Math Cycling Problem

Phil Holman said:

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" <[email hidden]> wrote in message
news:43c95880.0@entanet...

Quoted message said:

Phil Holman said:

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

Phil Holman said:

Quoted message said:

Quoted message said:

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 said:

Quoted message said:

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

Quoted message said:

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.

Donald Munro said:

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;while(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

Dan Connelly said:

s 333.522

I knew it was a track question.

--
E. Dronkert

"Dan Connelly" <d_j_c_o_n_n_e_l@i_e_e_e.o_r_g> wrote in message
news:[email hidden]...

Quoted message said:

Donald Munro said:

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;while(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

Donald Munro said:

Quoted message said:

Just plug it into Mathematica, Maple or Matlab instead.

Dan Connelly said:

% perl -e 'printf "%-8s %-8s %-8s\n","t","s","a";$dt=0.001;$v=37;$a=1;$vf=29;while(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

Neanderthal.

Phil Holman said:

A rider is traveling at 37 ft/sec and starts to coast.

There's yer mistake right there! :-)

-Mark

"Donald Munro" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Donald Munro said:

Quoted message said:

Just plug it into Mathematica, Maple or Matlab instead.

Dan Connelly said:

% perl -e 'printf "%-8s %-8s
%-8s\n","t","s","a";$dt=0.001;$v=37;$a=1;$vf=29;while(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

Neanderthal.

Here's the calculus solution posted to rbt. Probably has origins even
further to the left on the evolutionary scale.

"Greg Berchin" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

On Sat, 14 Jan 2006 11:28:38 -0800, "Phil Holman"

piholmanc@yourservice said:

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

(37²/29)-37 seconds [~10.2 seconds]

Quoted message said:

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