How do I measure duration in seconds in a shell script?

Using the time command, as others have suggested, is a good idea.

Another option is to use the magic built-in variable $SECONDS, which contains the number of seconds since the script started executing. You can say:

START_TIME=$SECONDS
dosomething
ELAPSED_TIME=$(($SECONDS - $START_TIME))

I think this is bash-specific, but since you're on Linux, I assume you're using bash.

Try following example:

START_TIME=$SECONDS
# do something
sleep 65

ELAPSED_TIME=$(($SECONDS - $START_TIME))

echo "$(($ELAPSED_TIME/60)) min $(($ELAPSED_TIME%60)) sec"    
#> 1 min 5 sec

Many of the answers mention $SECONDS, but that variable is actually even better than they realize:

Assignment to this variable resets the count to the value assigned, and the expanded value becomes the value assigned plus the number of seconds since the assignment.

This means you can simply query this variable directly at the end of your script to print the elapsed time:

#!/usr/bin/env bash

# Do stuff...

echo "Script finished in $SECONDS seconds."

You can also time smaller sections like so:

#!/usr/bin/env bash

# Do stuff

SECONDS=0

# Do timed stuff...

echo "Timed stuff finished in $SECONDS seconds."

Use the time command. time ls /bin.