# Sunday Times Teaser 2508 – No Title

### by Graham Smithers

#### Published October 17 2010 (link)

An Austin pootles along a country lane at 30mph; behind are a Bentley doing 40 and a Cortina doing 50. The Bentley and the Cortina brake simultaneously, decelerating at constant rates, while the Austin carries on at the same speed. The Bentley just avoids hitting the rear of the Austin and,  at exactly the same time, the Cortina just avoids a collision with the Bentley. The Bentley and the Cortina continue to decelerate at the same rates, and stop with a 45yd gap between them.

What was the gap between the Bentley and the Cortina at the moment they started to brake?

NOTE: When published by the Sunday Times, the text in red above was wrongly omitted.

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1. Let $$t = 0$$ when braking starts.  For cars A, B and C let the speeds be $$a$$, $$b$$ and $$c$$, the deceleration be $$0$$, $$g$$ and $$h$$, and the positions at time t, relative to C’s starting position, be $$x_a$$, $$x_b$$ and $$x_c$$.  At the start, let the distances between C and B be $$x$$ and that between B and A be $$y$$.    Hence:

 ____ $$x_a = x + y + a t$$ $$x_b = x + b t – g t^2 / 2$$ $$x_c = c t – h t^2 / 2$$

A and B just make contact (equal distance and speed at time $$t$$):

 ____ $$x+y+at=x+b t-g t^2 / 2$$ $$a=b – g t$$ ==> $$t=(b-a)/g$$,  $$2 g y= (b-a)^2$$

B and C just make contact (equal distance and speed at time $$t$$):

 ____ $$x + b t – g t^2 /2 = c t – h t^2 / 2$$ $$b – g t = c – h t$$ ==>  $$t=(c- b) / (h – g)$$,  $$2 x (h – g)=(c-b)^2$$

The two events above happen at the same time:

 ____ $$(b-a)/g=(c-b)/ (h – g)$$ ==>  $$(b – a)h=(c-a)g$$

B and C come to a halt at times $$t_b$$ and $$t_c$$ respectively at a distance $$d$$ apart:

 ____ $$d=(x+b t_b-g (t_b)^2/2)-(c t_c – h (t_c)^2/2)$$ $$b – g t_b=c-h t_c = 0$$ ==>  $$2(x – d) = c^2/h – b^2 / g$$

Eliminating $$g$$, $$h$$ and $$y$$ now gives:

 ____ $$x=(b/a-1)(c/a-1)d$$

Substituting $$a=30$$, $$b=40$$ and $$c=50$$ with $$d=45$$ yards gives $$x$$  as $$10$$ yards.