A runs $$1\frac{2}{3}$$ times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B reach it at the same time?

It is given that A runs $$1\frac{2}{3}$$ times as fast as B.

Let us assume B runs at a speed of 6 m/s, so A will run at a speed of 6*5/3 = 10 m/s

We know that A gives B a headstart of 80 m.

We need to find the time at which A and B reach at the same time at winning post.

The difference between their speed (A and B ) = 10-6 = 4 m/s

They will reach the winning post at the same time, when A will be able to reach at position of B by covering the headstart distance.

Total distance = 80 m

A's speed which is more than B to cover the distance of 80 m = 4 m/sec

Time it takes to cover the distance = 80/4 = 20 sec

So, the winning post will be reached by A and B together in 20 seconds.

A will cover the distance = 10*20 = 200 meter