Teaser 2508

[Cactus]

Sunday Times Teaser 2508 by John Owen
-------------------------------------------------------------

Cars A followed by B and then C are moving at 30, 40 and 50 mph
respectively. B and C start braking at the same time, each at a
constant rate, and just avoid collisions with the cars ahead of them.
They continue in the same way until they stop 45 yards apart. How far
apart were B and C when they started to brake?

Cactus

[Peter]

Hello All,

I have a problem with this Teaser. Something seems to be missing.
Let
a,b --> traveled distance of A,B at moment t = 0 of simultaneous
breaking
va, vb, vc --> speeds of A,B,C at t = 0
d --> constant deceleration of B,C
g = sb - sc --> gap from C to B after stopping (45 yards)
ta --> time when B reaches A (at speed va)
tb < tc --> stopping times of B and C.

I assume that C does not overtake B at any time. After elimination of
ta, tb, tc one finds there are 2 final equations for the 3 unknowns a,
b, d, namely

a - b = ((vb - va)^2)/2d and b = g + (vc^2 - vb^2)/2d

One sees that the answer for b is independent of va, but it depends on
the value of the deceleration d. I use the conversion 1 mile/hr =
1760/3600 yd/sec = 22/45 yd/sec. With a reasonable deceleration d = 5
yd/sec^2 one gets b ~ 66.5 yd, independent of va. Also a ~ 90.4 yd.

What did I miss? Are a and b supposed to be whole numbers of yards? Is
C allowed to overtake B (with relatively low speed) at the time when B
stopped, so that finally sc - sb = g > 0?

Peter

[Garry]

Hello Peter

I too have a difficulty, feeling that the problem is under-specified.
I don't see the relevance of A, as the spacing between A and B at the
time braking starts can be chosen to accommodate any rate of
deceleration by B. When B and C graze, they are at the same position
and at the same speed, but thereafter C falls behind so must have a
greater deceleration than B [i.e. I disagree with your single value
d]. So my unknowns are initial spacing of B and C, deceleration of B,
deceleration of C, and I could find only two equations (one from the
glancing contact of B and C, one from the final spacing of B and C)
with apparently no unique solution.

Garry

[Teasemaster]

Does this mean that B and C each has:
a) a different constant rate of deceleration, starting at the same
time but spatially separated?
b) the same constant speed after simultaneous (instant) deceleration
at the same time but spatially separated?

TM

[Garry]

Further to my post above, ONE solution would seem to be that B and C
have initial spacing the same as their final one, i.e. 45 yards (9/352
miles), and their respective decelerations are 281600/81 mphph and
440000/81 mphph. For these conditions, they are in the same position
with the same speed at time 9/1760 hours after braking begins. So
either I've missed something, or we have a case reminiscent of Teaser
2472 (boy falls in front of train).

Garry

[Garry]

Hello TM

From the phrase "just avoid collisions", I assume a). I've drawn a
position-time graph for B and C, that for B starting higher, i.e. with
a positive intercept on the vertical (position) axis. Time zero is
when both start braking. Both curves are inverted parabolas, touching
but not crossing, and turning into horizontal straight lines at their
apex.

Garry

[Harry]

Perhaps the missing bit of information is that both near-collisions
happened at the same time. Then, at the moment of braking, the gaps
between A and B and between B and C must both have been 10 yards.

[Peter]

Hello Garry,

A moment ago I read your first comment (and others). Many thanks for
pointing out my mistake of assuming the same deceleration d for B and
C. Obviously dc must exceed db in order that C can touch B (with the
same speed). I am going to revise my analysis later.

Peter

[Ben Alligin]

Hi folks,

The Sunday Times has done its usual trick of re-writing a teaser. I
can accept the change of tense from past to present (although I regret
that the word "pootling" no longer appears in the teaser), but they
have missed out a vital bit of the teaser. The original wording said:
"The Bentley just avoided hitting the rear of the Austin while, at the
same time, the Cortina just avoided a collision with the Bentley". I
hope that this helps.

John

[Cactus]

Thanks John,

I find it rather shocking that anyone in the Sunday Times thinks it is
acceptable to edit these teasers. At very least, I would have thought
it necessary to get both Victor Bryant and the original author to
endorse the proposed amendments. You have now had two of your teasers
marred by this so you have good reason to be displeased since many
readers may well think that this was a mistake on your part.

In fact I was rather hoping for a different resolution since,
_without_your_additional_constraint_, it is possible, although quite
difficult, to obtain an equation involving only the two distances (x,
y) between the cars when they start to brake and the final gap (g).
If we then insist that all these distances are integers in yards we
can show that x is given by:

x = k.(4.k + 3) 1, 45, 76 ... or
x = k.(4.k - 3) 1, 10, 27 ...

for any integer k. By setting a minor constraint on the answer, this
can hence be turned into a different teaser.

But I did not find it easy to obtain this relationship between x, y
and g - so maybe Peter can do his normal task of finding a simpler way
than I did :-)

Brian

[Guy]

John thanks for giving us the vital missing piece!

Assume Tx is the time the vehicle touch, all cars travelling at speed
Ua: you get the following
Tx=(Ua-Uc)/Fc
Tx=(Ua-Ub)/Fb

Hence Fb=((Ua-Ub)/(Ua-Uc))Fc (Eq1)

using S=UT+0.5FT^2 for cars B and C, and where x is the starting
distance between them gives:
x=(Uc-Ub)Tx+0.5(Fc-Fb)Tx^2 (Eq2)

using V^2=U^2 + 2FS at the end for cars B and C gives:
x=(-Uc^2/2Fc)+(Ub^2/2Fb)+45 (Eq3)

setting Eq2=Eq3 and substitute Eq1 should give the value x=10 yards
and Tx=4.09seconds.

[Peter]

Hello John,

thanks for providing the missing bit of the Teaser. So Harry's
suspicion was right and so is his solution.

For arbitrary velocities va < vb < vc one finds the initial positions
of A and B as

a = g*[(vb-va)/(vc-vb)]*[(vc-va)/va]^2
b = g*(vb-va)*(vc-va)/va^2

Peter

The deceleration of B and C are about 2.4 and 1.2 yd/sec^2

[Garry]

Hello all

I echo the sentiments about the inappropriateness of someone at the ST
deleting parts of the wording of the teaser. With the extra condition
we have a nice problem in the equations of motion under constant
acceleration, enabling the complete behaviour of the system to be
determined:

Distances measured from C's position at the time braking starts:

B's position then 10 yards <- required answer
A's position then 20 yards
Position of three-car contact 80 yards
C's standstill position 125 yards ) given 45 yard
B's standstill position 170 yards ) separation

Times elapsed from when braking starts:

Three-car contact 45/11 secs
C comes to standstill 225/22 secs
B comes to standstill 180/11 secs

Deceleration of B 22/9 mphps
Deceleration of C 44/9 mphps

Garry

[Peter]

Hello Brian,

I obtained the same relation as you presented, for the case of
velocities va = 30, vb = 40, vc = 50 mph, and gap g = 45 yd. One
obtains easily the general relations

2*(a-b) = ((vb-va)^2)/db, 2*b = ((vc-vb)^2)/(dc-db), g = b-(vc*vc/dc -
vb*vb/db)/2

Eliminating the decelerations db, dc one obtains the final gap g as
function of the initial distances a = CA, b = CB and of the initial
velocities. A rather clumsy formula! For the values given above one
obtains simply

g*a = 45*a = (4a-5b)^2, which reduces to a = 5*k^2, b = k*(4k -/+
3),

as you derived probably in the same way. With the + sign values k < 3
are invalid because one would get a < b.

The elimination of db and dc is somewhat easier if one uses the
specified velocities beforehand.

Peter

On 18 Oct, 10:50, Cactus <rieman...@gmail.com> wrote:

[Cactus]

Thanks Peter,

The best I did was to define p and q in terms of the velocities of the
cars (a, b, c):

p = b(c - b)/(b - a)^2
q = c(c - b)/(b - a)^2

and then show that the relationship between the inter car distances
(x, y) is:

(x - p y)^2 = g(x - (p - q)y)

Brian

[Ben Alligin]

Dr Bryant wasn't impressed with the changes to the original teaser:
"I'm horrified at the changes. The current sub-editor is very
sensitive about changes so I assume that some idiot changed it in her
absence. I don't know why the tense was changed or how anyone could
not believe that "while, at the same time" was relevant."

John