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Mathematics: Post your doubts here!

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i am tryin to do it :S:S r u sure they have the same velocity and r they travellin in the same line? can u pls show me the question itself
Unfortunately , there's no drawing for it :S . it's ok if you can't do it , i might ask my teacher about it and tell you later incha'allah .
 
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AsSalamoAlaikum Wr Wb


tan (A + B) = (tan A + tan B)/(1 - (tan A)(tan B))

from the info given, rearrange, and you get: x = tan A ; y = tan B

just substitute, and you get x + y / 1 - xy
WS!
Sorry I still don't get it. :(
If arctan A = x, that means tan x = A, right? :confused:
 
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Unfortunately , there's no drawing for it :S . it's ok if you can't do it , i might ask my teacher about it and tell you later incha'allah .
yeah please do tell me about it cuz its driving me crazy.

howa ante masrya 3ysha f masr ?
 
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Yeah I agree with you !
But the answer isn't right I guess :(
Have a look in the marking scheme !
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_ms_42.pdf

Sorry my bad, actually the answer scheme is correct. A continues to move upwards even as B remain on the floor.
Since the string has slackened, tension in string now =0 N and the velocity of A decreases from 1.6 m/s to 0m/s under the influence of gravity.
Let additional distance moved by A=s2
Then using v^2 =u^2 +2a(s2), we have 0^2 =1.6^2 -2(10)(s2) =====>s2=0.128m
Total distance travelled by A = 0.128+0.32 =0.448m (shown)

Hope this clears things up .Peace.
 

XPFMember

XPRS Moderator
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WS!
Sorry I still don't get it. :(
If arctan A = x, that means tan x = A, right? :confused:
Oops..! Yes you're right, that was a slip in my answer.

Well, then tell me what's your final answer.
I am 110% sure, if the question is as you mentioned, the answer in the book is wrong. Have ways to check that. :p
 
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Sorry my bad, actually the answer scheme is correct. A continues to move upwards even as B remain on the floor.
Since the string has slackened, tension in string now =0 N and the velocity of A decreases from 1.6 m/s to 0m/s under the influence of gravity.
Let additional distance moved by A=s2
Then using v^2 =u^2 +2a(s2), we have 0^2 =1.6^2 -2(10)(s2) =====>s2=0.128m
Total distance travelled by A = 0.128+0.32 =0.448m (shown)

Hope this clears things up .Peace.

No worries :) Thanks !
 
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