golgiapparatus31
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#1
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#1
https://imgur.com/a/hEdLoXi

I got v^2 = 28 - 3e^(-2x)
and terminal speed is sqrt(28)

answer is v = sqrt(3e^(-2x)+22) and terminal speed is sqrt(22)

Am I going crazy or is the solution wrong..?? Please help
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DFranklin
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#2
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#2
(Original post by golgiapparatus31)
https://imgur.com/a/hEdLoXi

I got v^2 = 28 - 3e^(-2x)
and terminal speed is sqrt(28)

answer is v = sqrt(3e^(-2x)+22) and terminal speed is sqrt(22)

Am I going crazy or is the solution wrong..?? Please help
I think it's your solution that's wrong - haven't done any calculations, but as it stands your solution has v increasing as x moves away from the origin when it should be decreasing.
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golgiapparatus31
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(Original post by DFranklin)
I think it's your solution that's wrong - haven't done any calculations, but as it stands your solution has v increasing as x moves away from the origin when it should be decreasing.
Thanks.. Not sure how you are getting that conclusion. Can you please explain more? I see a > 0 for all x


Here is my working.. Can you please see where I'm going wrong? Thank you for your time

a = 3e^(-2x)
vdv = 3e^(-2x) dx
(integrate)
1/2 v^2 = -3/2 e^(-2x) +c
use of when x =0, v=5
25/2 = -3/2 + c
so c = 14

so v^2 = 28 - 3 e^(-2x)
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mqb2766
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I think it's because the acceleration is towards the origin, so acceleration is negative.
Other than that, you seem to have done the right thing.
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DFranklin
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(Original post by golgiapparatus31)
Thanks.. Not sure how you are getting that conclusion. Can you please explain more? I see a > 0 for all x


Here is my working.. Can you please see where I'm going wrong? Thank you for your time

a = 3e^(-2x)
vdv = 3e^(-2x) dx
(integrate)
1/2 v^2 = -3/2 e^(-2x) +c
use of when x =0, v=5
25/2 = -3/2 + c
so c = 14

so v^2 = 28 - 3 e^(-2x)
Acceleration is towards the origin. (So if you take the initial 5m/s as being in direction of increasing x, the acceleration is -ve).
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MacsenT
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(Original post by DFranklin)
I think it's your solution that's wrong - haven't done any calculations, but as it stands your solution has v increasing as x moves away from the origin when it s
(Original post by DFranklin)
Acceleration is towards the origin. (So if you take the initial 5m/s as being in direction of increasing x, the acceleration is -ve).
v is 5m/s when x is negative, when x is negative the acceleration is positive...? if you were to draw a graph of acceleration by displacement the acceleration goes up exponentially for negative displacements ? youve all got me confused aswell now
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golgiapparatus31
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(Original post by mqb2766)
I think it's because the acceleration is towards the origin, so acceleration is negative.
Other than that, you seem to have done the right thing.
(Original post by DFranklin)
Acceleration is towards the origin. (So if you take the initial 5m/s as being in direction of increasing x, the acceleration is -ve).
Thanks!! I got the answer now

I'm also puzzled by the following question:
https://imgur.com/a/YBRFQY4
I'm unable to get the right answer.

What I did:

I used the boundary conditions to get the value of k:
so
2 = k(3pi/2)(-1)
so k = -4/(3pi)

now a = v dv/dx
at x = 5pi/2, v is -10/3

dv/dx = ksinx + kxcosx
I evaluate dv/dx at x = 5pi/2 as (-4/3pi)

so I get an answer of 40/(9pi) = 1.41 but the answer is 4.19

Please help

(Original post by MacsenT)
v is 5m/s when x is negative, when x is negative the acceleration is positive...? if you were to draw a graph of acceleration by displacement the acceleration goes up exponentially for negative displacements ? youve all got me confused aswell now
v is 5 m/s when x = 0, this is the boundary condition

The acceleration is towards the origin. So, as you go away from the origin (in direction of increasing x), the acceleration will be against the direction of your motion, so your velocity will reduce
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MacsenT
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(Original post by golgiapparatus31)
Thanks!! I got the answer now

I'm also puzzled by the following question:
https://imgur.com/a/YBRFQY4
I'm unable to get the right answer.

What I did:

I used the boundary conditions to get the value of k:
so
2 = k(3pi/2)(-1)
so k = -4/(3pi)

now a = v dv/dx
at x = 5pi/2, v is -10/3

dv/dx = ksinx + kxcosx
I evaluate dv/dx at x = 5pi/2 as (-4/3pi)

so I get an answer of 40/(9pi) = 1.41 but the answer is 4.19

Please help


v is 5 m/s when x = 0, this is the boundary condition

The acceleration is towards the origin. So, as you go away from the origin (in direction of increasing x), the acceleration will be against the direction of your motion, so your velocity will reduce
ah right yep my bad, hence 'terminal velocity' as x -> infinity
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mqb2766
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#9
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#9
(Original post by golgiapparatus31)
Thanks!! I got the answer now

I'm also puzzled by the following question:
https://imgur.com/a/YBRFQY4
I'm unable to get the right answer.

What I did:

I used the boundary conditions to get the value of k:
so
2 = k(3pi/2)(-1)
so k = -4/(3pi)

now a = v dv/dx
at x = 5pi/2, v is -10/3

dv/dx = ksinx + kxcosx
I evaluate dv/dx at x = 5pi/2 as (-4/3pi)

so I get an answer of 40/(9pi) = 1.41 but the answer is 4.19

Please help


v is 5 m/s when x = 0, this is the boundary condition

The acceleration is towards the origin. So, as you go away from the origin (in direction of increasing x), the acceleration will be against the direction of your motion, so your velocity will reduce
I'm getting the same as you.
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golgiapparatus31
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#10
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#10
(Original post by mqb2766)
I'm getting the same as you.
Thanks a lot!!
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