# M3 question....again

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A particle starts with speed 20ms^-1 and moves in a straight line. The particle is subject to a ******ation which is inititally 5ms^-2 and which increases uniformly with the distance moved, having a value of 11ms^-2 when the particle has moved 12m. Given that the particle has speed vms^-1 and has moved a distance of x, show that, while the particle is in motion

v.dv/dx = -(5+x/2) <=== easy peasy

Hence, or otherwise, calculate the distance moved by the particle in coming to rest.

Answer is 20m and I dont get that answer :'(

v.dv/dx = -(5+x/2) <=== easy peasy

Hence, or otherwise, calculate the distance moved by the particle in coming to rest.

Answer is 20m and I dont get that answer :'(

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I also have another question.

A rock climber of mass 80kg is attached to one end of a flexible rope of modulus 45000N and length 20m. She slips from an overhang which is 10m vertically below the point to which the other end of the rope is attached.

Calculate: the speed of the climber when the rope becomes taught (14ms^-1 and easy )

the maximum distance of the climber below the point where she slipped. 13m

the time from slipping to reaching her lowest point. 1.75s

A rock climber of mass 80kg is attached to one end of a flexible rope of modulus 45000N and length 20m. She slips from an overhang which is 10m vertically below the point to which the other end of the rope is attached.

Calculate: the speed of the climber when the rope becomes taught (14ms^-1 and easy )

the maximum distance of the climber below the point where she slipped. 13m

the time from slipping to reaching her lowest point. 1.75s

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#3

v.dv/dx = -(5+x/2)

∫ v dv = - ∫ 5 + 0.5x dx

0.5v² = -5x - 0.25x² + C

v=20 when x=0:

200 = 0 + C

So:

0.5v² = -5x - 0.25x² + 200

v=0:

-0.25x² - 5x + 200 = 0

x² + 20x - 800 = 0

(x-20)(x+40) = 0

x=20m

∫ v dv = - ∫ 5 + 0.5x dx

0.5v² = -5x - 0.25x² + C

v=20 when x=0:

200 = 0 + C

So:

0.5v² = -5x - 0.25x² + 200

v=0:

-0.25x² - 5x + 200 = 0

x² + 20x - 800 = 0

(x-20)(x+40) = 0

x=20m

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(Original post by

v.dv/dx = -(5+x/2)

∫ v dv = - ∫ 5 + 0.5x dx

0.5v² = -5x - 0.25x² + C

v=20 when x=0:

200 = 0 + C

So:

0.5v² = -5x - 0.25x² + 200

v=0:

-0.25x² - 5x + 200 = 0

x² + 20x - 800 = 0

(x-20)(x+40) = 0

x=20m

**dvs**)v.dv/dx = -(5+x/2)

∫ v dv = - ∫ 5 + 0.5x dx

0.5v² = -5x - 0.25x² + C

v=20 when x=0:

200 = 0 + C

So:

0.5v² = -5x - 0.25x² + 200

v=0:

-0.25x² - 5x + 200 = 0

x² + 20x - 800 = 0

(x-20)(x+40) = 0

x=20m

tx..:'(

∫ 0.5x dx does not equal x^2........but x^2/4

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#5

(Original post by

A rock climber of mass 80kg is attached to one end of a flexible rope of modulus 45000N and length 20m. She slips from an overhang which is 10m vertically below the point to which the other end of the rope is attached.

Calculate: the speed of the climber when the rope becomes taught (14ms^-1 and easy )

the maximum distance of the climber below the point where she slipped. 13m

**lgs98jonee**)A rock climber of mass 80kg is attached to one end of a flexible rope of modulus 45000N and length 20m. She slips from an overhang which is 10m vertically below the point to which the other end of the rope is attached.

Calculate: the speed of the climber when the rope becomes taught (14ms^-1 and easy )

the maximum distance of the climber below the point where she slipped. 13m

KE=0.5mv²=0.5(80)(14)²=7840

PE=mgh=80(9.8)x=784x

EPE=0

When she's at the lowest point:

KE=0

PE=0

EPE=(X/2l)e²=1125x²

Conservation of energy gives:

1125x²-784x-7840=0

x=3.01m [Using the quad. formula]

You want the distance from the point she fell, so add this to 10 (since that's when the string becomes taught) to get 13m.

the time from slipping to reaching her lowest point. 1.75s

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(Original post by

I've been trying for quite some time now, but I can't seem to get 1.75...

**dvs**)I've been trying for quite some time now, but I can't seem to get 1.75...

Ne1 else able to get the 1.75s?

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#7

Climber falls naturally for 13m, until when the rope just becomes taught, Then falls for a distance of x metres while the rope gets stretched.

Naturall fall.

s = ½gt²

13 = 4.9t² => t = 1.6288

t1 = 1.63 s

========

Constrained fall

Tension in rope is T = lambda*x/l

T = md, where d is the deceleration provided by the rope on the climber.

Effective acceleration on the climber is (g - d)

g - d = v.dv/dx

v.dv.dx = (g - d) = (g - T/m)

v.dv.dx = 9.8 - lambda.x/(lm)

v.dv/dx = 9.8 - 45000.x/(20.80)

v.dv/dx = 9.8 - 28.125x

integrating,

½v² = 9.8x -14.0625x² + c

v² = 19.6x - 28.125x² + c

at x=0, v = 14 => c = 196

v² = 19.6x - 28.125x² + 196

v = rt(196 + 19.6x - 28.125x²)

now to simplify the expression a bit,

196 + 19.6x - 28.125x² = -28.125(x² - 0.6969x - 6.9288)

196 + 19.6x - 28.125x² = 28.125{7.0902 - u²}, where u = x - 0.3484 giving du = dx

we can now write,

dx/dt = v = rt{k²(a² - u²)}, where a = rt(7.0902) = 2.6627 and k = rt(28.125) = 5.3033

du/dt = k.rt(a² - u²)

du/{k.rt(a² - u²)} = dt

(1/k)(1/a)arcsin(u/a) = t + c'

at t = 0, x = 0, giving u = -0.3484 => c' = -9.2925*10^(-3)

at x = 3.0112m, u = 2.6628 giving,

t = (1/5.3033)(1/2.6627).arcsin(2.6628/2.6627) + 9.2925*10^(-3)

t = (1/5.3033)(1/2.6627).arcsin(1) + 9.2925*10^(-3)

t = (1/5.3033)(1/2.6627).(pi/2) + 9.2925*10^(-3)

t = 0.1112 + 9.2925*10^(-3)

t = 0.1205

t2 = 0.12s

========

total time of fall is t = t1 + t2

t = 1.63 + 0.12

t = 1 75s

=======

Naturall fall.

s = ½gt²

13 = 4.9t² => t = 1.6288

t1 = 1.63 s

========

Constrained fall

Tension in rope is T = lambda*x/l

T = md, where d is the deceleration provided by the rope on the climber.

Effective acceleration on the climber is (g - d)

g - d = v.dv/dx

v.dv.dx = (g - d) = (g - T/m)

v.dv.dx = 9.8 - lambda.x/(lm)

v.dv/dx = 9.8 - 45000.x/(20.80)

v.dv/dx = 9.8 - 28.125x

integrating,

½v² = 9.8x -14.0625x² + c

v² = 19.6x - 28.125x² + c

at x=0, v = 14 => c = 196

v² = 19.6x - 28.125x² + 196

v = rt(196 + 19.6x - 28.125x²)

now to simplify the expression a bit,

196 + 19.6x - 28.125x² = -28.125(x² - 0.6969x - 6.9288)

196 + 19.6x - 28.125x² = 28.125{7.0902 - u²}, where u = x - 0.3484 giving du = dx

we can now write,

dx/dt = v = rt{k²(a² - u²)}, where a = rt(7.0902) = 2.6627 and k = rt(28.125) = 5.3033

du/dt = k.rt(a² - u²)

du/{k.rt(a² - u²)} = dt

(1/k)(1/a)arcsin(u/a) = t + c'

at t = 0, x = 0, giving u = -0.3484 => c' = -9.2925*10^(-3)

at x = 3.0112m, u = 2.6628 giving,

t = (1/5.3033)(1/2.6627).arcsin(2.6628/2.6627) + 9.2925*10^(-3)

t = (1/5.3033)(1/2.6627).arcsin(1) + 9.2925*10^(-3)

t = (1/5.3033)(1/2.6627).(pi/2) + 9.2925*10^(-3)

t = 0.1112 + 9.2925*10^(-3)

t = 0.1205

t2 = 0.12s

========

total time of fall is t = t1 + t2

t = 1.63 + 0.12

t = 1 75s

=======

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#8

Hmm, nicely done Fermat.

I approached it by first finding the time taken for free fall, then I went to see if the climber moves in SHM at some point and what the amplitude and period of that motion was. I obviously messed up somewhere!

I approached it by first finding the time taken for free fall, then I went to see if the climber moves in SHM at some point and what the amplitude and period of that motion was. I obviously messed up somewhere!

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tx every1. i have another question. I can do a,b,c but not d :'(

A particle P starts from rest at a fixed point O nad moves on a horizontal place in a straight line Ox. At time ts after leaving O, the acceleration of P is 15+4t-3t^2. The particle comes to instantaneous rest at the point A.

a. Show that P takes 5s to reach O.

b. Calculate the distance between O and A giving your answer to the nearest m.

The mass of P is 0.2kg. As P moves during the interval 0<=t<=5, calculate:

c. The greatest kinetic energy of P

d. the greatest value of the magnitude of the force exerteed on P giving your answer in newtons

b. 115m

c. 130J

d. 8N <=====this is the bit i cant do :'(

A particle P starts from rest at a fixed point O nad moves on a horizontal place in a straight line Ox. At time ts after leaving O, the acceleration of P is 15+4t-3t^2. The particle comes to instantaneous rest at the point A.

a. Show that P takes 5s to reach O.

b. Calculate the distance between O and A giving your answer to the nearest m.

The mass of P is 0.2kg. As P moves during the interval 0<=t<=5, calculate:

c. The greatest kinetic energy of P

d. the greatest value of the magnitude of the force exerteed on P giving your answer in newtons

b. 115m

c. 130J

d. 8N <=====this is the bit i cant do :'(

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#10

(Original post by

Hmm, nicely done Fermat.

I approached it by first finding the time taken for free fall, then I went to see if the climber moves in SHM at some point and what the amplitude and period of that motion was. I obviously messed up somewhere!

**dvs**)Hmm, nicely done Fermat.

I approached it by first finding the time taken for free fall, then I went to see if the climber moves in SHM at some point and what the amplitude and period of that motion was. I obviously messed up somewhere!

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Lol, i saw the mistake Fermat.....

You deleted before I could quote it though

4-6t=0

so t=1.5s? I dont think so

You deleted before I could quote it though

4-6t=0

so t=1.5s? I dont think so

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#13

DRAT

Force exerted on P = F,say where F = ma

F = m(15 + 4t - 3t²)

dF/dt = m(4 - 6t) = 0 for a turning point

4 - 6t = 0

t = 2/3 s

======

d²F/dt² = m(-6) = -ve => a max

@ t = 2/3

F = 0.2(15 + 8/3 - 3*(4/9))

F = 0.2(17.666 - 1.333)

F = 3.267 N

=========

Force exerted on P = F,say where F = ma

F = m(15 + 4t - 3t²)

dF/dt = m(4 - 6t) = 0 for a turning point

4 - 6t = 0

t = 2/3 s

======

d²F/dt² = m(-6) = -ve => a max

@ t = 2/3

F = 0.2(15 + 8/3 - 3*(4/9))

F = 0.2(17.666 - 1.333)

F = 3.267 N

=========

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#14

(Original post by

What about me newest question

**lgs98jonee**)What about me newest question

Oh well a bit late

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(Original post by

DRAT

Force exerted on P = F,say where F = ma

F = m(15 + 4t - 3t²)

dF/dt = m(4 - 6t) = 0 for a turning point

4 - 6t = 0

t = 2/3 s

======

d²F/dt² = m(-6) = -ve => a max

@ t = 2/3

F = 0.2(15 + 8/3 - 3*(4/9))

F = 0.2(17.666 - 1.333)

F = 3.267 N

=========

**Fermat**)DRAT

Force exerted on P = F,say where F = ma

F = m(15 + 4t - 3t²)

dF/dt = m(4 - 6t) = 0 for a turning point

4 - 6t = 0

t = 2/3 s

======

d²F/dt² = m(-6) = -ve => a max

@ t = 2/3

F = 0.2(15 + 8/3 - 3*(4/9))

F = 0.2(17.666 - 1.333)

F = 3.267 N

=========

Answer in the back of the book is 8N though...which is ridiculus. Are you sure this right cos if it is then there is a book mistake (which are not unheard of) and it also means that I actually got it right.

Heinamann just dont realise how much student's time their mistakes cost.

(Original post by

Max force => max acceleration

**JamesF**)Max force => max acceleration

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#16

(Original post by

That is what I got

Answer in the back of the book is 8N though...which is ridiculus. Are you sure this right cos if it is then there is a book mistake (which are not unheard of) and it also means that I actually got it right.

Heinamann just dont realise how much student's time their mistakes cost.

Yeh, thanks...I did do GCSE Physics...can you get the 8N answer though, as opposed to 3.3?

**lgs98jonee**)That is what I got

Answer in the back of the book is 8N though...which is ridiculus. Are you sure this right cos if it is then there is a book mistake (which are not unheard of) and it also means that I actually got it right.

Heinamann just dont realise how much student's time their mistakes cost.

Yeh, thanks...I did do GCSE Physics...can you get the 8N answer though, as opposed to 3.3?

Like jamesf said, max force => max accln

and max accln is 16 1/3, if you look at the eqn 15 + 4t - 3t².

You would need an accln of 40 m/s² to get a force of 8 N.

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(Original post by

I think it's a mistake in the book.

Like jamesf said, max force => max accln

and max accln is 16 1/3, if you look at the eqn 15 + 4t - 3t².

You would need an accln of 40 m/s² to get a force of 8 N.

**Fermat**)I think it's a mistake in the book.

Like jamesf said, max force => max accln

and max accln is 16 1/3, if you look at the eqn 15 + 4t - 3t².

You would need an accln of 40 m/s² to get a force of 8 N.

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#18

**lgs98jonee**)

That is what I got

Answer in the back of the book is 8N though...which is ridiculus. Are you sure this right cos if it is then there is a book mistake (which are not unheard of) and it also means that I actually got it right.

Heinamann just dont realise how much student's time their mistakes cost.

Yeh, thanks...I did do GCSE Physics...can you get the 8N answer though, as opposed to 3.3?

a = 15 + 20 - 75 = -40

ma = - 8N

|ma| = 8N

Aitch

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#19

**Fermat**)

DRAT

Force exerted on P = F,say where F = ma

F = m(15 + 4t - 3t²)

dF/dt = m(4 - 6t) = 0 for a turning point

4 - 6t = 0

t = 2/3 s

======

d²F/dt² = m(-6) = -ve => a max

@ t = 2/3

F = 0.2(15 + 8/3 - 3*(4/9))

F = 0.2(17.666 - 1.333)

F = 3.267 N

=========

[See post above this one]

Aitch

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#20

**Fermat**)

DRAT

Force exerted on P = F,say where F = ma

F = m(15 + 4t - 3t²)

dF/dt = m(4 - 6t) = 0 for a turning point

4 - 6t = 0

t = 2/3 s

======

d²F/dt² = m(-6) = -ve => a max

@ t = 2/3

F = 0.2(15 + 8/3 - 3*(4/9))

F = 0.2(17.666 - 1.333)

F = 3.267 N

=========

Then the word "magnitude" poked me in the eye. Looking at the curve, it's suddenly clear that, within the limits, the curve is furthest from y=0 at x=5.

Aitch

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