unix - Extract a line if it contains a word in a specified column -

i want extract line if contains word in specified column of textfile. how can on one-liner unix command it? maybe cat, echo, cut, grep several piples or something.

i have textfile looked format

#sentenceid<tab>sentence1<tab>sentence2<tab>other_unknown_number_of_columns<tab> ... 

an example of textfile looks this:

021348  english sentence coach .   c'est la phrase française avec l'entraîneur .   , there several nonsense columns these  . 923458  english sentence without word .   c'est une phrase d'une autre anglais sans le bus mot .  whatever foo bar    nonsense columns    2134234 $%^& 

the command should output if word looking coach in 2nd column:

021348  english sentence coach .   c'est la phrase française avec l'entraîneur .   , there several nonsense columns these  . 

i can python such, i'm looking unix command or one-liner:

outfile = open('out.txt') line in open('in.txt'):   if "coach" in line.split():     print>>outfile, line 

what this?

awk -f'\t' '{if($2 ~ "coach") print} your_file 

• -f'\t' --> makes delimiter tab.
• $2 ~ "coach" --> looks "coach" in second field.
• print $0 or print --> prints whole line.

edit

sudo_o has suggested following, shorter:

awk -f'\t' '$2~/coach/' file