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Mathematics: Post your doubts here!

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9(i)
Height is not given so it can be found by Volume / Base Area.
Base Area is 3x * x = 3x^2
Height is found to be 96/x^2
Total Area = 2(Base Area) + 2(Height * x) + 2(Height * 3x) = 6x^2 +192/x +576/x = 6x^2 + 768/x

thanks man!!!! :)
 
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Anyone here done the whole paper 1 "M/J 2014 - V12"?
Seriously it's hard and unfortunately I have the same variant tomorrow! Help! :(
 
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A car of mass 1100 kg starts from rest at O and travels along a road OAB. The section OA is straight,
of length 1760 m, and inclined to the horizontal with A at a height of 160m above the level of O. The
section AB is straight and horizontal (see diagram). While the car is moving the driving force of the
car is 1800N and the resistance to the car’s motion is 700 N. The speed of the car is v ms−1 when the
car has travelled a distance of x m from O.


(i) For the car’s motion from O to A, write down its increase in kinetic energy in terms of v and
its increase in potential energy in terms of x.

Increase in Potential Energy = 1000x . How ??
 
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