Mathematics: Post your doubts here!

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Haya Ahmed said:
How to find K value ??? please with explaination ... the ans is -11

2xfyUQ.jpg
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thats simple, if u derive that f(x), you get
d (f(x))/dx = -11/(x-4)^2 ======> k/(x-4)^2
which makes k = -11
 
RoOkaYya G

RoOkaYya G

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Wolfgangs said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_41.pdf

Q3 please
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i)
v=0
u= ?
a= -10
s=40+5 = 45
using v^2 = u^ 2 + 2*a*s
u^2= 900
u= 30
therefore speed of projection = 30 m/s

ii)
s=5
u=0
a=10
using s= ut + 0.5*a*t^2
5= 0+ 0.5*10*t
t=1
time above top of cliff = 1*2 = 2s
this is because the signal turns n falls back. so it takes as much time to fall back (same 5m) as it took to go up the 5 m above the cliff.

iii)
s= ?
u= 0
t= root 17
find the height reached by the signal (max height of signal)
s= ut+ 0.5*a*t^2
s= 0+ 0.5*10*( root 17/2) ^2
s= 21.25

time is divided by 2 then squared, ths is because root 17 is the total time tht the signal was abovet the top of tht cliff. means going up + coming down time

finding the speed:
use v^2 =u^2 +2as
v= 0
s=21.25 + 40 =61.25
a= -10
0= u^2 -2(10)(61.25)
u= root 1225 = 32 m/s
 
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Wolfgangs

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RoOkaYya G said:
jst shade the vt graph in part (i) form 20<t<26
s means distance means area under graph for vt graph
Click to expand...

According to the mark scheme, "For shading the triangle from t = 0 to t = 8, the trapezium from t = 8 to t = 20 and the trapezium from t = 20 to a value of t seen to be between 20 and 26". So, the area between 0-8 must be shaded too. But why? And how?
 
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