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droid
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#1
Report Thread starter 14 years ago
#1
Hey, I posted a question earlier about the same topic and that helped me do most of the questions after that, but this one has me stumped i can't seem to get the right answer, any help would again be brilliant, thanks.

A particle P of mass 1.5kg moves in a straight line through a fixed point O. At time t sec after passing through O the distance of p from O is x m and teh force acting on P has magnitude (3x+6)N and is directed away from O. given that P passes through O with speed 2root2 ms-1 calculate (A) te speed of P when x = 5 (b) the value of t when x= 20.

The answers are
(a)9.9
(b)1.7
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Aitch
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#2
Report 14 years ago
#2
(Original post by droid)
Hey, I posted a question earlier about the same topic and that helped me do most of the questions after that, but this one has me stumped i can't seem to get the right answer, any help would again be brilliant, thanks.

A particle P of mass 1.5kg moves in a straight line through a fixed point O. At time t sec after passing through O the distance of p from O is x m and teh force acting on P has magnitude (3x+6)N and is directed away from O. given that P passes through O with speed 2root2 ms-1 calculate (A) te speed of P when x = 5 (b) the value of t when x= 20.

The answers are
(a)9.9
(b)1.7
(a) F= 3x+6 = ma

m* 0.5 v^2 = int( 3x+6) = 1.5 x^2 + 6x+ c
mv^2 = 3x^2+ 12x+k
x=0, v^2 = 8 =>
1.5*8=k=12

x=5 =>
1.5 v^2=75+60+12=147
v^2 = 98
v = 9.9

Aitch

will think about B next!
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Aitch
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#3
Report 14 years ago
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(Original post by Aitch)
(a) F= 3x+6 = ma

m* 0.5 v^2 = int( 3x+6) = 1.5 x^2 + 6x+ c
mv^2 = 3x^2+ 12x+k
x=0, v^2 = 8 =>
1.5*8=k=12

x=5 =>
1.5 v^2=75+60+12=147
v^2 = 98
v = 9.9

Aitch

will think about B next!

v^2= 2(x+2)^2
v= dx/dt
so dt/dx = 1/(dx/dt)
int dt/dx = t = 1/root2 ln (x+2) + c
t=0 at x=0 so c = -(1/root2)ln2
x=20 => t = (1/root2)ln22-ln2 =1.69

Think this is correct - rushed it a bit .

Aitch
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Aitch
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#4
Report 14 years ago
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(Original post by Aitch)
v= 2(x+2)^2
v= dx/dt
so dt/dx = 1/(dx/dt)
int dt/dx = t = 1/root2 ln (x+2) + c
t=0 at x=0 so c = -(1/root2)ln2
x=20 => t = (1/root2)ln22-ln2 =1.69

Think this is correct - rushed it a bit .

Aitch
I have just corrected this above. I made a mistake in b.

OK now I think.
Aitch
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