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Home >> Advice >> Help! Maths

16.12.2006, 20:14 quote

Anonymous

Cidem wrote:
oi !! you trying to get me into trouble ??


tsk... Cidem, this is 'FLIRTbox', get with the jiggy man and do some serious flirtin'! Twisted Evil

 
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16.12.2006, 20:23 quote

Anonymous

Hee hee Loving your flirting ... you go ''''Girlfriend'''' Wink

 
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16.12.2006, 21:03 quote

Anonymous

cheekyarse wrote:
Hee hee Loving your flirting ... you go ''''Girlfriend'''' Wink

Hey babes, just LOVIN' the new hair do. Am I pleased to see you or is that a canoe in my pocket?! Wink Twisted Evil

 
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16.12.2006, 21:17 quote

Anonymous

toby wrote:
Gizzemoto wrote:
Any one in here good at maths, i need lots of help

1. Show how the area under the Time/velocity curve is equal to the distance traveled. A car traveling in a straigh line, accelerates uniformly from rest with a uniform acceleration of 2 m/s2 to a maximum speed of 30 m/s. It drives at this speed for 12 seconds and then decelerates to rest in a further 10 seconds. Find the time spent accelerating, the total distance covered, the magnitude of the deceleration??

Any ideas, solutions


Havent done anything like that for years..but I am trying to remember how to do it by "common sense"

1. The area "under" a graph is always the integral of it....

It is quite easy to understand if you imagine the following:

if you have a straight line with a disance of e.g. f(x)=5, then the area
of the rectangle between 0 and 10 would be 10*5,. so the integral of f(x)=5x. If the function is f(x)=x the integral is x^2...etc.

So if you have a function that tells you the velocity after a certain time (you need the acceleration for it), the area below the function is the integral of that function.

v = a * t (speed = acceleation * time).

Example: Acceleraion is 5m/s^2 this means after 1 second the speed is 5m/s and after 2 seconds the speend is 10 m/s etc....

v(t) = a*t. The integral would be 0.5 a*t^2 (oh no!..almost forgot how to integrate..quite shocking Confused )

The distance travelled is the speed * time.
If you want to compare it to the integral above, you need to get rid of the speed inside the equation.

Speed = a*t.

So distance travelled is v*t=a*t*t = a*t^2

So the area below is not the distance travelled, but half the distance travelled!

Q.E.D


2. just add the 3 different parts/areas under the graph together....to get the total distance travelled. The rest is done using the equations above..just play around with them until you have an equation that give you the result depending on the parameters you know.


Thats what I told them toby, but they don't want to listen Razz

 
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16.12.2006, 21:17 quote

Anonymous

toby wrote:
Gizzemoto wrote:
Any one in here good at maths, i need lots of help

1. Show how the area under the Time/velocity curve is equal to the distance traveled. A car traveling in a straigh line, accelerates uniformly from rest with a uniform acceleration of 2 m/s2 to a maximum speed of 30 m/s. It drives at this speed for 12 seconds and then decelerates to rest in a further 10 seconds. Find the time spent accelerating, the total distance covered, the magnitude of the deceleration??

Any ideas, solutions


Havent done anything like that for years..but I am trying to remember how to do it by "common sense"

1. The area "under" a graph is always the integral of it....

It is quite easy to understand if you imagine the following:

if you have a straight line with a disance of e.g. f(x)=5, then the area
of the rectangle between 0 and 10 would be 10*5,. so the integral of f(x)=5x. If the function is f(x)=x the integral is x^2...etc.

So if you have a function that tells you the velocity after a certain time (you need the acceleration for it), the area below the function is the integral of that function.

v = a * t (speed = acceleation * time).

Example: Acceleraion is 5m/s^2 this means after 1 second the speed is 5m/s and after 2 seconds the speend is 10 m/s etc....

v(t) = a*t. The integral would be 0.5 a*t^2 (oh no!..almost forgot how to integrate..quite shocking Confused )

The distance travelled is the speed * time.
If you want to compare it to the integral above, you need to get rid of the speed inside the equation.

Speed = a*t.

So distance travelled is v*t=a*t*t = a*t^2

So the area below is not the distance travelled, but half the distance travelled!

Q.E.D


2. just add the 3 different parts/areas under the graph together....to get the total distance travelled. The rest is done using the equations above..just play around with them until you have an equation that give you the result depending on the parameters you know.


Thats what I told them toby, but they don't want to listen Razz

 
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17.12.2006, 09:26 quote

Anonymous

Cheers for all the help guys and girls, ready for Q2?????

 
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