Math Cycling Problem

A rider is traveling at 37 ft/sec and starts to coast. The rider's
initial deceleration is 1 ft/sec^2 which varies with the square of the
velocity. How long will it take for the rider to decelerate from 37 to
29 ft/sec and how far will the rider have traveled?

Phil H

Phil Holman wrote:

A rider is traveling at 37 ft/sec and starts to coast. The rider's
initial deceleration is 1 ft/sec^2 which varies with the square of the
velocity. How long will it take for the rider to decelerate from 37 to
29 ft/sec and how far will the rider have traveled?

Just plug it into Excel for a mathematically inelegant but adequate
solution. It's actually a silly problem because, rather than giving us
a Cd and a frontal area, they've used a kludge to tell us that 37 ft/sec
happens to give a retardation of 1 ft/sec^2 based on the rider's shape
and size.

Zog The Undeniable wrote:
Phil Holman wrote:

A rider is traveling at 37 ft/sec and starts to coast. The rider's
initial deceleration is 1 ft/sec^2 which varies with the square of the
velocity. How long will it take for the rider to decelerate from 37 to
29 ft/sec and how far will the rider have traveled?

Just plug it into Excel for a mathematically inelegant but adequate
solution.

It's not that hard to do with pencil and paper. It's a
shame for us colonials to see that UK standards
have gone so far downhill.

It's actually a silly problem because, rather than giving us
a Cd and a frontal area, they've used a kludge to tell us that 37 ft/sec
happens to give a retardation of 1 ft/sec^2 based on the rider's shape
and size.

If one were measuring speed distance and time, as in a
coast down test, deceleration (I prefer that to retardation
because retard means something else in RBR) is the
directly derived parameter. Cd and A are a further inference
from that. And you also have to know the air density.

Ben

"Zog The Undeniable" wrote in message
news:43c95880.0@entanet...
Phil Holman wrote:

A rider is traveling at 37 ft/sec and starts to coast. The rider's
initial deceleration is 1 ft/sec^2 which varies with the square of
the
velocity. How long will it take for the rider to decelerate from 37
to
29 ft/sec and how far will the rider have traveled?

Just plug it into Excel for a mathematically inelegant but adequate
solution. It's actually a silly problem because, rather than giving
us a Cd and a frontal area, they've used a kludge to tell us that 37
ft/sec happens to give a retardation of 1 ft/sec^2 based on the
rider's shape and size.

Inelegant is correct, the expectation is to use calculus. For your
suggestion you'd need the mass of bike plus rider as well as the Cd and
frontal area, and then the rolling resistance and slope to make it even
more realistic. Who is "they" by the way?

Phil H

On Sat, 14 Jan 2006 11:28:38 -0800, "Phil Holman"
piholmanc@yourservice wrote:

How long will it take for the rider to decelerate from 37 to
29 ft/sec

(37²/29)-37 seconds [~10.2 seconds]

and how far will the rider have traveled?

37²*(ln|37²/29|-ln|37|) feet [~333.5 feet]

"Greg Berchin" wrote in message
...
On Sat, 14 Jan 2006 11:28:38 -0800, "Phil Holman"
piholmanc@yourservice wrote:

How long will it take for the rider to decelerate from 37 to
29 ft/sec

(37²/29)-37 seconds [~10.2 seconds]

and how far will the rider have traveled?

37²*(ln|37²/29|-ln|37|) feet [~333.5 feet]

Excellent.

a = kv^2
-1 = k(37)^2, so k = -1/37^2
so dv/dt = (-1/37^2)v^2

Separate the variables and take the antiderivative of both sides

Int[-37^2 dv/v^2] = Int dt
37^2(1/v) = t + c
Using (0, 37), c = 37
So t = (37^2/v) - 37
When v = 29, t = (37^2/29) - 37
t = 10.21 seconds

v = 37^2/(t + 37)
s = 37^2 ln[t + 37] + c
Using (0, 0), c = 37^2 ln(37)
So s = 37^2 {ln[t + 37] - ln(37)}
s = 37^2 ln [(t + 37)/37]
when t = 10.21, s = 37^2 ln[(10.21 + 37)/37]
s = 333.52 ft

Phil H

On Sat, 14 Jan 2006 16:57:33 -0800, "Phil Holman"
piholmanc@yourservice wrote:

Excellent.

What did I win? A new Waterford would be nice.

Phil Holman wrote:
A rider is traveling at 37 ft/sec and starts to coast. The rider's
initial deceleration is 1 ft/sec^2 which varies with the square of the
velocity. How long will it take for the rider to decelerate from 37 to
29 ft/sec and how far will the rider have traveled?

Zog The Undeniable wrote:
Just plug it into Excel for a mathematically inelegant but adequate
solution.

wrote:
It's not that hard to do with pencil and paper. It's a
shame for us colonials to see that UK standards
have gone so far downhill.

Just plug it into Mathematica, Maple or Matlab instead.

Dan Connelly wrote:
s 333.522

I knew it was a track question.

--
E. Dronkert

"Dan Connelly" wrote in message
m...
Donald Munro wrote:

Just plug it into Mathematica, Maple or Matlab instead.


% perl -e 'printf "%-8s %-8s
%-8s\n","t","s","a";$dt=0.001;$v=37;$a=1;$vf=29;whil e(1){$t+=$dt;$vold=$v;$v-=$dt*($a=(($v-$a*$dt/2)/37)**2);$s+=($v+$vold)/2*$dt;if
(($v=$vf)!=($vold=$vf)){printf "%-8g %-8g
%-8g\n",$t-$dt*($v-$vf)/($v-$vold),$s,$a;exit}}'
t s a
10.2069 333.522 0.614327

I should have excluded Dan from even attempting the solution :-)

1*29^2/37^2 = .614317.............was that a typo?

Phil H