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Where is Q 16?
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13bFor which one ?
I uploaded the same pic mistakenlyWhere is Q 16?
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Bingo!v^2 = u^2 +2as
12^2 = u^2 -2(5)(8)
144 = u^2 - 80
u^2 = 224
u = 14.995 = 15 ms^-1 Ans.
Wow :O HahahaBingo!
v^2 = u^2 +2as
12^2 = u^2 -2(5)(8)
144 = u^2 - 80
u^2 = 224
u = 14.995 = 15 ms^-1 Ans.
Thank you guysdeceleration is 5 means acceleration is -5
velocity (f) is 12
velocity (I) is 0
x coordinate is 8 meaning distance is 8
2as = v^2 - u^2
2(-5)(8) = 12^2 - u^2
-80 - 144 = -u^2
-224 = -u^2
u = 14.96 ~~ 15
dy/dx = 4 (6-2x)^-½
at x = 1, the gradient of the tangent to the curve is:
dy/dx ( at x=1 )= 4(6-2(1))^-½ = 2
gradient of normal = -1 / gradient of tangent
= -1/2
equ of normal at P,
y - 8 = -1/2 (x - 1)
2y + x = 17 -- (1)
When it cuts x-axis at Q, y=0, so x = 17
Q(17,0)
When it cuts y-axis at R, x = 0, y = 17/2
R (0, 17/2 )
mid point =( (17+0)/2 , (8.5+0)/2 ) = (17/2 , 17/4) Ans.
(b) integrate dy/dx
y = -4(6-2x)^½ +c
when x = 1, y = 8, so,
8 = -4(6-2)^½ +c
8= -4*2 + c
c = 16
so y = -4(6-2x)^½ +16
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