The Student Room Group
Reply 1
hussain
its a common graph with increasing/decreasing curves, (e.g. in kinematics) and with some questions in order to get the correct answer we must calcualte the area under the curve but how do we do this...??!

Count squares, thats what my teacher says, unless you know the equation then you could integrate i suppose but we don't do that in physics.
Reply 2
James K
Count squares, thats what my teacher says, unless you know the equation then you could integrate i suppose but we don't do that in physics.

...and calculate the area of each sq.
then area = area of each sq. * No of sqs. :tsr:
Or you could cut out the area of the curve from the paper and measure its mass. Then using the same type of paper, cut out a 1cm² and measure its mass. Then using the ratios, you can calculate the area under the curve.
Reply 4
endeavour
Or you could cut out the area of the curve from the paper and measure its mass. Then using the same type of paper, cut out a 1cm² and measure its mass. Then using the ratios, you can calculate the area under the curve.


i cant imagine doin that in the exam :biggrin: :rofl:

my teacher said count squares (including 1/2 squares) and the multiply by the value of 1 square, which is the value 1 sq on the x-axis * value of 1 sq on y-axis

hope that helps
sumitk87
i cant imagine doin that in the exam :biggrin: :rofl:

my teacher said count squares (including 1/2 squares) and the multiply by the value of 1 square, which is the value 1 sq on the x-axis * value of 1 sq on y-axis

hope that helps

Lol, but if a question is suggest how the area may be calculated, that method is a legitimate one!
Reply 6
endeavour
Lol, but if a question is suggest how the area may be calculated, that method is a legitimate one!

if that qs comes in my exam im gonna laugh, and come back and thank you for suggesting this idea, very nice idea- examiners would love that (use of kg m^-2 (mass per area) etc), but the thought havn to do that suppose in the physics prcatical exam makes me laugh haha
Reply 7
you can draw aline that makes a tringle so that you cut of same area and the area you let in and then use 1/2 bh to find the area or count squars and multiply them with the area they write that in the mark scheme
Reply 8
habosh
you can draw aline that makes a tringle so that you cut of same area and the area you let in and then use 1/2 bh to find the area or count squars and multiply them with the area they write that in the mark scheme

Besides square counting, try using rectangles, trapeziums and traingles. Although not 100% accurate, it'll get you close enough to the real answer.

Word to your mother.
Reply 9
If you have little time - a line of best fit. The mean would hopefully approximate the area. Alternatively, a series of five connected, logically-placed tangents?

Oldak.
Reply 10
Methods for physics:
Accurate method = count squares, work out what each square is worth => no.of squares x the value of each square
Method 2 = make trapeziums and triangles and add the area up.
Can't integrate with physics, you wont get no method marks.
Reply 11
nas7232
Methods for physics:
Accurate method = count squares, work out what each square is worth => no.of squares x the value of each square
Method 2 = make trapeziums and triangles and add the area up.
Can't integrate with physics, you wont get no method marks.


Really? Thats soo stupid, seeing its the most accurate way of doing it. :confused:
nas7232
Methods for physics:
Accurate method = count squares, work out what each square is worth => no.of squares x the value of each square
Method 2 = make trapeziums and triangles and add the area up.
Can't integrate with physics, you wont get no method marks.
:confused: If you CAN integrate the curve, and you show your workings, then they have to give you the marks - it's a legitimate, accurate method. Surely...

Besides, it's very unlikely that they will give you curve that you can actually integrate - either you won't know the equation of the curve, or it will be so time-consuming as to be impractical.

Alternatively, the question might simply say "without the use of calculus..."