,1,1,1) is going to be 1, right?
Now, your average acceleration is going to be your change in velocity over time. So that means if I started at 5mph, and I went up to 8mph over 10 hours, my change in velocity will 3mph.
This was done over 10 hours, so my change in velocity (3mph) over 10 hours will make my acceleration 3/10 or 0.3 repeating miles per hour squared.

- anonymous

oh okay so say a car has a velocity of 27 m.s and 23s later it increased to 96m.s the uniform acceleration would be
96m.s-27m.s/23s??

- anonymous

Remember to put parentheses around 96-27 or else your calculator will interpret it badly. But yes.
Also remember that this gives you the average acceleration, which is only equal to regular acceleration if the acceleration is uniform or constant.

- anonymous

okay, ill remember.
okay so the equation only "tells" you the regular acceleration if you have uniform acceleration?
sorry im just trying to understand it a bit better.

- anonymous

Yes. Think about it this way: your velocity v is actually a set of numbers telling you your velocity at any given time t. When I have a set of velocities, and I divide them by time, I will get the average velocity, right? Not the velocity at EVERY given point of time, just the average one.
It's like if I told you how much money I made over a period time. Let's say I made 10 dollars in 5 days. On average, I made 2 dollars per day if I divide 10/5. But I could've made 7 dollars the first day and 3 dollars the next four. Generally, though, I did make 2 dollars a day on average.
But imagine if I told you I made the same amount of money each day, i.e., a constant salary. Then I would have made 2 dollars a day, right?
So if I have a constant acceleration, average acceleration is going to be the actual acceleration.

- anonymous

oh okay i get it now. thank you.

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