The Maths and Science Thread - Collection of Problems and Facts

[TumTum]

1. What is the acceleration of a free falling object near the surface of the earth?

9.81 m/s^2


2. If the free falling object (of 64kg) is falling at a constant speed, what is the force applied by the wind resistance?

Constant speed, therefore upward force = downward force. That is, no resultant for or acceleration is acting on the body.
Therefore the wind resistance = weight of the object
= 64 * 9.81
= 627.84 N


3. If the wind resistance coefficient is 4kg/m, what speed is the object falling at?

Don't know the forumlas involved with coefficients of resistance.


4. Create a differential equation of motion of the object as it starts falling from rest.

Resultant force = Weight - Air Resistance

Air Resistance depends on Velocity for one thing, but I don't know what other variable.

dP/dt = mg - dv/dt if I use time for the second variable. So rate of change of momentum = mg - dv/dt, but dv/dt is just acceleration, so *shrug*.


5. Assuming 95% is close enough to 100%, how long will it take for the object to achieve full speed?

Don't think I can solve this without the above parts.

Good job so far

3. Air resistance, F=k(v^2). (F is relative to speed squared)

I told you k=4kg/m. If you put the units into the equation, you will notice you get F in kg.m/s^2 which is the same as N.

4. ma = mg-kv^2 (Force effective = m*a = Sum of all external forces)

5. You will notice you don't know 'a' or 'v' (however they are related). You need to rearrange the equation, and then solve it using a special technique for Non-Homogeneous DE's.

I am still giving you a chance to call you a legend in my signature

 

[Dutch]

Another question of reasoning:

A man goes out for a walk. He walks south one mile, east one mile, and north one mile, and ends up in the same place he started. He didn't start out at the north pole -- so where did he?

The point at the end of infinity (A valid term and concept in Projective Geometry)

 

[Kshitiz_Indian]

Equator?

TumTum added 1 Minutes and 11 Seconds later...



Ok but you forgot the force applied by the wind resistance. F=k(v^2)

Hence you cannot solve it using the constant acceleration formulas.

You never told me there was wind resistance.

 

[TumTum]

You never told me there was wind resistance.

Each new question continues on from the previous ones

 

[Varun]

Equator?

Well, he's 1 mile east of where he started, so unless he started one one of the latitudes or the equator, he can't have ended up where he started

The point at the end of infinity (A valid term and concept in Projective Geometry)

All three answers are wrong.

A HINT: It is not related to latitudes. Instead , to circumference of ____.

 

[TumTum]

All three answers are wrong.

A HINT: It is not related to latitudes. Instead , to circumference of ____.

Circle? :what

 

[ZoraxDoom]

Okay, therefore F=4(v^2)

4(v^2) = 627.84
v^2 = 156.96
v = 12.528 m/s


ma = mg - k(v^2)

we need v and time in the equation

so m.v/t = mg - kv^2

mv/t as it's basically v-u/t as the object is released from rest.

m.v/t = mg - kv^2

v = y
t = x

m.y/x = mg - ky^2

Yea, I think I'm stuck Didn't even get into the differentials yet, kinda not sure what to do here. Should go over my maths portion again sometime soon.

 

[Kshitiz_Indian]

Zorax, in the differential equation,

dv/dt = -g

modify it to say

dv/dt = -g + k(v^2)

Now you can solve this equation. Of course there will be a time when the velocity will be such that acceleration will be zero, which should also be shown by the resultant equation.

 

[TumTum]

LOL I can see that.

Use 95% of that Maximum speed to find the time, as t = infinity when the object will reach max speed (so never). You can imagine it will be like a y=-e^(-t)+Vmax graph, a horizontal asymptote at Vmax.

For non-homo (not gay :laugh) DE's, basically:
1- Make R = dx/dt, Hence 'a' or dx^2/d^2t = R^2. (dx^3/d^3t = R^3 etc...)
2- Rearrange to make R^2 + R + constant = 0 (quadratic equation)
Hint - You will not have R, as it will be squared anyway.
3- Solve for R.
4- y=-e^(-at)(cosb+isinb)+C where C=constant & R=a+ib (hint: b=0, as the solution is real )
5- Put in the initial conditions & solve for C. (y=0 m/s when t=0) (so basically C=12.528 m/s :laugh)
6- Now find the time when the speed is 95% of its max.

For #6, you can either use a graphics calculator or use natural log.

TumTum added 6 Minutes and 7 Seconds later...

Of course there will be a time when the velocity will be such that acceleration will be zero, which should also be shown by the resultant equation.

Velocity will only reach 12.528m/s when t=infinity.

That is the beauty of this question, the acceleration is never constant.

This can represent a real life situation where a person is sky diving with a parachute. He will get faster very quickly after jumping, but after a while he will still go faster and faster but will happen much slower. Acceleration is at its highest at the start, and slowly crawls to 0 as you reach but never reach the velocity threshold.

 

[Abhas]

All three answers are wrong.

A HINT: It is not related to latitudes. Instead , to circumference of ____.

Did he start 1 mile above the south pole?

 

[Varun]

Did he start 1 mile above the south pole?

Pretty Close. But not exactly correct. Instead he started 1 mile above a latitude. Is this hint enough?

 

[Kshitiz_Indian]

I have absolutely no idea but I'll have a guess he started 1 mile above the equator?

 

[Abhas]

I mile under the north pole?

 

[TumTum]

Magnetic North Pole? :confused:

 

[Abhas]

Ok, I think I got it.. He was 1 mile north of a latitude that was 1 mile in circumference.
He goes 1 mile south, to the latitude, and does 1 mile east, thus, goes around the latitude and reaches the same place. Then goes 1 mile north to the same place where he started.