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Mechanics problem

  • Thread starter suppy123
  • Start date
  • #1
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Homework Statement



A cyclist of mass m experiences a force due to wind resistance equal to -kv^2 when travelling at speed v on a horizontal road. Show that her speed halves in a distance of 0.69m/k if she stops pedalling.

Homework Equations




The Attempt at a Solution


ma = -kv^2
i m not sure if i m right.
 
Last edited:

Answers and Replies

  • #2
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ma = -kv^2
i m not sure if i m right.


well, you are right =)
So you are able to do it now?
 
  • #3
29
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lol :D but i don't get where to put 0.69m/k into the equation
and does it mean a = 0 when she stops pedalling?
 
  • #4
tiny-tim
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lol :D but i don't get where to put 0.69m/k into the equation
and does it mean a = 0 when she stops pedalling?
Hi suppy123! :smile:

erm … you don't put 0.69m/k into the equation, you get it out of the equation! :rolleyes:

And no, a = 0 only when v = 0.

Hint: a = dv/dt. :smile:
 
  • #5
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hi, tim ;)
ma = -kv^2
m(dv/dt) = -kv^2
m(dv/dt) = -k(d/t)^2
m(dv/dt) = -k d^2/t^2
m(dv/dt) * 1/k^2 = -k * d^2/t^2 * 1/k^2
m(dv/dt) / k^2 = -d/k * d/t^2
i got it out ;)
 
  • #6
tiny-tim
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oh suppy123 … a lot of that makes no sense at all! :redface:

(d/t) doesn't mean anything.

Nor does d^2/t^2 or d/k or d/t^2.
Hint: m(dv/dt) = -kv^2,

so mdv/v^2 = -kdt. :smile:
 
  • #7
29
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oh, does m in 0.69m/k means mass over k?
mdv/v^2 = -kdt
m/k = -dt/dv/v^2
m/k = -dt * v^2 /dv
 
  • #8
tiny-tim
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integrate

Hi suppy123! :smile:
mdv/v^2 = -kdt
m/k = -dt/dv/v^2
m/k = -dt * v^2 /dv
hmm … very good … but you're just juggling with fractions.

you need to integrate.

Hint: ∫mdv/v^2 = -∫kdt :smile:
 
  • #9
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oo, but how do i find out the "t" from k
or it is just -m/v = 0 ??
 
  • #10
tiny-tim
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oo, but how do i find out the "t" from k
or it is just -m/v = 0 ??
You find the "t" by integrating, until you get an equation in t.

Fist step: What is ∫kdt?
 
  • #11
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er,
k= -ma/v^2
k=-m(dv/dt)/v^2
∫kdt = -∫m(dv/dt)/v^2 dt = -∫m/v^2 dv = m/v
 
  • #12
tiny-tim
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∫kdt = -∫m(dv/dt)/v^2 dt = -∫m/v^2 dv = m/v
Yay! :biggrin:

Now integrate the left-hand side … what is ∫kdt? :smile:
 
  • #13
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∫mdv/v^2 = -m/v = -∫kdt
∫kdt= m/v
 
  • #14
tiny-tim
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∫kdt= m/v
Yes, I know!

And I know you know!

And you know I know you know! :rolleyes:

But … now integrate the left-hand side! :smile:
 
  • #15
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but i don't know the t
the left-hand side which is....?
 
  • #16
tiny-tim
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What who did next?

D'oh!

Hint:
K-k-k-katy beautiful Katy
You're the one and only girl that I adore
When the m-moonshine over the cowshed
I'll be waiting at the k-k-k=kitchen door
What is ∫kdt? :smile:
 
  • #17
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0????????
 
  • #18
tiny-tim
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… what kt did next …

What is ∫kdt? :smile:
∫kdt = kt + constant.

So the original ∫kdt = -∫m dv/v^2 becomes:

kt = m/v + constant. :smile:
 
  • #19
rl.bhat
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Since you want the relation between distance and velocity, try this one.
ma = -kv^2
dv/dt = -k/m*v^2
dv/ds*ds/dt = -k/m*v^2
dv/ds*v =-k/m*v^2
dv/ds = -k/m*v
dv/v = -k/m*ds
Now integrate. Put the initial condition to find the constant of integration.
 
  • #20
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- m/k = -vt+vc
m/k=vt-vc
 
  • #21
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Since you want the relation between distance and velocity, try this one.
ma = -kv^2
dv/dt = -k/m*v^2
dv/ds*ds/dt = -k/m*v^2
dv/ds*v =-k/m*v^2
dv/ds = -k/m*v
dv/v = -k/m*ds
Now integrate. Put the initial condition to find the constant of integration.
but it doesn't give any values except for the distance
 
  • #22
rl.bhat
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The integration becomes
ln(v) = -k/m*s + C
The initial condition is when s = 0 v = vo.
So ln(v) = -k/m*s + ln(vo)
Now find s when v = vo/2
 
  • #23
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ln(vo/2) = -k/m*s + ln(vo)
ln(vo)-ln(2)=-k/m*s + ln(vo)
ln(2)m / k = s
0.69m/k = s :)
 
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  • #24
Kurdt
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ln(vo/2) = -k/m*s + ln(vo)
ln(vo)-ln(2)=-k/m*s + ln(vo)
ln(2)m / k = s
And there is your answer. I bet you can guess what ln(2) is :wink:.
 
  • #25
29
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tim! i also want to know ur method :)
 

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