View Full Version : rate of change is 6x, actual distance travelled = 3x^2?

Silvercord

08-30-2003, 01:19 PM

Is that true?

I've got a car that's accelerating at 6m/sec but i want to know how much space it's actually displaced, working backwards 6m/sec is the derivative of 3m^2 so therefore 3x^2 is the actual amount the car has travelled?

This is sort of a homework question, except the homework question doesn't ask for how far the car has travelled (and this isnt' a calculus homework question)

Thantos

08-30-2003, 01:51 PM

Didn't feel like doing this the math way so I did using the spreadsheet method :)

I used multiple colums.

Col 1 was the number of secs passed (0 - 40).

Col 2 was the distance traveled in that second.

Col 3 was the sum of the distances of the time elapsed.

Col 4 was the quanity of 3 times the value of col 1 to the power of 2.

The values of col 3 and col 4 were not the same.

Working on a good formula for col 4 that makes it equal col 3.

Of course my brain hasn't awoken yet so might be a little bit

Thantos

08-30-2003, 02:03 PM

Ok provided that the previous was correct the following will make col 4 = col 3:

S= seconds passed

(S*6) * ( (S * .5) + .5)

Ok brain shutting down again.

confuted

08-30-2003, 02:22 PM

Physics, Silvercord?

Distance = d (meters)

velocity = v (meters/second)

acceleration = a (meters/second^2)

time = t (seconds)

D = v_i*t + (1/2)a*t^2

(v_i is v, subscript i for initial)

v_f = v_i + a*t

etc.

If you want to use some calculus...

the derivative (with respect to time) of the formula which gives you distance will give you velocity.

the derivative of the formula which gives you velocity will give you acceleration.

the definite integral of the velocity formula will give you distance

the definite integral of the acceleration formula will give you velocity.

I think I got all those right.

edit: Those formulas are only valid for speeds very small relative to c.

Zach L.

08-30-2003, 05:39 PM

Those are all valid, assuming constant acceleration.

>> edit: Those formulas are only valid for speeds very small relative to c.

Haha! Reminds me of a good joke.

A physicist came in to work, and recounted the story of his trip to his colleagues. He explains that he was on his way to work, and was pulled over by a cop. The cop came up to him and asked, "Sir, do you realize that you just ran a red light?" To this, the physicist replied, "No officer, it must have blue-shifted, so I thought it was green." One of the physicist's colleague then asked, "So did he give you a ticket for running the red light." The physicist replied, "No. He gave me a speeding ticket instead."

:D

confuted

08-30-2003, 06:02 PM

Good joke Zach.

The derivative/integral ones should be valid for non-constant accelerations as well, no?

Silvercord

08-30-2003, 09:29 PM

salem:

"after starting from rest, a car reaches a speed of 30m/sec in 5 sec, what is its average acceleration". The answer to that is 6m/sec

D = v_i*t + (1/2)a*t^2

If the acceleration is 6x then according to that equation distance is 3x^2 which is what I said.

Zach L.

08-30-2003, 11:10 PM

Originally posted by confuted

Good joke Zach.

The derivative/integral ones should be valid for non-constant accelerations as well, no?

Thanks. :D

Yeah... Sorry. I was just talking about the two formulas... I should have been more clear on that.

Zach L.

08-30-2003, 11:11 PM

Originally posted by Silvercord

If the acceleration is 6x then according to that equation distance is 3x^2 which is what I said.

Yep... it is.

Powered by vBulletin® Version 4.2.5 Copyright © 2019 vBulletin Solutions Inc. All rights reserved.