IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Reference Material => GCE AS & A2 Level => Pastpapers => Topic started by: Sue T on December 13, 2009, 01:45:22 pm
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does any1 have any notes for vectors? cuz they're really tricky! :o
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Vectors for AS physics are really easy....just work on sums in ur text and past papers...if u don't get any post them here, and we'll do it for u
I will look for notes on vectors anyway...try S-cool Physics...maybe that'll help
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http://id.mind.net/~zona/mstm/physics/mechanics/vectors/vectors.html
The base of vectors
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oh & if u want me to give u questions to do...I'll be glad to help and c how much you could do by yourself....lemme knw :)
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nid can u solve this 4 me i have an exam tommorow i wuld be relli thnkful
a student standin on a railway platform notices tth the first two carriages of an arriving train passes her in 2 sec and the nxt two in 2.4 s
the train is decelaratin uniformly.each carriage is 20m long when the train stops the student is opposite the last carriae how many carriages are there in the train
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When i get up
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Use s=ut+1/2at^2 twice
For the first two carriages
40=2u+2a
For the first for carraiges
80=4.4u+9.68a
Solve thes to get u=710/33
a=-50/33
then train takes s=(v^2-u^2)/2a=(710/33)^2/(2*50/33)=148.9
8 carrariages
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thank you so much
i'll study from the website u gave then if u can upload the questions it'll be great cuz i do need some practice :D
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Use s=ut+1/2at^2 twice
For the first two carriages
40=2u+2a
For the first for carraiges
80=4.4u+9.68a
Solve thes to get u=710/33
a=-50/33
then train takes s=(v^2-u^2)/2a=(710/33)^2/(2*50/33)=148.9
8 carrariages
Thanks alot sir