# Variable force

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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 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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#2

(Original post by

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

**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

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(Original post by

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.

**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.

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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#4

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.

Other than that, you seem to have done the right thing.

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#5

(Original post by

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)

**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)

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#6

(Original post by

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

**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

Acceleration is towards the origin. (So if you take the initial 5m/s as being in direction of increasing x, the acceleration is -ve).

**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).

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(Original post by

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.

**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.

**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).

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

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

**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

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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#8

(Original post by

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

**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

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#9

**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

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(Original post by

I'm getting the same as you.

**mqb2766**)I'm getting the same as you.

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