Anyone good at physics

M

Makdaddymac

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May 28, 2005

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Feb 1, 2009

#1

• Feb 1, 2009

• #1

i need help on some online HW! pm me
im desperate....

 

Mustang5L5

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#2

• Feb 1, 2009

• #2

What's the problem?

 

7upstang91

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#3

• Feb 1, 2009

• #3

haha I need help too!

 

FastDriver

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#4

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Physics used to be my bitch! Post up the problem and let me take a crack at it.

Chris

 

M

Makdaddymac

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#5

• Feb 1, 2009

• #5

fiy, if you help me i will continue to bother you until the end of the semester

In an historical movie, two knights on horseback start from rest 80.5 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.377 m/s2, while Sir Alfred's has a magnitude of 0.364 m/s2. Relative to Sir George's starting point, where do the knights collide?

its failr simple. using one of the kinematic equations, but idk what to do with two Accelrations(a)
i think i use: v^2 = v_0^2 + 2 ax

 

Mustang5L5

That is…until I whipped out my Bissell

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Feb 18, 2001

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#6

• Feb 1, 2009

• #6

Lol oh boy. That was an easy one in college..but i haven't touched those equations in 6 years. I need a refresher

 

FastDriver

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#7

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• #7

It can be solved using calculus.

The integral of acceleration is velocity, and the integral of velocity is distance.

Asg (acceleration of Sir George) = .377 m/s^2 * t^0
Vsg = 3.77t + 0t^0 (there is no initial velocity)
Dsg = (3.77/2) t^2 + 0t^1 +0t^0 (there is no initial velocity nor is there an initial distance)

Asa (acceleration of Sir Alfred) = -0.364 m/s^2 * t^0 (it's negative because he's moving in the opposite direction of Sir George)
Vsa = -0.364t
Dsa = (-0.364/2) t^2 + 80.5 (the initial distance is 80.5 meters away from Sir George)

Now, just set Dsg and Dsa equal to each other and solve for t, and you should come up with your answer in terms of time. You can then go back and plug the value for t into either distance equation and it should give you your answer in meters, and distance from Sir George's starting point. I'll let you work it out from here.

Yeah... I'm a dork (but at least I got something out of my engineering degree),

Chris

 

R

Rick James

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May 13, 2007

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Feb 1, 2009

#8

• Feb 1, 2009

• #8

FastDriver said:

It can be solved using calculus.

The integral of acceleration is velocity, and the integral of velocity is distance.

Asg (acceleration of Sir George) = .377 m/s^2 * t^0
Vsg = 3.77t + 0t^0 (there is no initial velocity)
Dsg = (3.77/2) t^2 + 0t^1 +0t^0 (there is no initial velocity nor is there an initial distance)

Asa (acceleration of Sir Alfred) = -0.364 m/s^2 * t^0 (it's negative because he's moving in the opposite direction of Sir George)
Vsa = -0.364t
Dsa = (-0.364/2) t^2 + 80.5 (the initial distance is 80.5 meters away from Sir George)

Now, just set Dsg and Dsa equal to each other and solve for t, and you should come up with your answer in terms of time. You can then go back and plug the value for t into either distance equation and it should give you your answer in meters, and distance from Sir George's starting point. I'll let you work it out from here.

Chris

Click to expand...

My eyes started bleeding from reading that.LOL

 

M

Makdaddymac

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Feb 1, 2009

#9

• Feb 1, 2009

• #9

ah typing this on the comp is so difficult.

how do i solve for t? can you set up the equation?

 

T1 3VOM

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#10

• Feb 2, 2009

• #10

Makdaddymac said:

ah typing this on the comp is so difficult.

how do i solve for t? can you set up the equation?

Click to expand...

You'd put the equation for Dsg on on side of the equal sign and Dsa on the other side because you're solving for the time (t) when the two meet. "t" will be the only variable in the equation. Solve for t.

Say Dsg = xt^2 + yt +z
and DSA = at^2 + bt + c

When you start solving for t you're looking for the spot where the two guys occupy the same spot, so, you want Dsg = Dsa. Another way of writing that is: xt^2 + yt +z = at^2 + bt + c

If you need someone to explain to you how to solve for t, you'll need to go back to your algebra book, because that is a very basic concept for much of what you'll be doing in your physics class. Not trying to be nasty... just saying.

However, if you read the question carefully, you will notice that in the first sentence, it states that this is part of a movie. So, the correct answer is, "They will meet at any time and distance other than, "t" and "d" because a movie cannot, by its nature, agree with physics."

 

Cobra912

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Feb 2, 2009

#11

• Feb 2, 2009

• #11

Makdaddymac said:

i need help on some online HW! pm me
im desperate....

Click to expand...

I waz loszt rihte afteer thiis...........nehver gotz paszed grade four

 

Strippt88

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Feb 2, 2009

#12

• Feb 2, 2009

• #12

Assuming George and Alfred were in fact jousting, they would not collide. All of your calculations were in error. The idea is to dismount the opposing rider, or strike a blow that would be counted in points, utilizing a breakable staff in most cases, so as not to injure the other rider. I don't believe they used physics at any point. And, I believe it is in the Stangnet rules that any and all physics type equations be PM'd, so as not to inflict permanant opthimological injury to those of us that have no f...ing clue what that was! Sheesh, i got to reading, and thought I was gonna go back in time.

 

T1 3VOM

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#13

• Feb 2, 2009

• #13

You gotta hit 88 mph first... and you need plutonium and 1.21 gigawatts of electricity.

 

M

Makdaddymac

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Feb 2, 2009

#14

• Feb 2, 2009

• #14

gotta hit 88mph first always for all physics problems!

 

FastDriver

I was uncomfortably high & wearing a helmet

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#15

• Feb 5, 2009

• #15

I'm sorry man. I didn't get your PM until today. I'm really busy trying to find a place to move into now that I'm back at Fort Benning.

As was said above, just set the two equations equal to each other, move all t's to one side of the equation and everything else to the other side of the equation, then solve for t. If you still can't figure this one out, then get someone you go to class with to start tutoring you, because algebra is a prerequisite for physics, and you will not be able to keep up as the course progresses without a good foundation in math.

Chris

 

Strippt88

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Cedar Lake, Indiana

Feb 5, 2009

#16

• Feb 5, 2009

• #16

Just curious, I just make roads(good thing we still need those) but, how does one apply this sort of equation? I mean, where can one use the ability to solve this sort of problem? Engineering would be my first guess, aeronautics, machining, like that? I would guess it'd pay a semi-load, cuz I don't know anyone that can do those sort of mathematical gymnastics. Or, maybe they just haven't told me they can, cuz I'm a bit simple, right?