Solve Cycling Force Problem: Mass, Resistance, Speed, Distance

[suppy123]

Homework Statement

A cyclist of mass m experiences a force due to wind resistance equal to -kv^2 when traveling 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.

 

[Raze2dust]

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

well, you are right =)
So you are able to do it now?

 

[suppy123]

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?

 

[tiny-tim]

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!

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

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

Hint: a = dv/dt.

 

[suppy123]

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 ;)

 

[tiny-tim]

oh suppy123 … a lot of that makes no sense at all!

(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.

 

[suppy123]

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

 

[tiny-tim]

integrate

Hi suppy123!

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

 

[suppy123]

oo, but how do i find out the "t" from k
or it is just -m/v = 0 ??

 

[tiny-tim]

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?

 

[suppy123]

er,
k= -ma/v^2
k=-m(dv/dt)/v^2
∫kdt = -∫m(dv/dt)/v^2 dt = -∫m/v^2 dv = m/v

 

[tiny-tim]

∫kdt = -∫m(dv/dt)/v^2 dt = -∫m/v^2 dv = m/v

Yay!

Now integrate the left-hand side … what is ∫kdt?

 

[suppy123]

∫mdv/v^2 = -m/v = -∫kdt
∫kdt= m/v

 

[tiny-tim]

∫kdt= m/v

Yes, I know!

And I know you know!

And you know I know you know!

But … now integrate the left-hand side!

 

[suppy123]

but i don't know the t
the left-hand side which is...?

 

[tiny-tim]

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? ​

 

[suppy123]

0?

 

[tiny-tim]

… what kt did next …

What is ∫kdt? ​

∫kdt = kt + constant.

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

kt = m/v + constant.

 

[rl.bhat]

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.

 

[suppy123]

- m/k = -vt+vc
m/k=vt-vc

 

[suppy123]

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

 

[rl.bhat]

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

 

[suppy123]

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 :)

 

[Kurdt]

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 .

 

[suppy123]

tim! i also want to know ur method :)

 

[suppy123]

And there is your answer. I bet you can guess what ln(2) is .

heh

 

[tiny-tim]

tim! i also want to know ur method :)

Hi suppy123!

kt = m/v + kC.

So v = m/k(t - C),

so dx/dt = m/k(t - C),

so dx = (m/k)dt/(t - C),

So x = … ?