Re: Time to redeem yourselves (maffs)
 

I read it all.

However, a ball rolled 40m takes 26 seconds. Calculations for how long it will take when rolled 80m with double the initial force... Er, KISS. Double the distance, but also double the force. Time remains the same. That aint maths, that's common sense!

20/20 btw.

Re: Time to redeem yourselves (maffs)
 

did it say initial force? i didn't see that, but still, force is not proportional to speed, which is what the question implied, an initial force would have caused the ball to accelerate at a particular rate, after a particular time it would reach it's initial speed, if you are saying the force is applied after the start time, the time when the ball crosses the beginning of the distance course, then the speed isn't constant (assuming they mean a frictionless environment, or a constant rate of deceleration) and you cannot calculate the time without knowing more details about the force applied and it's duration etc - if it is applied before the start time then it doesn't factor into the solution as you are only concerned about the ball's speed after the start time, but it still doesn't give you the ball's initial speed being double that of the first instance because you don't know if the force is applied for the same distance as in the first case, or the same time, which would alter the initial speed of the ball...

Re: Time to redeem yourselves (maffs)
 

Baph, see what you've done now....got TLW 'going' now :rolleyes: :p

By the way TLW, read it all and actually enjoyed reading it..... ;)



Ben

Re: Time to redeem yourselves (maffs)
 

Again, KISS. You're given limited details, so I worked on the theoretical assumptions that:
- In both cases, the force is applied, and at the exact time the ball moves is the time when the stopwatch starts.
- Given that they state 'double the force' you can assume in relative safety that the distance the force is applied remains constant - they also state this force is generated by rolling, implying that when the ball leaves your hand, there is no further force applying (other than perhaps friction due to gravity).
- In all cases (at least when I did A-level maths), relative variables are taken into consideration. In this case, a ball (approx diameter of 5cm?) is being rolled 1) 40m and 2) 80m. Mass of the ball is reasonably low (think tennis ball), therefore you can assume that friction/air resistance are both negligable and therefore everything is acting in a frictionless environment. If they wanted friction to be taken into account, they would of provided data for mass etc.

Given the above, KISS states that the speed will be double initially, and there is no friction. Therefore time remains constant.

Re: Time to redeem yourselves (maffs)
 

Bladder heck Baph, you have really gone and done it now, haven't you :o :D

I think I'm going to enjoy reading the musing from you both.....seeing as I too like the KISS principle myself ;)



Ben

Re: Time to redeem yourselves (maffs)
 

mechanical work is equal to the force applied multiplied by the distance it is applied for, and is measured in joules. assuming the ball starts at rest the energy it acquires from the work done on it is equal to half it's mass multiplied by it's velocity squared, it's kinetic energy

note the squared, if:

W = F . d = 1/2 .m . v^2

and the distance the force is applied for and the mass of the ball remain the same:

if you double the force applied, the end velocity of the ball increases but only proportional to twice the square of the velocity in the first instance

so double the force will not double the end velocity, which is what i was trying to get at...

Re: Time to redeem yourselves (maffs)
 

unfortunately i have to go to work now, but i'll be back this afternoon to continue:D

Re: Time to redeem yourselves (maffs)
 

KISS... F=ma. :confused:

EDIT: OK, just posting that was a little naughty. Given F=ma, in an environment where mass is negligable, lets call it 1. Therefore F=a. Double the force, double the accelleration... now feed that into s=d/t given the forumla a=delta(v)/delta(t).

But that's far too complex. I've been working with KISS for a long long time. I saw the question & insinctively knew the answer. I guess my brain is wired up different to yours TLW. :)

Re: Time to redeem yourselves (maffs)
 

yes, but when you accelerate you cover distance at an ever increasing rate

although the rate of acceleration is doubled if you double the force, if you are only allowed to accelerate for a set distance, twice the force will not equate to twice the final speed, because the time it takes to cover that set distance decreases exponentially when you accelerate more quickly

Re: Time to redeem yourselves (maffs)
 

none of you have girlfriends right? you have never seen a naked lady before? ;)

i bow my dopey head to you all. 15/20, i didnt even understand 18 and 19..!!:eek:

i do have a hangover, so that may expalin the high score!!