Answer to Question #252093 in Field Theory for dude

Question

A man has to run from his house to his neighbour’s house and back again before his favourite television show starts in 12 minutes. His neighbour lives 1.8 km away. How fast must he run on average to make it back in time to catch his show?

Answer 1

$$v=2\frac{s}{t}\\v=2\frac{1800}{12(60)}=5\frac{m}{s}$$