When (60-2t)(30+2t)=1800
i.e. when t=15 units
good attempt but ur answer is wrong
you havent specified that how did you get the equation
the correct answer is from the equation
(60-2t)(60-2t)=1800
Area of Triangle PQC=Half of PCxCQxSinC, where PC at any instant t is (60-2t) units and CQ is (30+2t) when Q is moving towards B. Since Q is initially the midpoint of BC and Q is moving towards C, the condition of the area will never be satisfied as we know the area of Triangle PQC will be less than half the area of ABC in that case. So my answer is from the fact that halfxPCxCQxSinC=halfxhalfxACxCBxSinC.
i.e. PCxCQxSinC= halfx60x60xSinC => (60-2t)(30+2t)=1800 =>4t2-60t=0
i.e. t=0 or t=15. Hence the answer.
so u cant take CQ as (30+2t) because DQ(where D is the mid-point of BC)is not equal 2t.
it is only at a particular time instant t when the velocity of particle Q becomes 2 units per second
you hav to use the concepts of differentiation and itegration here
Nikhil
I think you got 30+2t wrong as 2t is not the distance covered by Q (towards B) in t sec at variable speed. In fact in my opinion it should be (30+t) if you work it out using V=ft & S= 12 (ft2) and then apply ft=2 unit/sec the velocity attained after t sec.
So we get finally 900= 900-t2 which results in t=0 sec only.
That means there is no such situation when the problem (as given) can be accomplished.
sorry to everybody for the errors
sorry for the error
Dear Rekha Gautam;
You say that Q is moving towards A in that case the area of triangle can never be half of area of ABC. If you draw the diagram you will realise. In fact area of PQC will be approacing to ZERO as time goes on.
In my opinion you want to say "half of area of triangle "AQC" instead of ABC.
In that case there is a problem for which one try for solution.
Confirm if I am correct.