RunRyder RC
WATCH
 2 pages [ <<    <    ( 1 )     2     NEXT    >> ] 1329 views POST REPLY
HomeOff Topics › High school physics question
04-21-2011 12:47 AM  7 years agoPost 1
GyroFreak

rrProfessor

Orlando Florida ...28N 81W

My Posts: All  Forum  Topic

My high school and college physics memory is shot. Here is the question.
If an object is traveling at 164 ft/sec (50 mtrs/sec) and you apply 3 g's to stop it, how many feet (or meters) does it take to reach 0 velocity.

I think about the hereafter. I go somewhere to get something, then wonder what I'm here after ?

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 01:14 AM  7 years agoPost 2
ssmith512

rrKey Veteran

Indianapolis, IN USA

My Posts: All  Forum  Topic

Oh dear. Equations of motions. I think it is v^2 = u^2 + 2as

v=final velocity
u=intitial velocity
a=acceleration (g's)
s=distance

solving for "s" I get 140.08 feet

It's been a long time so I may be wrong

Steve

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 02:04 AM  7 years agoPost 3
rcjon

rrVeteran

Macon, GA

My Posts: All  Forum  Topic

That's what I got by:
v=a*t
solve for t

average velocity = (vi+vf)/2

distance = (average velocity) (time)

Ask your doctor if your heart is healthy enough for Radio Control Helicoptering.

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 06:57 AM  7 years agoPost 4
GyroFreak

rrProfessor

Orlando Florida ...28N 81W

My Posts: All  Forum  Topic

Thanks guys, I think I need to brush up on my old physics books. It's a question my grandson asked me. "If person is falling at terminal velocity and you could apply a constant 3 G force, how long would it take to stop safely in some type of crushable foam."
No, he isn't about to try this. He just likes to dabble in questions like this. So I told him I would check with some of my experts.
Paul

I think about the hereafter. I go somewhere to get something, then wonder what I'm here after ?

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 05:50 PM  7 years agoPost 5
fla heli boy

rrKey Veteran

cape coral, florida

My Posts: All  Forum  Topic

I am VERY rusty with physics, but it seems to me there has to be a place for MASS in that equation.....

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 05:55 PM  7 years agoPost 6
TrexRookie

rrKey Veteran

San Francisco, CA

My Posts: All  Forum  Topic

I'm not sure mass has anything to do with it... The question asks for distance. If this were a question of force then sure, but it doesn't look like this is the case. I may be wrong tho, so I'm not totally sure about that.. haha

+1 for obscure physics advice.

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 06:09 PM  7 years agoPost 7
smh20502

rrNovice

Houston, Texas

My Posts: All  Forum  Topic

Kind of a trick question since the g rating of a safe landing is very high (better than 30g's). Where is this variable?

Common sense isn't so common. Yes, there are stupid questions

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 06:48 PM  7 years agoPost 8
rcjon

rrVeteran

Macon, GA

My Posts: All  Forum  Topic

As Mr. Newton showed us, if acceleration is constant, mass does not matter.
Imagine two bowling balls traveling together. The would stop in the same distance. Now, with a small drop of CA, glue them together. Now you have an object with twice the mass. It would stop in the same distance.
More force will be needed to achieve the desired g, but the problem was stated with a constant g.

Ask your doctor if your heart is healthy enough for Radio Control Helicoptering.

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 07:08 PM  7 years agoPost 9
fla heli boy

rrKey Veteran

cape coral, florida

My Posts: All  Forum  Topic

OK - think about this. A bowling ball dropped next to a ball of the same size made of lightweight styrofoam. Surely mass HAS to matter if the same amount of force is being applied to stop/decelerate the objects.
A 10 ton truck going the same speed as a 2 ton car, will take longer to stop even though there is a higher amount of friction with the truck.
My physics is probably wrong, I'm trying to use common sense and as we all know, sometimes common sense has to take a back seat when it comes to physics.....

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 07:56 PM  7 years agoPost 10
MartyH

rrProfessor

USA

My Posts: All  Forum  Topic

I think mass is covered by the 3G's.

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 07:57 PM  7 years agoPost 11
rcjon

rrVeteran

Macon, GA

My Posts: All  Forum  Topic

Correct that the foam is not going to stop a light object in the same distance as a heavy object.
Force equals mass times acceleration. So more force.will be needed to accelerafe or deaccelerate a more massive object.

Ask your doctor if your heart is healthy enough for Radio Control Helicoptering.

PM  EMAIL  GALLERY  Attn:RR  Quote
04-21-2011 08:06 PM  7 years agoPost 12
bbaxter

rrApprentice

Central Illinois

My Posts: All  Forum  Topic

If you apply the same force to two objects, the lighter one will accelerate faster. However, the original question postulated the acceleration (deceleration) that was applied.

So, if you apply a 3G deceleration (negative acceleration) to both a truck and a car, if they are traveling at the same velocity, they'll take the same distance to stop. However, a greater force will be applied to the truck since force is proportional to mass and acceleration. F = MA.

Not counting any influence from the air, a bowling ball and a styrofoam ball will fall together because they are both in a constant-acceleration field. Gravity really isn't a force, it's an acceleration. Weight is the what we call the force we experience.

We use scales that measure force and have the readout calibrated to read mass, so they basically "take out" the acceleration of gravity.
If you use a scale and see that you weigh 100 Kg, you actually experience a force of 980 Newtons (100Kg x 9.8 m/s/s).

PM  EMAIL  Attn:RR  Quote
04-21-2011 08:43 PM  7 years agoPost 13
AirWolfRC

rrProfessor

42½ N, 83½ W

My Posts: All  Forum  Topic

Mass doesn't mattere here.
Only velocity since you state that the deceleration is 3 G's.
3G's is 3 x 32ft/sec² = 96 ft/sec².
Time to stop is 164 / 96 = 1.71 seconds.
Distance = 1/2 x A x T²
1/2 x 96 x 1.71² = 140 ft

PM  EMAIL  HOMEPAGE  GALLERY  Attn:RR  Quote
04-21-2011 11:59 PM  7 years agoPost 14
GyroFreak

rrProfessor

Orlando Florida ...28N 81W

My Posts: All  Forum  Topic

Thanks AirWolfRC, that is a very clear explanation of my stated question. Several others got the same answer, so thanks guys.
Paul

I think about the hereafter. I go somewhere to get something, then wonder what I'm here after ?

PM  EMAIL  GALLERY  Attn:RR  Quote
04-24-2011 05:38 PM  7 years agoPost 15
AirWolfRC

rrProfessor

42½ N, 83½ W

My Posts: All  Forum  Topic

There was a leap to the final formula I left previously so I'll explained it a bit more;

Mass doesn't mattere here.
Only velocity since you state that the deceleration is 3 G's.
3G's is 3 x 32ft/sec² = 96 ft/sec².
Time to stop is 164 / 96 = 1.71 seconds.

So, to be redundant, you decelerate at 96 ft² for 1.71 seconds
That's A x T which gives final (start) speed.

Average speed = 1/2 of that
Average speed = 1/2 x A x T

Distance = average speed x time
Distance = (1/2 x A x T) x T
Distance = 1/2 x A x T²
1/2 x 96 x 1.71² = 140 ft

PM  EMAIL  HOMEPAGE  GALLERY  Attn:RR  Quote
04-24-2011 06:16 PM  7 years agoPost 16
Justin Stuart (RIP)

rrMaster

Plano, Texas

My Posts: All  Forum  Topic

To do an approximation in your head:

Assume g is 10m/s^2. Initial velocity is 50m/s

Force is 3g or 30m/s^2. What this means it that the object will slow down at a rate of 30 meters per second every second.

After the first second, the object has decreased in velocity by 30m/s.

After slightly more than 2/3 of the 2nd second, the object has stopped.

So the time is approximately 1.7 seconds.

Distance is given by the average velocity multiplied by time. Initial velocity is 50m/s. Final velocity is 0m/s. So average velocity is 25m/s

25m/s x 1.7 seconds is equal to 25 + 12.5 + 2.5 + 2.5 = 42.5 meters

1 meter is equal to approximately 3.3 feet.

42.5 x 3.3 = 42.5 + 42.5 + 42.5 + 4.25+ 4.25 + 4.25 = 40 x 3 + 2.5 x 3 + 12.75 = ~140 feet.

Not an exact answer, but easier to do in your head than using a second degree polynomial equation: Vi^2 - Vi^2 +2at = 0

Avant RC
Scorpion Power Systems
Thunder Power RC
Kontronik Drives

PM  EMAIL  GALLERY  Attn:RR  Quote
04-24-2011 06:32 PM  7 years agoPost 17
AirWolfRC

rrProfessor

42½ N, 83½ W

My Posts: All  Forum  Topic

Do this in your head ? ! ? ! ?
25m/s x 1.7 seconds is equal to 25 + 12.5 + 2.5 + 2.5 = 42.5 meters

1 meter is equal to approximately 3.3 feet.

42.5 x 3.3 = 42.5 + 42.5 + 42.5 + 4.25+ 4.25 + 4.25 = 40 x 3 + 2.5 x 3 + 12.75 = ~140 feet.

PM  EMAIL  HOMEPAGE  GALLERY  Attn:RR  Quote
04-24-2011 06:38 PM  7 years agoPost 18
Justin Stuart (RIP)

rrMaster

Plano, Texas

My Posts: All  Forum  Topic

Do you have another way to multiply 42.5 x 3.3 in your head?

I look at it as 40 x 3 = 120

plus 2.5 x 3 = 7.5

plus 4.25 x 3 = 12.75

= ~140

Maybe there is a better way to do this (other than to break out a calculator which is sort-of cheating)?

Avant RC
Scorpion Power Systems
Thunder Power RC
Kontronik Drives

PM  EMAIL  GALLERY  Attn:RR  Quote
04-29-2011 05:11 AM  7 years agoPost 19
GyroFreak

rrProfessor

Orlando Florida ...28N 81W

My Posts: All  Forum  Topic

In my younger days many many years ago I was a slide rule cripple, meaning I couldn't do (or wouldn't do) any math in my head, even simple multiplication. (Course you still had to be able to add and subtract without the slide rule.) Now I am a handheld calculator cripple for the same reason.

I think about the hereafter. I go somewhere to get something, then wonder what I'm here after ?

PM  EMAIL  GALLERY  Attn:RR  Quote
04-29-2011 05:52 AM  7 years agoPost 20
Justin Stuart (RIP)

rrMaster

Plano, Texas

My Posts: All  Forum  Topic

Throw down your crutch and walk!!

Avant RC
Scorpion Power Systems
Thunder Power RC
Kontronik Drives

PM  EMAIL  GALLERY  Attn:RR  Quote
WATCH
 2 pages [ <<    <    ( 1 )     2     NEXT    >> ] 1329 views POST REPLY
HomeOff Topics › High school physics question
 Print TOPIC  Make Suggestion 

 8  Topic Subscribe

Monday, July 23 - 5:10 am - Copyright © 2000-2018 RunRyder   EMAILEnable Cookies

Login Here
 New Subscriptions 
 Buddies Online